- Tap to switch levels
- Numbers
- Measurement
- Data Analysis
- Geometry
- Ratio and Percentage
- Speed
- Algebra
- Others
- Integration

SELECT A SUBJECT

- Numbers

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 $(a+b+c+d)$.

10

11

12

13

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}

1

2

3

4

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*}

337

338

339

340

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 $(a+b+c+d)$.

4

5

6

7

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 $(a+b+c+d)$.

6

7

8

9

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 $(a+b+c+d)$.

10

11

12

13

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 $(a+b+c+d)$.

14

15

16

17

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 $(a+b+c+d)$.

17

18

19

20

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, $a+b+c+d=20$.

11

12

13

14

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.

13

14

15

16

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.

18

19

20

21

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.

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 $(a+b+c+d)$.

24

25

26

27

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, $a+b+c+d=24$.

19

20

21

22

Sorry. Please check the correct answer below.

Yay! Your are right.

24

25

26

27

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.

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 $(a+b+c+d)$.

26

27

28

29

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, $a+b+c+d=28$.

25

26

27

28

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.

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 $(a+b+c+d)$.

31

32

33

34

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 $(a+b+c+d)$.

36

37

38

39

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, $a+b+c+d=36$.

14

15

16

17

Sorry. Please check the correct answer below.

Yay! Your are right.

17

18

19

20

Sorry. Please check the correct answer below.

Yay! Your are right.

23

24

25

26

Sorry. Please check the correct answer below.

Yay! Your are right.

26

27

28

29

Sorry. Please check the correct answer below.

Yay! Your are right.

22

23

24

25

Sorry. Please check the correct answer below.

Yay! Your are right.

27

28

29

30

Sorry. Please check the correct answer below.

Yay! Your are right.

32

33

34

35

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.

29

30

31

32

Sorry. Please check the correct answer below.

Yay! Your are right.

34

35

36

37

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.

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 $(a+b+c+d)$.

38

39

40

41

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, $a+b+c+d=40$.

27

28

29

30

Sorry. Please check the correct answer below.

Yay! Your are right.

30

31

32

33

Sorry. Please check the correct answer below.

Yay! Your are right.

34

35

36

37

Sorry. Please check the correct answer below.

Yay! Your are right.

36

37

38

39

Sorry. Please check the correct answer below.

Yay! Your are right.

38

39

40

41

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.

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 $(a+b+c+d)$.

44

45

46

47

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, $a+b+c+d=44$.

37

38

39

40

Sorry. Please check the correct answer below.

Yay! Your are right.

43

44

45

46

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.

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 $(a+b+c+d)$.

48

49

50

51

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 $(a+b+c+d)$.

51

52

53

54

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, $a+b+c+d=52$.

18

19

20

21

Sorry. Please check the correct answer below.

Yay! Your are right.

23

24

25

26

Sorry. Please check the correct answer below.

Yay! Your are right.

26

27

28

29

Sorry. Please check the correct answer below.

Yay! Your are right.

30

31

32

33

Sorry. Please check the correct answer below.

Yay! Your are right.

28

29

30

31

Sorry. Please check the correct answer below.

Yay! Your are right.

30

31

32

33

Sorry. Please check the correct answer below.

Yay! Your are right.

36

37

38

39

Sorry. Please check the correct answer below.

Yay! Your are right.

35

36

37

38

Sorry. Please check the correct answer below.

Yay! Your are right.

40

41

42

43

Sorry. Please check the correct answer below.

Yay! Your are right.

41

42

43

44

Sorry. Please check the correct answer below.

Yay! Your are right.

29

30

31

32

Sorry. Please check the correct answer below.

Yay! Your are right.

35

36

37

38

Sorry. Please check the correct answer below.

Yay! Your are right.

40

41

42

43

Sorry. Please check the correct answer below.

Yay! Your are right.

40

41

42

43

Sorry. Please check the correct answer below.

Yay! Your are right.

42

43

44

45

Sorry. Please check the correct answer below.

Yay! Your are right.

47

48

49

50

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.

41

42

43

44

Sorry. Please check the correct answer below.

Yay! Your are right.

45

46

47

48

Sorry. Please check the correct answer below.

Yay! Your are right.

50

51

52

53

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.

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 $(a+b+c+d)$.

54

55

56

57

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, $a+b+c+d=56$.

36

37

38

39

Sorry. Please check the correct answer below.

Yay! Your are right.

40

41

42

43

Sorry. Please check the correct answer below.

Yay! Your are right.

43

44

45

46

Sorry. Please check the correct answer below.

Yay! Your are right.

42

43

44

45

Sorry. Please check the correct answer below.

Yay! Your are right.

45

46

47

48

Sorry. Please check the correct answer below.

Yay! Your are right.

51

52

53

54

Sorry. Please check the correct answer below.

Yay! Your are right.

46

47

48

49

Sorry. Please check the correct answer below.

Yay! Your are right.

51

52

53

54

Sorry. Please check the correct answer below.

Yay! Your are right.

53

54

55

56

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.

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 $(a+b+c+d)$.

59

60

61

62

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, $a+b+c+d=60$.

51

52

53

54

Sorry. Please check the correct answer below.

Yay! Your are right.

56

57

58

59

Sorry. Please check the correct answer below.

Yay! Your are right.

59

60

61

62

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.

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 $(a+b+c+d)$.

61

62

63

64

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 $(a+b+c+d)$.

66

67

68

69

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 $(a+b+c+d)$.

6

7

8

9

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 $(a+b+c+d)$.

13

14

15

16

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 $(a+b+c+d)$.

22

23

24

25

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 $(a+b+c+d)$.

29

30

31

32

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, $a+b+c+d=32$.

14

15

16

17

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.

23

24

25

26

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.

32

33

34

35

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.

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 $(a+b+c+d)$.

40

41

42

43

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, $a+b+c+d=40$.

32

33

34

35

Sorry. Please check the correct answer below.

Yay! Your are right.

40

41

42

43

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.

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 $(a+b+c+d)$.

48

49

50

51

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, $a+b+c+d=48$.

46

47

48

49

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.

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 $(a+b+c+d)$.

54

55

56

57

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 $(a+b+c+d)$.

61

62

63

64

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, $a+b+c+d=64$.

23

24

25

26

Sorry. Please check the correct answer below.

Yay! Your are right.

32

33

34

35

Sorry. Please check the correct answer below.

Yay! Your are right.

40

41

42

43

Sorry. Please check the correct answer below.

Yay! Your are right.

45

46

47

48

Sorry. Please check the correct answer below.

Yay! Your are right.

39

40

41

42

Sorry. Please check the correct answer below.

Yay! Your are right.

46

47

48

49

Sorry. Please check the correct answer below.

Yay! Your are right.

55

56

57

58

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.

56

57

58

59

Sorry. Please check the correct answer below.

Yay! Your are right.

64

65

66

67

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.

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 $(a+b+c+d)$.

72

73

74

75

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, $a+b+c+d=72$.

46

47

48

49

Sorry. Please check the correct answer below.

Yay! Your are right.

Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added. \begin{align*} 28\times 4 &= 112 \\