 Lesson: Playing with Numbers
In each of the questions 1 to 17, out of the four options, only one is correct.
Question 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 b c d
(c)
Solution:
A four-digit number abdc is written as 1000a + 100b + 10d + c
Question 2
Generalised form of a two-digit number xy is:
(a) x + y
(b) 10x + y
(c) 10x y
(d) 10y + x
(b)
Solution:
A two digit number xy is written as 10x + y
Question 3
The usual form of 1000a + 10b + c is:
(a) abc
(b) abco
(c) aobc
(d) aboc
(c) Solution:
Since 1000a + 10b + c = 1000a + 100 b + c so the usual form of
1000a + 10b + c is aobc
Question 4
Let abc be a three - digit number, where a and c are different.
Then abc cba is not divisible by:
(a) 9
(b) 11
(c) 13
(d) 33
(c)
Solution:
abc cba = 100a + 10b + c 100c 10b a = 99a 99c = 11  (a c)
Which is always divisible by 11, 9, 33 and 99.
This will be divisible by 13 provided a c is divisible by 13, which is
not possible.
Therefore, abc cba is not divisible by 13.
Question 5
The sum of all the numbers formed by using the digits x, y and z of the number xyz
is divisible by:
(a) 11
(b) 33
(c) 37
(d) 64
(c)
Solution:
Given,
xyz = 100x + 10y + z
yxz = 100y + 10x + z
yzx = 100y + 10z + x
zyx = 100z + 10y + x
zxy = 100z + 10x + y
Sum of these numbers = 222 (x + y + z) = 37  x + y + z),
which is divisible by 37.
The sum of all the numbers formed by using the digits x, y and z of the number xyz
Is divisible by 37. Question 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
(c)
Solution:
aabb = 1000a + 100a + 10b + b = 11 (100a + b),
Which is always divisible by 11.
Since, the number is divisible by 55; it is divisible by both 11 and 5.
This means that b has to be either 5 or 0.
Question 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
(d)
Solution:
abc + bca + cab
Can be written as:
100a + 10b + c + 100b + 10c + a + 100c + 10a + b
= 111a + 111b + 111c
=111(a + b + c) = 37  (a + b + c)
So, 111(a + b + c) is not divisible by 9.
This number will be divisible by 9, if it is further given that a + b + c is divisible by 3.
Therefore, abc + bca + cab is not divisible by 9.
Question 8
A four-digit number 4ab5 is divisible by 55. Then the value of b a is:
(a) 0
(b) 1
(c) 4
(d) 5
(b) Solution:
The four digit numbers 4ab5 is divisible by 55; hence, it is divisible by both
11 and 5.
For all vales of a and b the number is divisible by 5as the digit at the units
Place is 5.
Since it divisible by 11, we must have b + 4 a 5 = b a 1
Divisible by 11, which means that (b a) must be 1.
Question 9
If abc is a three digit number, then the number abc a b c is
divisible by:
(a) 9
(b) 90
(c) 10
(d) 11
(a)
Solution:
If the number is abc, then,
abc a b c = 100a + 10b + c a b c = 99a + 9b
= 9 (11a + b), Which is divisible by 9.
Question 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
(d)
Solution:
Let the three digit number be abc.
So,
abcabc = 100000a + 10000b + 1000c + 100a + 10b + c
= 1000100a + 10010b + 1001c
= 1001 (100a + 10b + 1)
Which is divisible by all of 7, 11, 13 and 1001.
The correct option is (d).
Hence ,
the 6 digit number formed by repeating a three-digit number is divisible by 1001. Question 11
If the sum of the digits of a number is divisible by three, then the number is always
divisible by:
(a) 2
(b) 3
(c) 6
(d) 9
(b)
Solution:
If the sum of the digits of a number is divisible by three, then the numberis always
divisible by 3.
For example, let us consider a three digit number.
abc = 100a + 10b + c = 99a + 9b + a + b + c If it is further given
that a + b + c is divisible by 3, then a + b + c = 3n, for some natural
number n and abc = 3 (33a + 3b + n), which is clearly divisible by 3.
Question 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
(a)
Solution:
Since, z is an odd digit, z can be either be1 or 3 or 5, because if z becomes 7 or
9, x + y + z = 6 will be impossible. Further, if z = 1, then
x + y = 5, if z = 3, then x + y = 3, if z = 5, then x + y = 1,
The possible values of
(x, y) = (1, 4), (2, 3), (3, 2), (4,1), (1, 2), (2,1), (1, 0) .
Which will be an odd multiple of 3 for all possible values of x and y.
Question 13
If 5A + B3 = 65, then the values of A and B are:
(a) A = 2, B = 3
(b) A = 3, B = 2
(c) A = 2, B = 1
(d) A = 1, B = 2
(c) Solution:
We have,
There are two letters A and B whose values we have to find.
The addition of A and 3is giving a number whose ones digit is 5.
This is possible only when digit A is 2. In that case, the addition of 2 and 3will
give 5.
Clearly, A is 2 as 2 + 3 = 5.
In the next step,
Clearly, B is 1as 5 + 1 = 6.
So the puzzle can be solved as shown below.
Hence, A and B are 2 and1 respectively.
Question 14
If A3 + 8B = 150, then the value of A + B is:
(a) 13
(b) 12
(c) 17
(d) 15
(a)
Solution:
We have,
There are two letters A and B whose values we have to find. The addition of 3 and B
is giving a number whose ones digit is 0.
This is possible only when digit B is 7. In that case, the addition of 3 and 7 will
give 10.
Clearly, B is 7 as 3 + 7 = 10.
In the next step,
This is possible only when the digit A is 6. In that case, the addition of 1,6 and 8
will give 15.
Clearly, A is 6 as 1 + 6 + 8 = 15.
So the puzzle can be solved as shown below.
Hence, A and B are 6 and 7 respectively.
So, A + B = 6 + 7 = 13.
Question 15
If 5A  then the value of A is:
(a) 3
(b) 6
(c) 7
(d) 9 (c)
Solution:
Given,
Since A is at the unit’s place in both the numbers 5A and A,
A number with one’s digit 9.
A is either 3 or 7.
Taking A as 3, we get 53 3 = 159;talking a as 7,we get 57 7 = 399.
Therefore, A = 7.
Question 16
If 6A B = A8B, then the value of A B is.
(a) 2
(b) 2
(c) 3
(d) 3
(a)
Solution:
We have,
This means that the product of A and B is giving a number with unit’s digit B.
There are many possible combinations of numbers to give such a result.
The possible values of B are 2,3,4,5, 6, and 8.
Taking B as 8, we get A = 6, this does not satisfy the second step i.e.
6 B + 4 ≠ A8.
Taking B as 6, we get A = 6, this does not satisfy the second step i.e.
6 B + 3 ≠ A8.
Taking B as 5, we get A = 1 or 3 or 5 or 7 or 9.
For all these values, the second step does not get satisfied.
In the same manner B = 2does not satisfy the second step and its solutions.
Now, taking B = 3, A is 1.
Clearly, A = 1and B = 3
Hence A and B are 1 and 3 respectively.
A B = 1 3 = 2
Question 17
Which of the following numbers is divisible by 99?
(a) 913462
(b) 114345
(c) 135792
(d) 3572406
(b) Solution:
For a number to be divisible by 99, it must be divisible by both 11 and 9.
A number is divisible by 11 if the difference between the sum of the digits at its
odd places and that of digits at its even places is either
0 or divisible by 11.
A number is divisible by 9 if the sum of its digits is divisible by 9.
The given number is 913462.
Sum of its digits = 9 + 1 + 3 + 4 + 6 + 2 = 25, which is not divisible by 9.
Given number is114345.
Sum of its digits = 1 + 1 + 4 + 3 + 4 + 5 = 18, which is divisible by 9 and
the difference of sums of its digits at odd places and even places is 9 9 =0.
It is divisible by 11.
Therefore, 114345 is divisible by 99.
Given number is135792.
Sum of its digits = 1 + 3 + 5 + 7 + 9 + 2 = 27, which is divisible by 9.
But, difference of sums of digits at odd places and even places (15 12 = 13) is
not divisible by 11.
135792 is not divisible by 99.
Given number is 3572406.
Sum of its digits = 3 + 5 + 7 + 2 + 4 + 0 + 6 = 27, which is divisible by 9.
But, difference of sums of digits at odd places and even places (20 7 = 13) is
not divisible by 11.
3572406 is not divisible by 99.
114345 is divisible by 99.
In questions 18 to 33, fill in the blanks to make the statements true.
Question 18
3134673 is divisible by 3 and _____________.
9
Solution:
Given number is 3134673.
Sum of its digits = 3 + 1 + 3 + 4 + 6 + 7 + 3 = 27.
27 is divisible by 3 and 9.
Question 19
20x3 is a multiple of 3 if the digit x is _____________or _____________or
_____________.
1, 4, 7 Solution:
20x3 is multiple of 3 if
Is a multiple of 3
Is a multiple of 3
………… -------- (i)
But, x is a digit of the number 20x3. S0, x can take values 0, 1, 2, 39.
can take values 5, 6, 7, 8... 14. ------- (ii)
From (i) and (ii), we get
Hence x can take values 1 or, 4 or, 7.
Question 20
3x5 is divisible by 9 if the digit x is _____________.
1
Solution:
3x5 is divisible by 9 if 3 + x + 5 is a multiple of 9 is a multiple of 9.
But, x is a digit. So, it can take values 0, 1, 2, 3…, 9.
And hence,
8 x
can take values 8, 9, 10, 11..., 17. ------- (ii)
From (i) and (ii), we get
Hence,
1x
Question 21
The sum of a twodigit number and the number obtained by reversing the digits is
always divisible by _____________.
11
Solution:
The sum of a twodigit number and the number obtained by reversing the digits is
always divisible by 11 as.
Question 22
The difference of a twodigit number and the number obtained by reversing its
digits is always divisible by _____________.
9
Solution:
The difference between a 2-digit number and the number obtained by reversing its
digits is always divisible by 9 as. Question 23
The difference of three-digit number and the number obtained by putting the digits
in reverse order is always divisible by 9 and _____________.
11
Solution:
Any 3-digit number can be represented as:
100A + 10B + C
The number, when reversed is:
100C + 10B + A
The difference:
(100A + 10B + C) (100C + 10B + A) = 99A 99C = 99(A C)
Since the result is equal to 99 times some integer, the result must be divisible by
99, 9 and 11.
Question 24
If
2
8
B
AB
A
then A = _____________and B = _____________.
A = 6,B = 3
Solution:
There are two letters A and B whose values we have to find.
The addition of B and B is giving a number whose ones digit is A.
Also, there is a second A.
The above sum will be possible only when digit B is 3.
In that case, the addition of 3and 3 will give A.
Clearly, A is 6 as 3 + 3 = 6.
In the next step, 2 + A = 8 as 2 + 6 = 8
So the puzzle can be solved as shown below.
23
63
86
Hence, A and B are 6 and 3 respectively. Question 25
If
96
AB
B
then A = _____________ and B = _____________.
A = 2,B = 4 or A = 1, B = 6
Solution:
We have,
96
AB
B
This means that the product of B with itself is either 6 or it has unit’s digit as 6.
The possible values of B are 4 and 6 whose product with itself is a number having 6
at unit’s place.
Taking B = 4 we have
4
4
96
A
Clearly, A = 2
24
4
96
Taking B = 6 we have
6
6
96
A
Clearly, A = 1in this case.
Hence A and B are 2 and 4 respectively or 1 and 6 respectively. Question 26
If
1
49
B
B
B
then B = _____________.
7
Solution:
We have,
1
49
B
B
B
There is a letter B whose value we have to find.
This means that the product of B and 1 isa number whose ones digits is B.
If
is put in the given equation then 11 1 11 doesn’t satisfy the equation.
If
2B
is put in the given equation then 21 2 = 42 doesn’t satisfy the equation.
If
3B
put is in the given equation then 31 3 = 93 doesn’t satisfy the equation.
If
4B
is put in the given equation then 41 4 = 164 doesn’t satisfy the equation.
If
5B
is put in the given equation then 51 5 = 255 doesn’t satisfy the equation.
If
6B
is put in the given equation then 61 6 = 366 doesn’t satisfy the equation.
If
7B
is put in the given equation then 71 7 = 497 satisfies the equation.
71 7 = 497
Hence B = 7
Question 27
1 x 35 is divisible by 9 if x = _____________.
09or
Solution:
1 x 35 is divisible by 9 if
1 + x + 3 + 5 is a multiple of 9
9 + x is a multiple of 9
9 + x = 0 or 9 or 18 ………… -------- (i)
But, x is a digit. So, it can take values 0, 1, 2, 3… 9.
And hence, 9 + x can take values 9, 10, 11... 18. ------- (ii)
From (i) and (ii), we get
9 + x = 9 or 18
x = 0 or 9
Hence, x = 0 or 9 Question 28
A four-digit number abcd is divisible by 11, if d + b =_____________ or
_____________.
a + c or a + c + 11 or a + c 11
Solution:
A number is divisible by 11 if the difference between the sum of the digits at its
odd places and that of digits at its even places is either 0 or divisible by 11.
A four-digit number abcd is divisible by 11, if d + b = a + c or a + c + 11 or a + c 11
as we have to keep difference of d + b and a + c divisible by 11 keeping in mind
that the value of d + b or a + c cannot exceed 18.
Question 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 ___________.
11
Solution:
Let’s write a 3-digit number abc as
100a + 10b + c = 99a + 11b + (a b + c) = 11 (9a + b) + (a b + c)
If, (a + c b), i.e., the difference between the sum of the digits at its odd places
and that of digits at the even places is either divisible by 11 or 0, then the number
abc will be divisible by 11.
Similarly let’s write a 4-digit number abcd as
1000a + 100b + 10c + d
= (1001a + 99b + 11c) (a b + c d)
= 11 (91a + 9b + c) + [(b + d) (a + c)]
If, [(b + d) (a + c)], i.e., the difference between the sum of the digits at its
odd places and that of digits at the even places is either divisible by 11 or 0, then
the number abc will be divisible by 11.
Therefore, 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 11.
Question 30
If a 3-digit number abc is divisible by 11, then _____________ is either 0 or
multiple of 11.
(a + c) b Solution:
Let’s write the 3-digit number abc as.
If the number abc is divisible by 11, then (a + c b), i.e., the difference between
the sum of the digits at its odd places and that of digits at the even places should
either be divisible by 11 or 0.
If a 3-digit number abc is divisible by 11, then (a + c) b is either 0 or multiple
of 11.
Question 31
If A 3 = 1A then A=_____________.
5
Solution:
We have,
A 3 = 1 A ---------------- (i)
A 3 is a number with ones digit A itself.
This is possible only when A 5.
If A 5 is put in the equation (i) then 5  satisfy the equation.
5 
Hence A 5
Question 32
If B B = AB, then either A = 2, B = 5 or A = _____________
B
_____________.
3, 6
Solution:
We have,
B B = AB ------------------------------ (i)
B can be either 5 or 6.
Putting the value of
2A
and
5B
in equation (i), we get.
Putting the value of
3A
and
6B
in equation (i), we get.
6 6 = 36
Hence, either A = 2, B = 5 or A = 3, B = 6. Question 33
If the digit 1 is placed after a 2-digit number whose tens digit is t and ones digit is
u, the new number is _____________.
t u1
Solution:
If the digit 1 is placed after a 2-digit number whose tens digit is t and ones digit is
u, the new number is tu1.
State whether the statements given in questions 34 to 44 are true (T) or false (F):
Question 34
A two-digit number ab is always divisible by 2 if b is an even number.
True
Solution:
A two-digit number ab is always divisible by 2 if b is an even number.
If b is an even number, then b must be a multiple of 2.
There for b can be written as 2n where n = 0 to 4 and n is a natural number.
Therefore, ab = 10a + b = 10a + 2n = 2 (5a + n), which is divisible by 2.
Question 35
A three-digit number abc is divisible by 5 if c is an even number.
False
Solution:
A three-digit number abc is divisible by 5 if c is an even number.
This statement is false,
e.g; let the three digit numbers be 224, 246, 268, 286…
In each case the last digit of each number, though, is an even number.
Yet the numbers are not divisible by 5.
Any number is divisible by 5 if its ones digit is either 0 or 5.
Question 36
A four-digit number abcd is divisible by 4 if ab is divisible by 4.
False Solution:
A four-digit number abcd is divisible by 4 if ab is divisible by 4.
This statement is false,
e.g; let four digit number be 4467.
Here ab is divisible by 4, but the number 4467 is not divisible by 4.
Any number is divisible by 4 if the number formed by last two digits is divisible
by 4.
Question 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.
True
Solution:
A three-digit number abc is divisible by 6 if c is an even number and a + b + c
is a multiple of 3.if c is even, then c can be written as 2k for k = 0 to 4 and k being
a natural number.
We can write abc = 100a + 10b + c = 100a + 10b + 2k.
If c is an even number, then abc is divisible by 2. Further, a + b+ c is a multiple
of 3.
Therefore, the number is divisible by 3.
If a number is divisible by both 2 and 3, it is divisible by 6.
Hence the given Statement is true.
Question 38
Number of the form 3N + 2 will leave remainder 2 when divided by 3.
True
Solution:
Number of the form 3N + 2 will leave remainder 2 when divided by 3 is true.
Question 39
Number 7N + 1will leave remainder 1 when divided by 7.
True
Solution:
Number 7N + 1will leave remainder 1 when divided by 7.
Question 40 If a number a is divisible by b, then it must be divisible by each factor of b.
True
Solution:
If a number a is divisible by b, then it must be divisible by each factor of b.
Question 41
If AB 4 = 192, then A + B = 7.
False
Solution:
We have,
A + B =
192
4
= 48
Hence A = 4 and B = 8. So, A + B = 4 + 8 = 12.
So, the given statement is false.
Question 42
If AB + 7C = 102, where B ≠ 0, C ≠ 0, then A + B + C = 14.
True
Solution:
We have,
7
1 0 2
AB
C
There are three letters A, B and C whose values we have to find.
The addition of B and C is giving a number whose ones digit is 2.
This is possible when digit B plus C is 12.
In that case, 1 will be the carry-forward to the next step.
In the next step,
So, A = 2
Therefore A + B + C = A + (B + C) = 2 + 12 =14 Question 43
If 213x 27 is divisible by 9, then the value of x is 0.
False
Solution:
Since 213x 27 is divisible by 9 is a multiple of 9.
15 + x is a multiple of 9
15 + x = 0 or 9 or 18 ………… -------- (i)
But, x is a digit. So, it can take values 0, 1, 2, 3 9.
And hence,
15 x
can take values 15, 16, 17, 18... 24. ------- (ii)
if
0,x
then 15 + x = 15 which is not divisible by 9.
Hence,
0.x
So the given statement is false.
Question 44
If N ÷ 5 leaves remainder 3 and N ÷ 2 leaves remainder 0, then N ÷ 10 leaves
remainder 4.
False
Solution:
If N ÷ 5 leaves remainder 3 and N ÷ 2 leaves remainder 0, then
10N
leaves
remainder 4.
Let’s take N = 18.
Does not satisfy the statesmen.
So, the Statement is False.
Solve the following:
Question 45
Find the least value that must be given to number a so that the number 91876a2 is
divisible by 8.
3
Solution:
If the last three digits of a whole number are divisible by 8, then the entire number
is divisible by 8.
For 91876a2 to be divisible by 8, 6 a2 is divisible by 8.
Or, 6 a2 = 600 +10a + 2 = 8(75) + 10a + 2 is divisible by 8.
Or,
10 2a
is divisible by 8.
For this, a has to be either 3 or 7.
The least value of a is 3. Question 46
If
1
6
p
p
Q
where Q P = 3, then find the values of P and Q.
6 and 9
Solution:
We have,
1
6
p
p
Q
There are two letters p and Q whose values we have to find.
This means that the product of P with itself is a number whose ones digits is 6.
P can be either 6 or 4.
In the next step, for
6P
, we have.
Satisfies this as 3 + 6 1 9
Therefore Q = 9.
For
4,P
we have.
1 + P 1 = Q. Q = 5satisfies this as 1 + 4 1 = 5.
Therefore Q = 5.
But in this case Q p is not 3.
So the puzzle can be solved as shown below.
16
6
96
Hence, the values of P and Q are 6 and 9 respectively.
Question 47
If 1AB + CCA = 697 and there is no carryover in addition, find the value of
A + B + C.
12
Solution:
We have,
1AB + CCA = 697 i.e.
1
6 9 7
AB
C C A
There are three letters A, B and C whose values we have to find. Since there is no carry-over, addition in unit’s place, ten’s place or hundred’s place
can be done in any order. Let’s take the addition in the hundred’s place.
1 + C = 6
C = 5
Taking the value of
5C
for the addition in the ten’s place, we get
A + 5 =9
A = 4
Taking the value of
4A
for the addition in the unit’s place, we get
B = 3
Therefore, A + B + C = 5 + 4 + 3 = 12
Question 48
A five-digit number AABAA is divisible by 33.
Write all the numbers of this form.
33033, 66066, 99099
Solution:
AABAA must be divisible by 11 and 3.
If a number is divisible by 11, then 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 11.
If a number is divisible by 3, then the sum of its digits is divisible by 3.
Since, AABAA to be divisible by 11 therefore (A + B + A) (A + A) = B
should be divisible by 11.
So B is bound to be 0.
Since, AABAA to be divisible by 3, therefore, A + A + B + A + A = 4A + B
must be divisible by 3.
Since B = 0 , 4A should be divisible by 3.
Therefore, A could be 3, 6 or 9.
Question 49
Find the value of the letters in each of the following questions.
AA
AA
X A Z
.
A = 9, Z = 8, X = 1 Solution:
There are three letters A, X and Z whose values we have to find.
The addition of A and A is a number whose ones digit is Z .In the second step
AA
gives A, a number whose ones digit is A.
Keeping in mind the placement of r A’s, the above addition is possible only when
digit A is 9.
In that case, the addition of 9 and 9 will give 18 and thus, 1 will be the carry- over
for the next step.
In the next step, a number with ones digit A.
Clearly, A is 9 as 9 + 9 = 18 and 1 will be a carry- over for the next step.
In the next step,
1 x
So the puzzle can be solved as shown below.
Clearly, Z is 8.
Hence, A, X and Z are 9, 1 and 8 respectively.
Question 50
85
4
3
A
BC
A = 8, B = 1, C = 3
Solution:
There are three letters A, B and C whose values we have to find.
The addition of 5 and A is a number whose ones digit is 3.
This is possible only when digit A is 8.
In that case, the addition of A (8) and 5 will give 13 and thus, 1 will be the
carry - over for the next step.
In the next step,
1 + 8 + 4 = 13
So the puzzle can be solved as shown below. 85
4
3
A
BC
Clearly, C is 3 and B is 1
Hence, A, B and C are 8, 1 and 3 respectively.
Question 51
6
8
2
B
A
CA
A = 6, B = 7, C = 1
Solution:
There are three letters A, B and C whose values we have to find.
The addition of 6 and A is a number whose ones digits is 2.
This is possible only when the digit A is 6.
In that case, the addition of 6 and A will give 12 and thus, 1 will be the
carry- over for the next step.
In the next step,
1 + B + 8 = a number with ones digit A.
Clearly, B is 7 as 1 + 7 + 8 = 16 and 1 will be a carry-over for the next step.
In the next step,
So the puzzle can be solved as shown below.
76
86
1 6 2
Hence, the values of A, B and C are 6, 7 and 1 respectively.
Question 52
1
82
BA
A B A
B A = 6, B = 9
Solution:
There are two letters A and B whose values we have to find.
The addition of A and A is a number whose ones digits is 2.
This is possible only when the digit A is 6. In that case, the addition of 6 and 6
will give 12 and thus, 1 will be the carry-over for the next step.
In the next step,
A number with unit’s digit B.
Clearly, B is 9 as 1 + 9 + 9 = 19 and 1 will be a carry-over for the next step.
In the next step,
1 + 1 + A = 8 Clearly, A is 6 as1 + 1 + 6 = 8
So the puzzle can be solved as shown below.
1 9 6
6 9 6
8 9 2
Hence, the values of A and B are 6 and 9 respectively.
Question 53
1 3 0
C B A
C B A
A
A = 5, B = 6, C = 7
Solution:
There are three letters A, B and C whose values we have to find.
The addition of A and A is a number whose ones digits is 0.
This is possible only when digit A is 5.
In that case, the addition of 5 and 5 will give 10 and thus, 1 will be the carry-over
for the next step.
In the next step,
1 + B + B = a number with unit’s digit as 3.
Clearly, B is 6 as 1 + 6 + 6 = 13 and 1 will be the carry-over for the next step.
In the next step,
1 + C + C = a number with the unit’s digit A.
1 + C + C = a number with the unit’s digit 5.
Clearly, C is 7 as 1 + 7 + 7 = 15 and 1 will be the carry-over for the next step.
In the next step,
1 1
So the puzzle can be solved as shown below. 7 6 5
7 6 5
1 5 3 0
Hence, the values of A, B and C are 5, 6 and 7 respectively.
Question 54
38
B A A
B A A
A
A = 9, B = 1
Solution:
There are two letters A and B whose values we have to find.
The addition of A and A is a number whose ones digits is 8.
This happens only for
4A
and
9.A
If
4A
, then the sum is4 + 4 = 8, in next step A + A = a number with ones
digit.
A then 4 + 4 = 8 which makes
8A
too, which is not possible.
We reject this possibility.
If
9.A
In that case, the addition of 9 and 9 will give 18 and thus, 1 will be the carry-over
for the next step.
In the next step,
1 + A + A = a number with unit’s digit A.
Clearly, A is 9 as 1 + 9 + 9 = 19 and 1 will be carry-over for the next step.
In the next step,
Clearly, B is 1 as 1 + 1 + 1 = 3
So the puzzle can be solved as shown below.
1 9 9
1 9 9
3 9 8
Hence, the values of A and B are 9 and 1 respectively. Question 55
01
10
1 0 8
AB
AB
B
A = 8, B = 9
Solution:
There are two letters A and B whose values we have to find.
The addition of B and B is a number whose ones digits is 8.
This happens only for B = 4 and B = 9.
If
4,B
then the sum is 4 + 4 = 8, in next step 1 + A = a number with one’s
digit 0 then 1 + 9 = 10 which makes A = 9.
In next step 1 + 0 + 0 = 1 and then in next step A +1 must give B but 9 + 1
is not 4.
We reject this possibility.
The addition of 9 and 9 will give 18 and thus, 1 will be the carry-over for the
next step.
In the next step,
A number with unit’s digit as 0
Clearly, A is 8 as 1 + 1 + 8 = 10 and 1 will be carry-over for the next step.
In the next step,
1 + 0 + 0 = 1In the next step,
A + 1 = B as 8 + 1 9
So the puzzle can be solved as shown below.
8 0 1 9
1 0 8 9
9 1 0 8
Hence, the values of A and B are 8 and 9 respectively.
Question 56
6
68
AB
C
A = 7, B = 8, C = 4 Solution:
We have,
6
68
AB
C
There are three letters A, B and C whose values we have to find.
This means that the product of B and 6 is a number whose ones digits is 8.
Since
8B
satisfies B 6 = a number with ones digit 8 as 8 
Therefore
8.B
4 will be the carry-over for the next step.
In the next step the product of A and 6 must give a number with ones digit 6.
Since
7A
, satisfies Therefore A = 7
4 will be the carry-over for the next step.
In the next step.
C = 4
So the puzzle can be solved as shown below.
6
68
AB
C
Hence, the values of A, B and C are 7, 8 and 4 respectively.
Question 57
6
AB
AB
AB
A = 2, B = 5
Solution:
Here, we have to find the values of A and B.
Since ones digit of
BB
is B. Therefore B = 0 or B =5
Now, AB AB = 6 AB ---------------------- (i)
The square of a two digit number is a three digit number.
So, A can take values 1, 2 and 3.
We find that A = 2, B = 5 satisfies equation (i).
i.e. 25 25 = 625.
A = 2 and B = 5
No other pair of values of A and B satisfy the equation (i)
Hence, A = 2 and B = 5 Question 58
AA
A
C A B
9, 1, 8A B C
Solution:
We have,
AA
A
C A B
There are three letters A, B and C whose values we have to find.
This means that the product of A and A is a number whose ones digits is B fulfils
the requirement.
8 will be the carry-over for the next step. In the next step the product of A and A
gives a number whose ones digits is A.
Since
9A
satisfies 8 + A A  A as 8 + 9 9 = 89.
8 will be the carry-over for the next step.
In the next step.
8C
So the puzzle can be solved as shown below.
99
9
8 9 1
Hence, the values of A, B and C are 9, 1 and 8 respectively.
Question 59
7
45
AB
B
A = 7, B = 2
Solution:
There are two letters A and B whose values we have to find.
The subtraction of 7 from B is giving a number whose ones digits is 5.
This is possible only when digit B is 2. In that case, the subtraction of 7 from 12
will give 5 and thus, 1 will be the borrow from the next step.
A number with unit’s digit 5. Clearly, 12 7 = 5, hence B is 2.
In the next step,
A B 1 = 4
Clearly, A is 7 as 7 2 1 = 4
So the puzzle can be solved as shown below.
72
27
45
Hence, the values of A and B are7and 2 respectively.
Question 60
8
5
4 8 8
A B C
A B C
D
A = 7, B = 2, C = 3, D = 1
Solution:
There are four letters A, B, C and D whose values we have to find.
The subtraction of 5 from C is a number whose ones digits is 8.
This is possible only when digit C is 3.
In that case, the subtraction of 5 from C will give 8 and thus, 1 will be the borrow
from the next step.
Clearly, 13 5 = 8. C = 3.
In the next step,
A number with unit’s digit 8.
i.e. B 4 = a number with unit’s digit 8.
The subtraction of 4from B is giving a number whose ones digits is 8.
This is possible only when digit B is 2.
In that case, the subtraction of 4from B will give 8 and thus, 1 will be the borrow
for the current step to the next step.
Clearly, 12 4 = 8, hence
2B
.
In the next step,
A B 1 = a number with unit’s digit4.
A 2 1 = a number with unit’s digit4.
The subtraction of 3 from A is giving a number whose ones digits is 4.
This is possible only when digit A is 7.
In that case, the subtraction of 2 from A will give 4 and thus,
A 2 1 = a number with unit’s digit4.
Clearly, A is 7 as 7 2 1 = 4
In the next step,
8 7 = 1 So the puzzle can be solved as shown below.
8 7 2 3
7 2 3 5
1 4 8 8
Hence, the values of A, B, C and D are 7, 2, 3and 1 respectively.
Question 61
If 2A7 ÷ A = 33, then find the value of A.
9
Solution:
Given
2A7 ÷ A = 33,
(200 + 10A + 7) ÷ A =33
200 + 10A + 7 = 33A
Hence,
9A
Question 62
212 x 5 is a multiple of 3 and 11.
Find the value of x.
8
Solution:
If a number is divisible by 11, then 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 11.
If a number is divisible by 3, then the sum of its digits is divisible by 3.
It is given that 212 x 5 is multiple of 3 and 11
Is a multiple of 3 and 9 x 1 is divisible by 11.
10 + x is a multiple of 3 and 8 x is divisible by11.
Hence x can take the value of 8.
Question 63
Find the value of k where 31k 2 is divisible by 6.
k is either 0 or 3 or 6 or 9 Solution:
Since 31k 2 is divisible by 6
2y 10 is a multiple of 3
= 6 + k is a multiple of 3
Hence, k is either 0 or 3or 6 or 9
Question 64
1y3y6 is divisible by 11. Find the value of y.
5
Solution:
If a number is divisible by 11, then 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 11.
Since 1y3y6 is divisible by 11.
2y 10 is a multiple of 11.
This is possible only if 2y 10 = 0
Or,
5y
Hence, y is 5
Question 65
756 x is a multiple of 11, find the value of x.
8
Solution:
If a number is divisible by 11, then 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 11.
It is given that 756 x is multiple of 11.
5 + x 13 is a multiple of 11.
This is possible if 5 + x 13 = 0
Or, x = 8
Hence x is 8
Question 66
A three-digit number 2a3 is added to the number 326 to give a three-digit number
5b9 which is divisible by 9.
Find the value of
–.ba
2 Solution:
We have,
A three digit number 2a3 which is added to the number 326 giving a three digit
number 5b9 which is divisible by 9.
23
3 2 6
59
a
b
5 + b + 9 is divisible by 9, therefore b must be 4. Hence, a must be 2.
So, b a = 4 2 = 2
Question 67
Let E = 3, B = 7 and A =4
Find the other digits in the sum
B A S E
B A L L
G A M E S
S = 8, L = 5, M = 9, G = 1
Solution:
Putting the values of E, B and A in the sum
7 4 3
74
43
S
LL
G M S
There are four letters S, L, M and G whose values we have to find.
The addition of 3 and L is a number whose ones digits is S.
There are two possibilities; 3 + L = S or 3 + L = 10 + S
In the second step.
The addition of S and L is a number whose ones digits is 3.
Here again there are four possibilities;
S + L = 3 or S + L + 1 = 3 or S + L = 10 +3 or S + L + 1 = 10 + 3
Considering that each of the digits are from 0 to 9 and the maximum value of the
addition of two digits will be 18and keeping in mind the placement of L and S, the
above addition is possible only for.
5L
and S = 8. In the next step.
SL
a number with ones digit 3.
8 + 5 = 13 and 1 will be the carry-over for the next step.
In the next step,
1 4 4
a number with the units digit M. M = 9 as 1 + 4 + 4 = 9.
In the next step,
7 7 
a number with the units digit 4 and as 7 + 7 = 14 and 1 will be a
carry-over for the next step.
In the next step, So the puzzle can be solved as shown below.
7 4 8 3
7 4 5 5
1 4 9 3 8
Hence, the values of S, L, M and G are 8, 5, 9 and 1 respectively.
Question 68
Let D = 3, L =7 and A =8.
Find the other digits in the sum
M A D
AS
A
B U L L
S = 6, M = 9, B = 1, U = 0
Solution:
Putting the values of D, L and A in the given sum
83
8
8
77
M
S
BU
There are four letters S, M, B and U whose values we have to find.
The addition of 3, M and 8 is giving a number whose ones digits is 7.
Here 1will be the carry-over to the next step as
88
has been shown as 17.
This happens only for S = 6as 3 + 6 + 8 = 17 and 1 will be the carry-over for
the next step.
In the next step,
1 + A + A = a number with ones digit 7.
Clearly, A is 8 as 1 + 8 + 8 = 17 and 1 will be the carry-over for the next step.
In the next step,
1 + M = BU
So the puzzle can be solved as shown below.
983
86
8
1 0 7 7
Hence, the values of S, M, B and U are 6, 9, 1and 0 respectively Question 69
If from a two-digit number, we subtract the number formed by reversing its digits
then the result so obtained is a perfect cube.
How many such numbers are possible?
Write all of them.
96, 85, 74, 63, 52, 41, 30
Solution:
Case 1: If the tens digit is larger than the ones digit (that is,
ab
) then
(10a + b) (10b + a) = 10a + b 10b a
= 9a 9b = 9 (a b).
Case 2: If the ones digit is larger than the tens digit (that is,
ba
) then,
(10b + a) (10a + b) = 9 (b a).
Since, the difference is a perfect cube, in the above two cases,
(b a) or (a b) = 3
Case 3: If
, the difference = 0, which is not a perfect cube.
Hence, the numbers are 96, 85, 74, 63, 52, 41and30.
Question 70
Work out the following multiplication.
1 2 3 4 5 6 7 9
9
─────────
Use the result to answer the following questions.
(a) What will be12345679 45?
(b) What will be12345679 63?
(c) By what number should 12345679 be multiplied to get
888888888?
(d) By what number should 12345679 be multiplied to get
999999999?
(a) 555555555
(b) 777777777
(c) 72
(d) 81
Solution:
Given,
1 2 3 4 5 6 7 9
9
─────────
The product of 9 9 = 81so, a number whose ones digit is 1 and 8 will the
carry-over for the next step. In the next step
9 so, a number whose tens digit is 1and 7 will the
carry-over for the next step.
In the next step
9 6 + 7 = 54 + 7 = 61 so, a number whose hundredths digit is 1and 6 will
the carry-over for the next step.
In the next step
9 5 + 6 = 45 + 6 = 51 so, a number whose thousands digit is 1and 5 will
the carry-over for the next step.
In the next step
9 4 + 5 = 36 + 5 = 41 so, a number whose ten thousands digit is 1 and 4
will the carry-over for the next step.
In the next step
9 3 + 4= 27 + 4 = 31 so, a number whose lakhs digit is 1and 3 will the
carry-over for the next step.
In the next step
9 2 + 3 = 18 + 3 = 21 so, a number whose ten lakhs digit is 1and 2 will the
carry-over for the next step.
In the next step
9 1 + 2 = 9 + 2 =11 so, a number whose crores digit is 1 and 1 will
carry-over for the next step.
In the next step 1
Multiplication of 12345679 and 9 is 111111111
12345679 45 = 111111111 5 = 555555555
12345679 63 = 111111111 7 = 777777777
12345679 72 = 111111111 8 = 888888888
12345679 81 = 111111111 9 = 999999999
Question 71
Find the value of the letters in each of the following:
6
PQ
Q Q Q
(ii)
2
1
18
LM
LM
M
(i) P = 7, Q = 4
(ii) M = 7, L = 4 Solution:
(i) We have,
6
PQ
Q Q Q
There are two letters P and Q whose values we have to find.
This means that the product of Q and 6 is a number whose ones digits is Q.
The possible values of Q are 2, 4, 6 and 8.
Putting these values, one by one we find that Q = 4 satisfies Q 6 = a number
with unit’s digit Q as 4 6 = 24.Therefore Q = 4.
2 will be the carry-over for the next step. In the next step,
Since P = 7 satisfies 2 + P 6 = a number with ones digit Q as 2 + 7 6 = 44.
4 will be the carry-over for the next step.
In the next step Q = 4
So the puzzle can be solved as shown below.
74
6
4 4 4
Hence, the values of P and Q are 7 and 4 respectively
(ii)
2
1
18
LM
LM
M
There are two letters L and M whose values we have to find.
The addition of M and 1 is giving a number whose ones digits is 8.
This happens only for
7.M
In that case, the addition of 7 and 1 will give 8 as 7 + 1 = 8.
In the next step L and M when added gives a number with unit’s digit1.
This happens only for
4L
.
In that case, the addition of 4 and 7 will give 11.
Clearly, L is 4 as 4 + 7 = 11 and 1 will be carry-over for the next step.
In the next step,
1 + 2 + L = M
Clearly, L is 4 as 1 + 2 + 4 = 7
So the puzzle can be solved as shown below.
2 4 7
4 7 1
7 1 8
Hence, the values of L and M are 4 and 7 respectively Question 72
If 148101B095 is divisible by 33, find the value of B.
4
Solution:
Since 148101B095 is divisible by 33, it is divisible by both 11 and 3.
If a number is divisible by 11, then 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 11.
1 + 4 + 8 + 1 + 0 + 1 + B + 0 + 9 + 5 is a multiple of 3
29 + B is a multiple of 3. Also, 18 + B 11 is divisible by 11
Hence, B is 4.
Question 73
If 123123A4 is divisible by 11, find the value of A.
4
Solution:
If a number is divisible by 11, then 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 11.
Since 123123A4 is divisible by 11,
6 + A 10 = A 4 is divisible by 11
This means A 4 = 0 or A = 4
Hence, A is 4.
Question 74
If 56x32y is divisible by 18, find the least value of y. 