DIVISIBILITY RULES - Solved Problems - 2
Divisibility by 11: Solved Problems on Divisibility Rules.
14) What is the least value of x such that 4876x is divisible by
11?
a) 1
b)
2
c) 3
d)
4
Answer: C
Explanation:
A number is divisible by 11 when the difference between the
sum of digits at even places and odd places is 0 or a multiple of 11
The given number is 4876x.
(Sum of digits at EVEN places) – (sum of digits at ODD
places)=0
(6
+ 8 ) - ( x + 7 + 4 ) = 0
14
- ( x + 11 ) = 0
Here the value of x must be 3.
Divisibility Rules Solved Problem – Example for
divisibility by 11
15) If M183 is divisible by 11, find the value of the
smallest natural number M.
a)
5 b)
6
` c)
7
d)
9
e)8
Answer: C
Explanation:
A number is divisible by 11 when the difference between the sum of
digits at even places and odd places is 0 or a multiple of 11
The given number is M183.
(Sum of digits at EVEN places) – (sum of digits at ODD
places)=0
(8
+ M )- (3+1)= 0
(8
+ M ) - 4 = 0
Here the value of M must be 7.
Divisibility Rules
Solved Problem: How to know if a number is divisible by 11 or not.
16) What is the smallest number that must be added to 8261955 to
obtain a sum that is divisible by 11?
a) 1
b)2
c)3
d)4
e)5
Answer: B
Explanation:
For divisibility by 11, the difference of sums of digits
at even and odd places must be either zero or divisible by 11.
For 8261955, Difference =(8+6+9+5) -(2+1+5)=28-8=20.
The unit digit is in an odd place. So we add 2 to the number
=> 8261955 +2 = 8261957
Now, (8+6+9+7) -(2+1+5)=30-8=22 => 22 is a multiple of 11
and hence 8261957 is also divisible by 11.
Divisibility
by 11: Solved examples
17) What is the smallest
number that should be added to 935127 to make it exactly divisible by 11?
a) 5 b)
2 c) 3 d) 4
Answer: A
Explanation:
For a number to be divisible by 11,
the difference between the sum of its alternate digits must be a multiple of
11.
Here, the alternating sum of digits in
935127 is (9 - 3 + 5 - 1 + 2 - 7) = 5.
To make it divisible by 11, the
smallest number to add is 5.
Therefore, 5 must be added to 935127
to make it divisible by 11.
Divisibility Rules Solved Problems for SSC - Divisibility by 11
18) Find the least value of * for which 6291*548 is divisible by 11.
a)5 b)8 c)3 d)7
Answer: B
Explanations :
Let the digit in place of * be x.
(Sum of digits at odd places from right) – (Sum of digits at
even places from right) =
(4 + x + 9 + 6) - (8 +
5+1+2) = (19 + x) - 16, which should be divisible by 11.
(3+x) should be a multiple of 11 => x = 8
Divisibility by Any Composite Number
19) For a number to be divisible by 88 it should be:
a) Divisible by 8.
b) Divisible by 8 and 22. c) Divisible by 8 and 11.
d) Divisible by 4,22 e)
can’t say
Answer: C
Explanation:
For a number to be divisible by 88, the number must be
divisible by 8 and 11.
Write 88 as a product of two
factors: 22,2
11,8
44,2
Of these pairs, 11 and 8 are co-primes. So the number must be
divisible by 8 and 11
Therefore, the least value for x to make
389628x4 divisible by 16 is 6.
Divisibility by 12: Number
System Solved Problems
20) What is the smallest positive
integer value for the variable a for which the number 6248a is divisible by 12?
a) 4 b)6 c)0 d)8
Answer: A
Explanation :
For a number to be divisible by 12, it
must be divisible by both 3 and 4.
Divisibility by 3:
To check divisibility by 3, find the sum
of the digits: (6+2+4+8+a)
The sum of the digits is 20+a.
(20+a) should also be divisible by 3.
a
can be 1 4 or 7
Divisibility by 4:
For divisibility by 4, check if the
number formed by the last two digits,8a is divisible by 4.
The last two digits form the number 8a.
For divisibility by 4, this number should be divisible by 4. Then a can be 0 or
4 or 8
Combining both rules, find the smallest
positive integer value for a that satisfies both conditions is 4
Solved
Problems on Divisibility Test – Divisibility by 15
21) What is the smallest positive
integer value for x such that the number formed by the digits 72652x is
divisible by 15?
a)3 b)4 c)0 d)5
Solution:
To check divisibility by 15, the number
must be divisible by both 3 and 5.
For divisibility by 3, find the sum of all
digits:
7+2+6+5+2+x = 22+x.
For divisibility by 3, (22+x) should
also be divisible by 3.
x can be 2, 5 and 8
Divisibility by 5:
For divisibility by 5, the last digit
should be either 0 or 5.
So x can be o or 5
Combining both rules, the smallest
positive integer value for x that satisfies both conditions is 5.
Divisibility
by 16: Solved Questions on Divisibility Tests 16
22) Find the least value for the
variable x to make the number 389628x4 divisible by 16.
a)1 b)3 c)2 d)8
Answer: C
Explanation:
For a number to be divisible by 16, the
number formed by its last four digits must be divisible by 16. In this case,
the last four digits are 28x4
Divisibility by 16: Check if 28x4 is
divisible by 16.
The smallest positive integer value for
x is 2, making 28x4=2864 divisible by 16.
Divisibility
Test by 18: Solved Problems on Divisibility Rules by 18 for SSC CHSL
23) Find the smallest
positive integer P such that the number 7743P is divisible by 18.
a)6 b)4 c)8 d)2
Answer: A
Explanation:
A number is divisible by
18 if and only if it is divisible by both 2 and 9.
Divisibility by 2 :
For a number to be divisible by 2, its unit digit
must be even.
Since 7743P ends
in P, it is to be divisible by 2, P must be even.
P can be 0 or 2 or 4 or
6 or 8
Divisibility by 9:
The sum of the digits of a number must
be divisible by 9 for the entire number to be divisible by 9.
Let the sum of the
digits of 7743P => 7 + 7 + 4 +
3 + x = 21 + x.
To make the sum
divisible by 9, we need to find the smallest positive integer P such that
21+P is divisible by 9.
The next multiple of 9
is 27. Therefore P=6
Now, combining both conditions, the
smallest positive integer P is 6, making 77436 divisible by 18.
Divisibility
Rules for Composite Numbers – Divisibility by 22
24) Find the smallest positive
integer x such that the number 5947x is divisible by 22.
a)4 b)6 c)8 d)5
Answer: B
Explanation:
A number is divisible by
22 if and only if it is divisible by both 2 and 11.
The number will be
divisible by 2 if its last digit is even.
Therefore, x must be 0
or 2 or 4 or 6 or 8
The divisibility rule
for 11: A number is divisible by 11 if the difference between the sum of the
digits in the odd positions and the sum of the digits in the even positions is
either 0 or a multiple of 11.
In the case of 5947x,
the sum of the digits in the odd positions is x + 4 + 5 = x + 9, and the sum of
the digits in the even positions is 7+3+5 = 15.
Therefore, we have the
equation 15-(x+9) = 0 or 11.
Since x must be a
positive integer, x = 6 is the smallest value that satisfies the divisibility
by both 2 and 9
Divisibility by 22 :
Divisibility by Composite Numbers
25) In the six-digit
number 46778x, what is the value of x that makes the number divisible by 22?
a)0 b)2 c)6 d)8
Answer :
Explanation :
A number is divisible by
22 if and only if it is divisible by both 2 and 11.
Divisibility by 11:
The difference between the alternating
sum of the digits of a number must be divisible by 11 for the entire number to
be divisible by 11.
In our case, the alternating sum of the
digits is
(8+7+4) - (x+7+6) = 0 or 11
19-(x+13)=0 =>x−6 to be divisible by
11, x must be 6, making x−6=0.
Divisibility by 2:
For a number to be divisible by 2, its
unit digit must be even. Since x is in the unit's place of 45358x, x must be even. As we found in the first
step, x=6, which is indeed even.
Therefore, the value of x that makes the
six-digit number 45358x divisible by 22 is x=6.
