SELECT A SUBJECT
Mark fills in each circle with a number from $1$, $2$, $3$, $\dots$, $8$, such that the sum of numbers at all corners of any triangle is 12. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 12\times 4 &= 48 \\ 1+2+3+\cdots + 7 &= 36 \\ 48-36 &= 12 \end{align*} Thus, $a+b+c+d=12$. Try, $1+2+3+6=12$
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 12\times 4 &= 48 \\ 1+2+3+\cdots + 7 &= 36 \\ 48-36 &= 12 \end{align*} Thus, $a+b+c+d=12$. Try, $1+2+3+6=12$
Find the remainder when the $2017^{\text{th}}$ term of the following sequence is divided by 5. \begin{equation} 1,\, 2,\, 4,\, 7,\, 11,\, 16,\, 22,\, 29,\, \dots \end{equation}
Sorry. Please check the correct answer below.
Yay! Your are right.
Observe that pattern of remainders: \begin{equation} \begin{array}{ccccccccccc} n & 1 & 2 & 4 & 7 & 1 & 1 & 2 & 2 & 3 & 4 \\ & & & & & 1 & 6 & 2 & 9 & 7 & 6 \\\\ R & 1 & 2 & 4 & 2 & 1 & 1 & 2 & 4 & 2 & 1 \end{array} \end{equation} The pattern recurs: 1, 2, 4, 2, 1, $\dots$.
How many numbers are the same in both number patterns below? \begin{align*} 1,\ 3,\ 5,\ 7,\ 9,\ \dots ,\ 2015,\ 2017 \\ 1,\ 4,\ 7,\ 10,\ \dots ,\ 2014,\ 2017 \\ \end{align*}
Sorry. Please check the correct answer below.
Differences are 2 and 3 in $1^{\text{st}}$ and $2^{\text{nd}}$ sequences respectively. We are looking for 1, 7, 13, 19, 25, $\dots$ in both. Difference is 6 here, $(2\times 3 = 6)$. \begin{align*} (2017-1) \div 6 &= 336 \\ 336+1 &= 337 \end{align*}
Yay! Your are right.
Differences are 2 and 3 in $1^{\text{st}}$ and $2^{\text{nd}}$ sequences respectively. We are looking for 1, 7, 13, 19, 25, $\dots$ in both. Difference is 6 here, $(2\times 3 = 6)$. \begin{align*} (2017-1) \div 6 &= 336 \\ 336+1 &= 337 \end{align*}
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 10. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 10\times 4 &= 40 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 40-36 &= 4 \end{align*} Thus, $a+b+c+d=4$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 10\times 4 &= 40 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 40-36 &= 4 \end{align*} Thus, $a+b+c+d=4$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 11. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 11\times 4 &= 44 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 44-36 &= 8 \end{align*} Thus, $a+b+c+d=8$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 11\times 4 &= 44 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 44-36 &= 8 \end{align*} Thus, $a+b+c+d=8$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 12. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 12\times 4 &= 48 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 48-36 &= 12 \end{align*} Thus, $a+b+c+d=12$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 12\times 4 &= 48 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 48-36 &= 12 \end{align*} Thus, $a+b+c+d=12$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 13. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 13\times 4 &= 52 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 52-36 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 13\times 4 &= 52 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 52-36 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 14. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 12. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 12\times 4 &= 48 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 48-36 &= 12 \end{align*} Thus, $a+b+c+d=12$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 12\times 4 &= 48 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 48-36 &= 12 \end{align*} Thus, $a+b+c+d=12$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 13. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 13\times 4 &= 52 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 52-36 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 13\times 4 &= 52 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 52-36 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 14. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 15. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 14. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 15. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 16. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 16. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 17. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 13. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 13\times 4 &= 52 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 52-36 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 13\times 4 &= 52 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 52-36 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 14. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 15. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 16. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 15. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 16. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 17. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 17. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 19. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 16. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 17. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 19. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 19. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 21. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 22. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 14. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 14\times 4 &= 56 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 56-36 &= 20 \end{align*} Thus, $a+b+c+d=20$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 15. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 15\times 4 &= 60 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 60-36 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 16. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 17. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 16. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 16\times 4 &= 64 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 64-36 &= 28 \end{align*} Thus, $a+b+c+d=28$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 17. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 19. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 17. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 17\times 4 &= 68 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 68-36 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 19. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 19. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 21. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 21. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 22. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 23. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 23\times 4 &= 92 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 92-36 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 23\times 4 &= 92 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 92-36 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 72-36 &= 36 \end{align*} Thus, $a+b+c+d=36$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 19. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 19\times 4 &= 76 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 76-36 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 80-36 &= 44 \end{align*} Thus, $a+b+c+d=44$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 21. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 22. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 21. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 21\times 4 &= 84 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 84-36 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 22. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 23. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 23\times 4 &= 92 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 92-36 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 23\times 4 &= 92 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 92-36 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 24. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 96-36 &= 60 \end{align*} Thus, $a+b+c+d=60$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 96-36 &= 60 \end{align*} Thus, $a+b+c+d=60$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 22. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 88-36 &= 52 \end{align*} Thus, $a+b+c+d=52$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 23. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 23\times 4 &= 92 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 92-36 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 23\times 4 &= 92 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 92-36 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 24. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 96-36 &= 60 \end{align*} Thus, $a+b+c+d=60$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 96-36 &= 60 \end{align*} Thus, $a+b+c+d=60$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 25. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 25\times 4 &= 100 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 100-36 &= 64 \end{align*} Thus, $a+b+c+d=64$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 25\times 4 &= 100 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 100-36 &= 64 \end{align*} Thus, $a+b+c+d=64$.
Mark fills in each circle with a number from $ 1 $, $ 2 $, $ 3 $, $\dots$, $ 8 $, such that the sum of numbers at all corners of any triangle is 26. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 104-36 &= 68 \end{align*} Thus, $a+b+c+d=68$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 2 + 3+\cdots + 8 &= 36 \\ 104-36 &= 68 \end{align*} Thus, $a+b+c+d=68$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 18. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 72-64 &= 8 \end{align*} Thus, $a+b+c+d=8$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 18\times 4 &= 72 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 72-64 &= 8 \end{align*} Thus, $a+b+c+d=8$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 80-64 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 80-64 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 22. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 88-64 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 88-64 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 24. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 96-64 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 96-64 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 20. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 80-64 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 20\times 4 &= 80 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 80-64 &= 16 \end{align*} Thus, $a+b+c+d=16$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 22. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 88-64 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 88-64 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 24. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 96-64 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 96-64 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 26. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 104-64 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 104-64 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 24. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 96-64 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 96-64 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 26. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 104-64 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 104-64 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 28. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 28. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 30. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 30\times 4 &= 120 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 120-64 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 30\times 4 &= 120 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 120-64 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 32. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 32\times 4 &= 128 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 128-64 &= 64 \end{align*} Thus, $a+b+c+d=64$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 32\times 4 &= 128 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 128-64 &= 64 \end{align*} Thus, $a+b+c+d=64$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 22. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 88-64 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 22\times 4 &= 88 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 88-64 &= 24 \end{align*} Thus, $a+b+c+d=24$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 24. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 96-64 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 24\times 4 &= 96 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 96-64 &= 32 \end{align*} Thus, $a+b+c+d=32$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 26. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 104-64 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 104-64 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 28. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 26. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 104-64 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 26\times 4 &= 104 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 104-64 &= 40 \end{align*} Thus, $a+b+c+d=40$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 28. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 30. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 30\times 4 &= 120 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 120-64 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 30\times 4 &= 120 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 120-64 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 30. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 30\times 4 &= 120 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 120-64 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 30\times 4 &= 120 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 120-64 &= 56 \end{align*} Thus, $a+b+c+d=56$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 32. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 32\times 4 &= 128 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 128-64 &= 64 \end{align*} Thus, $a+b+c+d=64$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 32\times 4 &= 128 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 128-64 &= 64 \end{align*} Thus, $a+b+c+d=64$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 34. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 34\times 4 &= 136 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 136-64 &= 72 \end{align*} Thus, $a+b+c+d=72$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 34\times 4 &= 136 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 136-64 &= 72 \end{align*} Thus, $a+b+c+d=72$.
Mark fills in each circle with a number from $ 1 $, $ 3 $, $ 5 $, $\dots$, $ 15 $, such that the sum of numbers at all corners of any triangle is 28. Find $(a+b+c+d)$.
Sorry. Please check the correct answer below.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\ 1 + 3 + 5+\cdots + 15 &= 64 \\ 112-64 &= 48 \end{align*} Thus, $a+b+c+d=48$.
Yay! Your are right.
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\