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NCERT Exemplar for Class 8 Maths Solutions Chapter 13 Playing With Numbers

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Last updated date: 25th May 2024
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Class 8 Maths NCERT Exemplar Solutions Chapter 13 Playing With Numbers

Free PDF download of NCERT Exemplar for Class 8 Maths Chapter 13 - Playing With Numbers Numbers solved by expert Maths teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 13 - Playing With Numbers exercise questions with solutions to help you to revise the complete syllabus and score more marks in your examinations.


Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 8 Science, Maths solutions, and solutions of other subjects. You can also download NCERT Solutions for Class 8 Maths to help you to revise the complete syllabus and score more marks in your examinations.

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Access NCERT Exemplar Solutions for Class 12 Mathematics Chapter 13 - Playing with Numbers

In examples 1 to 4, out of four options, only one is correct. Write the correct answer.

1. Generalized form of a three-digit number xyz is

(a) x + y + z 

(b) 100x + 10y + z 

(c) 100z + 10y + x 

(d) 100y + 10x + z

Ans: The correct answer is (b).


2. The usual form of 100a + b + 10c is 

(a) abc 

(b) cab 

(c) bac 

(d) acb

Ans: The correct answer is (d).

Because, Here a is in the hundreds place, c is in tens place, and b is in ones place.


3. If 5 × A = CA then the values of A and C are

(a) A = 5, C = 1 

(b) A = 4, C = 2

(c) A = 5, C = 2 

(d) A = 2, C = 5

Ans: The correct answer is (c).

Here 5 × A = CA where A is in ones place and C is in tens place

Therefore 5A = 10 × C + A

4A = 10C 

2A = 5C 

Hence A = 5 and C = 2


4. If 5 A + 25 is equal to B 2, then the value of A + B is

(a) 15 

(b) 10 

(c) 8 

(d) 7

Ans: The correct answer is (a).

5A = 5 × 10 + A, as A is in ones place and 5 is in tens place

B2 = B  × 10 + 2, as 2 is in ones place and B is in tens place

\[ \Rightarrow \] 5A + 25 = B2

A number should be substituted for A, so that on the RHS ones place will be 2

Add 7 in place of A,

\[ \Rightarrow \] 57 + 25 = 82

Hence A = 7, B = 8

So, A + B = 7 + 8 = 15


In Examples 5 To 7, Fillin the Blanks to Make the Statements True.

5. The number ab – ba where a and b are digits and a > b is divisible by ________.

Ans: The difference between a two-digit number and the result of reversing its digits is always divisible by 9.


6. When written in the usual form 100a + 10c + 9 is equal to ________.

Ans: ac9

a is in 100s place, c is in 10s place, 9 is in ones place


7. If AB × B = 9B, then A = _________, B = _________.

Ans: 9,1

9 1 1 = 9 1

9 = 9


In examples 8 to 10, state whether the statements are true (T) or false (F).

8. If abc, cab, bca are three digit numbers formed by the digits a, b, and c then the sum of these numbers is always divisible by 37.

Ans: True

Given abc + cab + bca can be written in expanded form,

\[ \Rightarrow \] (100 a + 10 b + c) + (100 c + 10 a + b) + (100 b + 10 c + a)

\[ \Rightarrow \] (100 a + 100 c + 100 b) + (10 b + 10 a + 10 c) + (c + b + a)

\[ \Rightarrow \] 100(a + c + b) + 10(b + a + c) + (c + b + a)

\[ \Rightarrow \] (100 + 10 + 1) (a + b + c)

\[ \Rightarrow \] (111) (a + b + c) = 37       (because 111 = 3 37)


9. Let ab be a two-digit number, then ab + ba is divisible by 9.

Ans: False

Given ab + ba can be written in expanded form,

\[ \Rightarrow \] (10 a + b) + (10 b + a)

\[ \Rightarrow \] 10(a + b) + (b + a)

\[ \Rightarrow \] (10 + 1)(a + b)

\[ \Rightarrow \] 11(a + b)


10. If a number is divisible by 2 and 4, then it will be divisible by 8.

Ans: False

12, 20, 28 that can be divided by 2 and 4, but is not divisible by 8.


11. A three-digit number 42x is divisible by 9. Find the value of x.

Ans:  Since 42x is divisible by 9, the sum of its digits, i.e.

4 + 2 + x must be divisible by 9.

i.e. 6 + x is divisible by 9

i.e. 6 + x = 9 or 18, _____.

Since x is a digit, therefore 6 + x = 9 or, x = 3


12. Find the value of A and B if

\[\begin{align}   & \text{4 1 A} \\  & \underline{\text{+ B 4}} \\  & \text{5  1 2} \\ \end{align}\]

Ans:  From one's column A + 4 gives a number whose digit is 2. So, A = 8. The value of B can be obtained by solving 2 + B is a number whose one digit is 1. So,B = 9.

\[ \begin{align}   & \text{4 1 8} \\   & \underline{\text{+ 9 4}} \\  & \text{5  1 2} \\ 

\end{align}\]


13. Suppose that the division x ÷ 5 leaves a remainder 4 and the division x ÷ 2 leaves a remainder 1. Find the ones digit of x.

Ans:  Since x ÷ 5 leaves a remainder 4, so one digit of x can be 4 or 9. Also, since x ÷ 2 leaves a remainder 1, so one digit must be 9 only.


14. If 756x is divisible by 11, where x is a digit find the value of x

Ans:  For a number to be divisible by 11 

Sum of digit in odd places- Sum of digit in even places = multiple of 11

Sum of digit in odd places= 7+6

Sum of digit in odd places= 5+x 

Sum of digit in odd places- Sum of digit in even places =13-(5+x)=8-x

Now (x – 8) should be equal to 0 or a multiple of 11 (i.e. 11, 22,33, ..., etc.)

x – 8 = 0

x = 8 or x – 8 = 11

x = 11 + 8 = 19

Since x is a digit so it can take values from 0 – 9

Hence x = 8

Required number is 7568.


1.Generalised form of a four-digit number abdc is

(a) 1000a + 100b + 10c + d 

(b) 1000a + 100c + 10b + d

(c) 1000a + 100b + 10d + c 

(d) a  \[ \times \] b  \[ \times \]  c  \[ \times \]  d

Ans:  (c) A number is expressed as the sum of the products of its digits with their respective place values.

abdc is written in generalised form as 1000a + 100b + 10d + c.i.e. abdc = 1000a + 100b + 10d + c


2. Generalised form of a two-digit number xy is

(a)x + y 

(b)10x + y 

(c)10x-y 

(d)10y+x

Ans:  (b)The two-digit number xy is represented as the sum of the product of its digits with their respective place value.


3. The usual form of 1000a + 10b + c is

(a) abc 

(b) abc0 

(c) a0be 

(d) ab0c

Ans:  (c) A number is expressed as the sum of the products of its digits with their respective place values.

We can write it as 1000  \[ \times \]  a + 100  \[ \times \]  0 + 10  \[ \times \]  b + c i.e. a0bc, which is the usual form.


4. Let abc be a three-digit number. Then, abc – cba is not divisible by

(a) 9 

(b) 11 

(c) 18 

(d) 33

Ans:  (c) Given, abc is a three-digit number.

Then, abc = 100a + 10b + c

and cba = 100c + 10b + a

abc -eba = (100a + 10b + c)- (100c + 10b + a)

= 100a – a + 10b – 10b + c -100

= 99a – 99c = 99 (a -c)

= abc-cba is divisible by 99.

\[ \Rightarrow \] abc – cba is divisible by 9,11,33, but it is not divisible by 18.


5. The sum of all the numbers formed by the digits x, y, and z of the number xyz is divisible by 

(a) 11 

(b) 33 

(c) 37 

(d) 74

Ans:  (c) We have, xyz + yzx + zxy

= (100x + 10y + z) + (100y + 10 z+ x) + (100z+ 10x + y) …(i)

= 100x + 10x + x + 10y + 100y + y + z+ 100z+ 10Z

= 111x + 111y + 111z = 111 (x + y + z)

= 3 \[ \times \] (37)  \[ \times \]  (x + y + z)

Hence, Eq. (i) is divisible by 37, but not divisible by 11,33 and 74.


6. A four-digit number aabb is divisible by 55. Then, possible value(s) of b is/are  

(a) 0 and 2 

(b) 2 and 5 

(c) 0 and 5 

(d) 7

Ans:  (c) It is given that aabb is divisible by 55. Then, it is also divisible by 5.

Now, if a number is divisible by 5, then its unit digit is either 0 or 5. Hence, the possible values of b are 0 and 5.


7. Let abc be a three-digit number. Then, abc + bca + cab is not divisible by

(a) a + b + c 

(b) 3 

(c) 37 

(d) 9

Ans:  (d) We know that the sum of three-digit numbers taken in cyclic order can be written as 111 (a + b + c).

i.e. abc + pea + cab = \[ 3 \times \]  \[ 37 \times \] (a + b + c).Hence, the sum is divisible by 3, 37 and (a + b + c) but not divisible by 9.


8. A four-digit number 4ob5 is divisible by 55. Then, the value of b-a is

(a) 0 

(b) 1 

(c) 4 

(d) 5

Ans:  (b) Given, a four-digit number 4ab5 is divisible by 55. Then, it is also divisible by 11.

The difference of sum of its digits in odd places and sum of its digits in even places is either 0 or multiple of 11.i.e. (4 + b) – (a + 5) is 0 or a multiple of 11, if 

4+b-a-5 = 0

 \[ \Rightarrow \] b-a = 1


9. If abc is a three-digit number, then number abc -a-b-c is divisible by

(a) 9 

(b) 90 

(c)10 

(d) 11

Ans:  (d) We have, abc= 100a + 10b+c

abc – a – b-c = (100a + 10b + c)-a – b-c = 100a – a + 10b – b = 99a + 9b = 9(11a + b)

Hence, the given number abc – a – b-c is divisible by 9.


10. A six-digit number is formed by repeating a three-digit number. For example, 256256, 678678, etc. Any number of this form is divisible by 

(a) 7 only 

(b) 11 only 

(c) 13 only 

(d) 1001

Ans:  (d) A six-digit number which is formed by repeating a three-digit number is = abcabc.

 abcabc = 100000a + 10000b + 1000c + 100a + 10b + c 

= a(100000 + 100) + b(10000 + 10) + c (1000 + 1) 

= 1001 (a × 100 + b × 10 + c) 

Therefore, A six-digit number which is formed by repeating a three-digit number is divisible by 1001


11. If the sum of digits of a number is divisible by three, then the number is always divisible by 

(a) 2 

(b) 3 

(c) 6 

(d) 9

Ans:  (b) A six-digit number which is formed by repeating a three-digit number is = abcabc. 

Abcabc = 100000a + 10000b + 1000c + 100a + 10b + c

= a(100000 + 100) + b(10000 + 10) + c (1000 + 1) 

= 1001 (a × 100 + b × 10 + c) 

Therefore, A six-digit number which is formed by repeating a three-digit number is divisible by 1001. Hence it is divisible by 3


12. If x +y + z = 6 and z is an odd digit, then the three-digit number xyz is

(a) an odd multiple of 3 

(b) an odd multiple of 6 

(c) an even multiple of 3 

(d) an even multiple of 9

Ans: (a) We have, x + y+ z= 6 and z is an odd digit. Since the sum of the digits is divisible by 3, it will also be divisible by 2 and 3 but the unit digit is odd, so it is divisible by 3 only. Hence, the number is an odd multiple of 3.


13 If 5A + 53 = 65, then the values of A and B is 

(a) A = 2,8 = 3 

(b) A = 3,8 = 2 

(c)A = 2,8 = 1 

(d)A=1,8 = 2

Ans:  given, 5 A + B 3 = 65

 \[ \begin{bmatrix} 5 & A \\ +B & 3 \\ 6 & 5 \end{bmatrix} \]

In the first or 1’s column, A + 3 = 5 A is a digit so the value is taken from 0 to 9. If we take the maximum value 9, Then, \[ A + 3 = 9 + 3 = 12 \ne 5 \] 

Therefore, A = 5 -3 = 2;

In the second or 10’s column, If A = 2 , Hence there is no carry from A + 3. Therefore, B = 6  5 = 1 Hence, A =2 and B = 1.


14. If A3 + 8B = 150, then the value of A + B is 

(a) 13 

(b) 12 

(c) 17 

(d) 15

Ans:  (a) We have, A3+ 8B = 150

Here, 3 + B = 0, so 3 + B is a two-digit number whose unit’s digit is zero.

 3+B = 10  \[ \Rightarrow ~ \] B = 7

: Now, considering ten’s column, A+ 8 + 1 = 15

= A + 9 = 15

\[ \Rightarrow ~ \] A = 6

Hence, A+B=6+7 = 13


15. If 5A x A = 399, then the value of A is

(a) 3 

(b) 6 

(c) 7 

(d) 9

Ans:  (c) We have, 5A \[ \times \] A = 399

Here, \[ A \times  A= 9 \] i.e. \[ A \times A \] is the number 9 or a number whose unit’s digit is 9. Thus, the number whose product with itself produces a two-digit number having its unit’s digit as 9 is 7.

i.e. \[ A  \times  A = 49 \] 

\[ \Rightarrow ~ \]  A=7

Now, \[ 5 \times  A + 4 = 39 \]

\[ \Rightarrow ~ 5 \times \] 7+4 = 39

So, A satisfies the product.

Hence, the value of A is 7.


16. If 6A \[ \times  B \Rightarrow ~ \] 488, then the value of A-B is

(a)-2 

(b) 2 

(c) -3 

(d) 3

Ans:  (c) Given,\[ 6A \times B = A86 \]

Let us assume, A = 1 and S = 3 Then, LHS = 61 $\times $ 3 = 183 and RHS = 183 Thus, our assumption is true. A-6 = 1-3=-2


17. Which of the following numbers is divisible by 99?

(a) 913462 

(b) 114345 

(c) 135792 

(d) 3572406

Ans:  (b) Given a number is divisible by 99.

Now, going through the options, we observe that the number (b) is divisible by 9 and 11 both as the sum of digits of the number is divisible by 9 and sum of digits at odd places = sum of digits at even places.


Fill in the Blanks

In questions 18 to 33, fill in the blanks to make the statements true.


18. 3134673 is divisible by 3 and  _______.

Ans:  9

3134673 is divisible by 3 and 9 as the sum of the digits, 3+1+3+4+6+7 + 3 = 27 is divisible by both 3 and 9.


19. 20×3 is a multiple of 3, if the digit x is   _______ or _______ or_______.

Ans:  1,4,7

We know that, if a number is a multiple of 3, then the sum of its digits is again a multiple of 3, i.e. 2+0+x+3 is a multiple of 3.

x + 5 = 0, 3, 6, 9, 12, 15 But, x is a digit of the number 20 x 3.

x can take values 0, 1,2, 3,……..9.

\[ \Rightarrow ~ \] x + 5 = 6 or 9 or 12

Hence, x = 1 or4 or 7


20. 3×5 is divisible by 9, if the digit x is _______.

Ans:  1

Since the number 3x5 is divisible by 9, then the sum of its digits is also divisible by 9. i.e. 3 + x + 5 is divisible by 9.

\[ \Rightarrow ~ \] x + 8 can take values 9,18, 27,…

But x is a digit of the number 3x5, so x = 1.


21. The sum of a two-digit number and the number obtained by reversing the digits is always divisible by_______.

Ans:  11

Let ab be any two-digit number, then the number obtained by reversing its digits is ba.

Now, ab + ba = (10a + b) + (10b + a) = 11a + 11b = 11(a + b)

Hence, ab + ba is always divisible by 11 and (a + b).


22. The difference between a two-digit number and the number obtained by reversing its digits is always divisible by_______.

Ans:  9

Let ab be any two-digit number, then we have

ab – ba = (10a + b)- (10b + a)

\[ \Rightarrow ~ \] 9a – 9b = 9(a – b)

Hence, ab – ba is always divisible by 9 and (a – b).


23. The difference between the three-digit number and the number obtained by putting the digits in reverse order is always divisible by 9 and_______.

Ans:  11

Let abc be a three-digit number, then we have

abc -cba = (100a + 10b + c)- (100c + 10b+ a) ; = (100a – a) + (c – 10Cc)

= 99a – 99c = 99(a -c)

= 9 \[ \times  11 \times  (a – c) \]

Hence, abc – cba is always divisible by 9,11 and (a-c).


24.If +\[ \frac{\overset{\text{2}}{\mathop{\text{A}}}\,\overset{\text{B}}{\mathop{\text{  B}}}\,}{\text{8   A}}\] then A = ______ and B = ______ 

Ans:  A=6 and B = 3

In the question B = 0 to 9. 

For B = 0, A =0 (not fit in 10’s column) 

For B =3, A =6 (Fit in 10’s column)

 Hence, A=6 and B = 3

              

25. If \[\frac{\overset{\text{A}}{\mathop{\times }}\,\overset{\text{ B}}{\mathop{\text{  B}}}\,}{\text{9   6}}\] then A = ______ and B = ______

Ans:  A=2 and B = 4
From the question, B × B = 6

Therefore, the value of B is 4 or 6. 

For, B = 4, 

A × B + 1= 9

Hence, A=2 and B = 4


26. If \[\frac{\overset{\text{B}}{\mathop{\text{ }\!\!\times\!\!\text{ }}}\,\overset{\text{  1}}{\mathop{\text{  B}}}\,}{\text{4   9B}}\] then A = ______ and B = ______

Ans:  B = 7

From the question,

B × 1 = 1

\[ \Rightarrow \] B × B = 49 Hence B = 7


27.1 × 35 is divisible by 9 if x =_______.

Ans: A number is divisible by 9 if the sum of its digits is also divisible by 9. Therefore, 1 + x +3 + 5 =a number multiple of 9.

x + 9 = 0, 9,18,... According to the question it is a single digit number. Hence, x = 0.


28. A four-digit number abed is divisible by 11, if d + b = _______ or _______.

Ans:  a + c,12(a + c)

We know that a number is divisible by 11, if the difference between the sum of digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

Hence, abcd is divisible by 11, if (d + b)- (a + c) = 0,11,22, 33,…

\[ \Rightarrow ~ \] d + b = a + c or d + b = 12(a + c)


29. A number is divisible by 11, if the difference between the sum of digits at its odd places and that of digits at the even places is either 0 or divisible by _______.

Ans:  11

By test of divisibility by 1,1, we know that a number is divisible by 11, if the sum of digits at odd places and even places are equal or differ by a number, which is divisible by 11.


30. If a three-digit number abc is divisible by 11, then _______ is either 0 or multiple of 11.

Ans:  (a+c)-b

Since, abc is divisible by 11, the difference of sum of its digits at odd places and that of even places is either zero or multiple of 11, i.e. (a + c) – b is either zero or multiple of 11.


31.If A × 3 = 1A, then A = ______.

Ans:

From the question, 

3 × A = a two digit number with unit digit A.

A = 0 to 9. 

Only 5 is a number whose product with 3 is a two digit number with unit digit itself (5). 

Hence, A = 5


32. If B × B = AB, then either A = 2, B = 5 or A = ______, B = ______.

Ans: 3, 6

From the question, B × B is a two-digit number with unit digit B.

B = 0 to 9 

Only 5 and 6 are the number whose product with itself is a two-digit number with unit digit itself


33. If the digit 1 is placed after a two-digit number whose ten’s is t and one’s digit is u, the new number is _______.

Ans:  tu1

Given, a two-digit number whose one’s digit is u in and tens digit isu. If the digit 1 is placed after this number, then the next number will be tu1.


True/False

In questions 34 to 44, state whether the given statements are True or False.

34. A two-digit number ab is always divisible by 2, if b is an even number.

Ans:  True

By the test of divisibility by 2, we know that a number is divisible by 2, if its unit’s digit is even.


35. A three-digit number abc is divisible by 5, if c is an even number.

Ans:  False

By the test of divisibility by 5, we know that if a number is divisible by 5, then its one’s digit will be either 0 or 5, i.e. the numbers ending with the digits 0 or 5 are divisible by 5.


36 A four-digit abcd is divisible by 4, if ab is divisible by 4.

Ans:  False

As we know, if a number is divisible by 4, then the number formed by its digits in unit’s and ten’s place is divisible by 4.


37. A three-digit number abc is divisible by 6, if c is an even number and a + b + c is a multiple of 3.

Ans:  True

If a number is divisible by 6, then it is divisible by both 2 and 3. Since, abc is divisible by 6, it is also divisible by 2 and 3. Therefore, c is an even number and the sum of digits is divisible by 3, i.e. multiple of 3.


38. Number of the form 3N + 2 will leave the remainder 2 when divided by 3.

Ans:  True

Let x = 3N + 2. Then, it can be written as.

x = (a multiple of 3) + 2

i.e. x is a number which is 2 more than a multiple of 3

i.e. x is a number, which when divided by 3, leaves the remainder 2.


39. Number 7N+1 will leave remainder 1 when divided by 7.

Ans:  True

Given, a number of the form 7N + 1 = x (say)

Here, we observe that x is a number which is one more than a multiple of 7. i.e. when x is divided by 7, it leaves the remainder 1.


40. If a number a is divisible by b, then it must be divisible by each factor of b.

Ans:  True

Given, a is divisible by b.

Let b = \[p_{1} \times p_{2}\], where  p1 and p2 are primes.

Since, a is divisible by b, a is a multiple of b

i.e. a = mb

\[ \Rightarrow  a = m (p_{1})(p_{2}) \]

or a= \[cp_{2}\]=\[dp_{1}\], where \[ c = mp_{1}\], \[d = mp_{2}\]

\[ \Rightarrow \] a is a multiple of \[p_{1}\] as well as \[p_{2}\].

Hence, a is divisible by each factor b.


41. If AB × 4 = 192, then A + B = 7.

Ans: False

From the question,

 B + 4 is a two-digit number with unit digit 2. 

Therefore, B = 3 or 8. For B =3, (A × 4) + 1  19 

Hence B =8 and A = 4 Then A + B = 12.


42. If AB + 7C = 102, where B ≠ 0, C ≠ 0, then A + B + C = 14

Ans: True

From the question, B + C = 2 or a two-digit number with unit digit 2 and less than 14.

Therefore, B = 1, 5 ,7 and C =1 , 5, 7

B =1 and C =1 is not suitable for the question.

Hence, For B= 5, C =7 and A = 2 For B =7, C=5 and A = 2

 

43. If 213×27 is divisible by 9, then the value of x is 0.

Ans: False

Given, 213x27 is divisible by 9, so the sum of its digits is also divisible by 9.

i.e. 2+1+ 3+ x +2 +7 =0, 9,18, 27, 36,…

\[ \Rightarrow \] x+ 15= 0,9,18,27,36,…

\[ \Rightarrow \] x + 15 = 18 [x is a digit of a number]

\[ \Rightarrow \] x= 3


44. If N ÷ 5 leaves remainder 3 and N ÷ 2 leaves remainder 0, then N ÷ 10 leaves remainder 4

Ans: \[N  \div 5  = Remainder 3 \]

\[ N \div 2  = Remainder 0 \]

\[N \div 10  = Remainder 4 \]

N=5n+3   Where, n =0, 1, 2, 3, ....

According to the question, N \[ \div \] 2  leaves remainder 0. 

Therefore N is an even number and n =1, 3, 5, 7, ... If we put the values of n in equation N=5n+3,

Then, N = 8, 18, 28, …

If we divide these values of N by 10 then it leaves the remainder 8.


45. Find the least value that must be given to number a, so that the number 91876a2 is divisible by 8.

Ans:  Given, 91876a2 is divisible by 8.

Since, we know that, if a number is divisible by 8, then the number formed by the last 3 digits is divisible by 8.

So, 6a2 is divisible by 8.

Here, a can take values from 0 to 9.

For a = 0, 602 is not divisible by 8.

For a = 1, 612, which is not divisible by 8.

For a = 3, 632 is divisible by 8.

Hence, the minimum value of a is 3 to make 91876a2 divisible by 8.


46. If  \[\frac{\overset{\text{1}}{\mathop{\text{ }\!\!\times\!\!\text{ }}}\,\overset{\text{  P}}{\mathop{\text{  P}}}\,}{\text{Q   6}}\] then where Q – P = 3, then find the values of P and Q

Ans: Given, Q-P=3
From the question, P$\times $P=6   

Therefore, P = 4 or 6 

If, P =4, Q =5 not suitable for the question.

Hence, P =6 and Q = 9


47. If 1AB + CCA = 697 and there is no carryover, in addition, find the value of A + B + C.

Ans: According to the question there is no carry-over in addition.

1 + C = 6 \[ \Rightarrow \] C = 5 

A + C = 9 \[ \Rightarrow \] A = 4

Now, B + A = 7 \[ \Rightarrow \] B = 3 

Hence A + B + C = 12


48. A five-digit number AABAA is divisible by 33. Write all the numbers of this form.

Ans:  Given, the number of the form AABAA is divisible by 33. Then, it is also divisible by 3 and11, as if a number a is divisible by b, then it is also divisible by each factor of b.

Since AABAA is divisible by 3, the sum of its digits is also divisible by 3.

 i.e. 4 + 4 + 8 + A +4 = 0,3, 6,9…

or 4A + 8 = 0, 3, 6 9,… …(i)

From Eq. (i), we have

Further, the given number is also divisible by 11, therefore 

(2A + 8) – 2A = 0,11,22,…

Therefore, B = 0 (single digit)

A = 3, 6, 9, ..... 

Hence numbers are 33066, 66066, 99099.


49. Find the value of the letters in each of the following questions

If  \[\frac{\overset{\text{   A}}{\mathop{\text{+A}}}\,\overset{\text{  A}}{\mathop{\text{  A}}}\,}{\underline{\text{XA   2}}}\]

Ans: From the 1’s column A = 0 to 9.

From the 10’s column

A = 5 to 9 (The A + A is two digit number) 

A = 5 to 8 does not satisfy the 10’s column 

Hence, A =9 and then Z =8 and X =1.


50.   \[\frac{\overset{\text{   8}}{\mathop{\text{+4}}}\,\overset{\text{  5}}{\mathop{\text{  A}}}\,}{\underline{\text{BC   3}}}\]

Ans: From the 1’s column, 

The value of 5 + A is a two digit number whose last digit is 3. 

Therefore, 5 + A = 13 \[ \Rightarrow \] A = 8 

From the column 10’s, 

10B + C =8 + 4 + 1

\[ \Rightarrow \]10B + C = 13 =10 × 1 + 3 

Hence, A =8, B =1 and C =3


51.  \[\frac{\overset{\text{   B}}{\mathop{\text{+8}}}\,\overset{\text{  6}}{\mathop{\text{  A}}}\,}{\underline{\text{CA   2}}}\]

Ans: from the 1’s column,

6 + A = 12 \[ \Rightarrow \] A = 6 

From the 10’s column, 

CA = B + 8 + 1 

C6 = B + 9 

From the above equation, B + 9 is a number with unit digit 6.

B =7 C =1 

Hence, A =6, B =7 and C =1. 


52. \[+\frac{\overset{\text{1}}{\mathop{\text{A}}}\,\text{ }\overset{\text{B}}{\mathop{\text{B}}}\,\overset{\text{  A}}{\mathop{\text{  A}}}\,}{\underline{\text{8   A    2}}}\] 

Ans: from the 1’s column, 6 + A = 12 \[ \Rightarrow \] A = 6 

From the 10’s column, 

B + B + 1 =B 

The above condition is fulfilled when B = 9. 

From the 100’s column, 

A +1 + 1 = 8 

\[ \Rightarrow \] A = 6 = Satisfy the 1’s column value. 

Hence, A =6, B = 9.


53. \[\frac{\text{+}\overset{\text{C}}{\mathop{\text{C}}}\,\text{ }\overset{\text{B}}{\mathop{\text{B}}}\,\overset{\text{  A}}{\mathop{\text{  A}}}\,}{\underline{\text{ 1 8  3    0}}}\]

Ans: From the 1’s column,

A + A = 0 \[ \Rightarrow \] A = 0 or 5 

From the 10’s column, 

B + B + 1 =3 

The above condition is fulfilled when B = 1,6. 

From the 100’s column, 

C + C + 1 = 1A 

A = 0 does not satisfy the condition of 10’s column. 

Therefore, A = 5 

C + C =15 -1 \[ \Rightarrow \] C = 7

B = 1 does not satisfy the condition of 10’s column. 

Therefore, B = 6 Hence, A = 5, B = 6 and C = 7


54. \[+\frac{\overset{\text{B}}{\mathop{\text{B}}}\,\text{ }\overset{\text{A}}{\mathop{\text{A}}}\,\overset{\text{  A}}{\mathop{\text{  A}}}\,}{\underline{\text{3  A    8}}}\]

Ans: From the 1’s column, 

A + A =8 \[ \Rightarrow \]  A = 9 (because A + A is not a single digit number)

From the 100’s column,

B + B + 1 =3 

\[ \Rightarrow \] B = 1

Hence, A = 9 and B = 1


55. \[\frac{\text{+}\overset{\text{A}}{\mathop{\text{1}}}\,\text{ }\overset{\text{0}}{\mathop{\text{0}}}\,\overset{\text{1}}{\mathop{\text{ A}}}\,\overset{\text{ B}}{\mathop{\text{ B}}}\,}{\underline{\text{ B  1  0  8}}}\]

Ans: from the 1’s column, B + B = 8

\[ \Rightarrow \] B = 9 (B +B not satisfy the 1000’s column)

From the 10’s column,

A + 1 + 1 =0 

\[ \Rightarrow \] A = 8 

Hence, A = 8 and B = 9 


56. \[\frac{\overset{\text{   A}}{\mathop{\times \text{   }}}\,\overset{\text{ B}}{\mathop{\text{  6}}}\,}{\underline{\text{C 6  8}}}\]

Ans: 6 B is a number with unit digit 8. 

Therefore, B = 3 or 8 

For B = 3 equation \[ A \times 6+1=C6\] is not satisfied.

 For B = 8 equation \[ A \times 6+1=C6 \]   is satisfied.

 Hence, A = 7 and B = 8.


57. \[\frac{\overset{\text{   A}}{\mathop{\text{ }\!\!\times\!\!\text{  A }}}\,\overset{\text{ B}}{\mathop{\text{B}}}\,}{\underline{\text{6 A  B}}}\]

Ans: Given, \[ AB \times AB = 6AB\] ... (i)

 i.e., \[ B \times B \] is a number with unit digit B

 So B= 1 or 5 \[ since, B \ne 0, else AB \times A \ne 6A \] 

A gain, \[  AB \times AB = 6AB  \]

The square of a two digit number is a three digit number.

Therefore, A can be 1, 2 and 3.

If A = 1, 2, 3 and B = 1, equation (i) is not valid. 

For, A = 1, B = 5 equation (i) is not valid. 

Therefore, A = 2, B = 5 equation is valid.

 Hence, A = 2, B =5.


58. \[\frac{\overset{\text{   A}}{\mathop{\text{  }\!\!\times\!\!\text{ }}}\,\overset{\text{ A}}{\mathop{\text{A}}}\,}{\underline{\text{C A  B}}}\]

Ans:  \[ AA \times \] A = three-digit number, that has unit digit B. 

Therefore, A can be 4-9 

And A = 0, 1, 2, 3 give a single digit or a two-digit number. 

As the ten’s digit of the product is A itself. 

Therefore, A$\ne $ 4, 5, 6, 7, and 8.

 Hence, A = 9, B = 1, C = 8. 


59. \[\frac{\text{-}\overset{\text{   A}}{\mathop{\text{ B}}}\,\overset{\text{ B}}{\mathop{\text{7}}}\,}{\underline{\text{   4  5}}}\]

Ans:  In the ones column= B – 7 = 5 

As, 12 – 7 = 5 so B =2 


60. \[\frac{\text{-}\overset{\text{8}}{\mathop{\text{A}}}\,\text{ }\overset{\text{A}} {\mathop{\text{A}}}\,\overset{\text{B}}{\mathop{\text{ C}}}\,\overset{\text{ C}}{\mathop{\text{ 5}}}\,}{\underline{\text{   D  4  8  8}}}\]

Ans:  The ones column = C – 5 = 8 

We know, 13 – 5 = 8 therefore C = 3

 In the ten’s column, B – (C + 1) = 8  

B = 8 + C + 1

\[ \Rightarrow \]B = 8 + 3 +1 

\[\Rightarrow ~~\] B = 12 

Therefore, B = 2

 In the hundred’s column 

A – (B + 1) = 4  

\[ \Rightarrow \] A = 4 + B + 1 

\[ \Rightarrow \] A = 4 + 2 + 1 =7 

In the thousand’s column, 

8 – A = D  

\[ \Rightarrow \] 8 – 7 = D  

\[ \Rightarrow \] D = 1

Thus, A = 7, B = 2, C = 3 and D = 1. 


61. If 27÷A = 33, then find the value of A

Ans:  We observe that 4 x 3 can never be a single-digit number 2, so 4 x 3 must be a two-digit number, whose ten’s digit is 2 and unit’s digit is the number less than or equal to 4. Therefore, the value of 4 can be 9, as the values of 4 from 1 to 8 do not fit.


62. 212 × 5 is a multiple of 3 and 11. Find the value of x.

Ans:  Since, 212 x 5 is a multiple of 3,

2 +1 + 2 +x+5 = 0, 3, 6,9,12,15,18,

\[ \Rightarrow \]10 + x = 0, 3, 6,………..

\[ \Rightarrow \] x =2, 5, 8 …(i)

Again, 212×5 is a multiple of 11, (2 + 2 + 5) — (1 + x) = 0,11,22, 33

\[ \Rightarrow \] 8 – x = 0,11,22,…

\[ \Rightarrow \] x = 8 …(ii)

From Eqs. (i) and (ii), we have ‘

x = 8


63. Find the value of k, where 31K2 is divisible by 6.

Ans:  Given, 31k2 is divisible by 6. Then, it is also divisible by 2 and 3 both.

Now, 31K2 is divisible by 3, the sum of its digits is a multiple of 3.

i.e. 3+ 1 + k + 2 = 0, 3, 6, 9,12,…

\[ \Rightarrow \] k+ 6 = 0, 3,6, 9,12

\[ \Rightarrow \] k = 0 or 3, 6, 9

 

64. 1y3y6 is divisible by 11. Find the value of y.

Ans:  It is given that 1y3y6 is divisible by 11.

Then, we have (1 + 3 + 6) – (y + y) = 0,11,22,…

\[ \Rightarrow \] 10-2y= 0,11,22,…

\[ \Rightarrow \]10-2y = 0 [other values give a negative number]

\[ \Rightarrow \] 2y = 10

\[ \Rightarrow \] y= 5


65. 756x is a multiple of 11, find the value of x.

Ans:  We are given that 756x is a multiple of 11. Then, we have to find the value of x.

Since, 756x is divisible by 11, then (7 + 6) – (5 + x) is a multiple of 11,

i.e. 8-x = 0,11,22,…

\[ \Rightarrow \] 8- x = 0

\[ \Rightarrow \] x = 8


66. A three-digit number 2 a 3 is added to the number 326 to give a three-digit number 5b9 which is divisible by 9. Find the value of b –a.

Ans: \[ \frac{+\overset{\text{ 2}}{\mathop{\text{3}}}\,\overset{\text{ a}}{\mathop{2}}\,\overset{\text{ 3}}{\mathop{6}}\,}{\underline{\text{ 4 b 6}}}\]

From 1’s column, 

3 + 6 = 9

From 10’s Column,

2 + 3 = 5 

From the above conditions, 

a + 2 is a single digit number b.

a + 2 = b

\[ \Rightarrow \] b -a = 2. 


67. Let E = 3, B = 7 and A = 4. Find the other digits in the sum

\[\frac{+\overset{\text{B}}{\mathop{\text{B}}}\,\text{ }\overset{\text{A}}{\mathop{\text{A}}}\,\overset{\text{S}}{\mathop{\text{ L}}}\,\overset{\text{ E}}{\mathop{\text{ L}}}\,}{\underline{\text{   G  AM  E  S}}}\]

Ans: From 1’s column

3 + L = S

\[ \Rightarrow \] S -L = 3 ....(1)

From 10’s column, S + L =3 ....(2) 

Solving equation (1) and (2) 

S = 3 and L =0 

From 100’s column 

4 + 4 =M 

\[ \Rightarrow \] M = 8 

From 1000’s column, 

7 + 7 =G4

\[ \Rightarrow \] G =1

Hence, L = 0, M = 8 and G = 1


68. Let D = 3, L = 7 and A = 8. Find the other digits in the sum

\[ \begin{align} & \text{M A D} \\ & \text{ + A S} \\ & \underline{\text{ + A }} \\ & \underline{\text{B U L L}} \\ \end{align} \]

Ans: From the 1’s column 

3 + S +8 is a two digit number with a unit digit of 7. 

Therefore, S = 6 

From 10’s column,

2A + 1 =16 + 1 =7 and carry 1 

From 100’s column,

M + 1 is a 2 digit number. 

Therefore, M = 9 Now, M + 1 = 9 + 1 =10 B =1, U = 0 

Hence, S =6, M =9, B = 1 and U = 0 


69. If from a two-digit number, we subtract the number formed by reversing its digits then the result obtained is a perfect cube. How many such numbers are possible? Write all of them.

Ans:  Let ab be any two-digit number. Then, the digit formed by reversing it digits is ba.

Now, ab-ba = (10a+b)-(10b +a)

=(10a-a)+(b-10b)

= 9a – 9b = 9(a – b)

Further, since ab-ba is a perfect cube and is a multiple of 9.

.-.The possible value of a – b is 3.

i.e. a = b + 3

Here, b can take a value from 0 to 6.

Hence, possible numbers are as follows.

For b = 0, a = 3, i.e. 30

For b = 1, a = 4, i.e. 41

For b =2, a = 5, i.e. 52

For b = 3, a = 6, i.e. 63

For b = 4, a = 7, i.e. 74

For b = 5,a = 8, i.e. 85

For b = 6, a = 9, i.e. 96


70. Work out the following multiplication.

\[\begin{align}  & \text{123456789} \\ & \underline{\text{           }\!\!\times\!\!\text{     9}} \\   & \underline{\text{        }} \\  \end{align}\]

Use the result to answer the following questions.

(a) What will be 12345679 × 45? 

(b) What will be 12345679 × 63? 

(c) By what number should 12345679 be multiplied to get 888888888? 

(d) By what number should 12345679 be multiplied to get 999999999?    

Ans:

\[\begin{align}& \text{ 123456789} \\ & \text{ }\underline{\text{           }\!\!\times\!\!\text{     9}} \\  & 1111111111 \\ \end{align}\]

(a) Solution of the multiplication,

\[\begin{align} & \text{ 123456789} \\  & \text{ }\underline{\text{           }\!\!\times\!\!\text{  4 5}} \\  & 5555555555 \\  \end{align}\]

(b) Solution of the multiplication, 

\[\begin{align}& \text{ 123456789} \\ & \text{ }\underline{\text{           }\!\!\times\!\!\text{  6 3}} \\  & 7777777777 \\ \end{align}\]

(c) Solution of the multiplication,  

\[\begin{align}& \text{ 123456789} \\  & \text{ }\underline{\text{           }\!\!\times\!\!\text{  7 2}} \\  & 8888888888 \\ \end{align}\]

(d) Solution of the multiplication, 

\[\begin{align} & \text{ 123456789} \\ & \text{ }\underline{\text{           }\!\!\times\!\!\text{  8 1}} \\  & 9999999999 \\ \end{align}\]


71) Find the value of the letters in each of the following:

(i) \[\begin{align}& \text{   P Q} \\ & \underline{\text{    }\!\!\times\!\!\text{   6   }} \\ & \underline{\text{Q  Q  Q}} \\ \end{align}\]

Ans:  \[\begin{align}& \text{ P Q} \\ & \underline{\text{    }\!\!\times\!\!\text{   6   }} \\ & \underline{\text{Q  Q  Q}} \\ \end{align}\]

From the first column, \[6 \times P+1 = Q\].

Therefore, Q = 2, 4, 6 or 8 Equation

\[6 \times P+1\]   is satisfied when Q = 4.

After solving \[6 \times P+1 \]  

 P = 7 and Q = 4 

(ii) \[\begin{align} & \text{    2 L M} \\ & \underline{\text{+  L M 1}} \\ & \underline{\text{     M 1 8}} \\ \end{align}\] 

Ans:   \[\begin{align}& \text{    2 L M} \\ & \underline{\text{+  L M 1}} \\ & \underline{\text{     M 1 8}} \\ \end{align}\]

From the first column, M + 1 = 8

Therefore, M = 7

From the second column, L + M = a number with unit digit 1.

Therefore, L = 4. 

Hence L = 4 and M =7


72. If 148101B095 is divisible by 33, find the value of B.

Ans:  Given that the number 148101S095 is divisible by 33, therefore it is also divisible by 3 and 11 both.

Now, the number is divisible by 3, its sum of digits is a multiple of 3. i.e. 1 + 4+ 8+1 + 0+1 + B+ 0+ 9+ 5 is a multiple of 3.

29 + B = 0, 3, 6,9,…

B=1,4,7 …(i)

Also, given number is divisible by 11, therefore

(1+ 8 + 0+ B + 9)-(4 + 1+ 1+ 0 + 5)=0, 11,22, …

(18 + B) -11 = 0,11,22

B+ 7 = 0,11,22

B+7 = 11

B = 4 …(ii)

From Eqs. (i) and (ii), we have B = 4


73. If 123123A4 is divisible by 11, find the value of A.

Ans: Given, 12312344 is divisible by 11, then we have (1 + 3 + 2 + 4) – (2 + 1 + 3 + 4) is a multiple of 11.

i.e. (6+ 4)-10 =0,11,22,,..

A-4 = 0,11,22,…

A-4 = 0 [A is a digit of the given number]

A = 4


74. If 56 x 32y is divisible by 18, find the least value of y.

Ans:  It is given that the number 56 x 32y is divisible by 18. Then, it is also divisible by each factor of 18.

Thus, it is divisible by 2 as well as 3.

Now, the number is divisible by-2, its unit’s digit must be an even number that is 0, 2,4, 6, Therefore, the least value of y is 0.

Again, the number is divisible by 3 also, the sum of its digits is a multiple of 3. i.e. 5 + 6 + x + 3 + 2 + y is a multiple of 3

16 + x + y = 0, 3, 6, 9,…

16+ x = 18

x = 2, which is the least value of x.


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Vedantu provides NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers solutions in PDF format, which students may use to prepare for their exams. These examples were created by our subject experts following the CBSE syllabus (2024-25) and are required by the board for 8th grade. One of the most crucial chapters for Class 8 Maths is Playing With Numbers. Students should answer NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers, to comprehend the topics of this chapter and to do well on the exam.


In NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers, Vedantu presents sample answers for various types of issues. Let's look at the subjects covered in Chapter 13 of the 8th-grade Maths curriculum. Students can also practice for the exam by solving Solutions for 8th Standard Maths, as well as exemplars.


The Following Topics are Covered in NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers:

  1. Number Representation in a Generalized Form

  2. Taking part in number games

  3. Divisibility tests by tens, fives, nines, and threes.

NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers covers a variety of topics, including divisibility, obtaining numbers from conditions, and the remainder theorem, among others.


There are a total of 74 questions in Chapter 13 that are all different and relevant. All of these problems are founded on the chapter's key principles and have been well expressed through theory explanations, examples, diagrams, and other means. Multiple-choice, fill-in-the-blank, true-false, and other types of questions are included in the practice.

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NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers addresses all of the questions and provides detailed explanations and step-by-step reasoning.

4. What are the different kinds of questions in NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers?

There are a total of seventy-four questions in NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers.  Each of these questions is different and pertains to the themes taught in the chapter.


The first activity in the chapter consists of multiple-choice questions. To deduce the answer to the question, you must carefully evaluate it and conduct the relevant operations or apply the reasoning learned in the chapter.


The other activity consists of fill-in-the-blank problems based on observations. There are logical and mathematical problems that must be answered, with the solution filling in the blanks provided with the problem. There are also true or false problems. Again, you must assess the offered assertions and provide your findings as to whether they are true or false based on the chapter's important themes.


Other problems, such as question 71, present you with an additional challenge using numerals and alphabets. The appropriate substitution of the letters with a digit must be determined.

5. What are the advantages of NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers?

Vedantu's  NCERT Exemplar Class 8 Maths Solutions Chapter 13 Playing With Numbers strives to answer all of your questions in one spot. We make certain that when you use our study materials, you will not have to strain to find multiple resources for reference. Vedantu seeks to create a learning environment that is stress-free, beneficial, productive, and effective. No obstacle will be able to stop you from getting excellent grades in your tests if you use our methods and put up the effort.


All of the base concepts have been addressed in NCERT Exemplar Solutions for Class 8 Maths Chapter 13 – Playing with Numbers. These NCERT Exemplar Class 8 Maths Exemplar Problems Solutions cover all of the exercises in the chapter and will assist you to prepare for your examinations. These answers will guarantee that you understand the chapter and that you review all of the problems in the exercise regularly.