Divisibility Rules - Solved Examples - 3
Divisibility by 24:
Divisibility Rules for Composite Numbers
26) What is the value of
the smallest positive integer x such that the number 78413x2 is divisible by 24?
a)8 b)6 c)5 d)2
Answer: C
Explanation :
A number is divisible by
24 if and only if it is divisible by both 3 and 8.
The divisibility rule
for 3 :
A number is divisible by
3 if the sum of its digits is divisible by 3.
In the case of 78413x2,
the sum of the digits is (7 + 8 + 4 + 1 +
3 + x + 2) = 25 + x.
For 25 + x to be
divisible by 3 => x must be 2 or 5 or
8
The divisibility rule
for 8 :
A number is divisible by
8 if the number formed by the last 3 digits of the number is divisible by 8.
78413x2=> The last 3
digits are 3x2.
3x2 must be divisible by
8. => Substituting the values x = 2 or 5 or 8
322 is not divisible by
8.
382 is not divisible by
8.
352 is not divisible by
8
5 is the smallest value
that satisfies the both divisibility rules of 3 and 8
Therefore, the smallest
positive integer x such that the number 78413x2 is divisible by 24 is x = 5.
GMAT Problem Solving
Questions on Divisibility Rule 36 :
27) What is the greatest
possible value of ‘a’ if the number 52346a7b is divisible by 36, where a and b
are positive integers?
a)7 b)2 c)4 c)5
Answer: A
Explanation :
A number is divisible by
36 if and only if it is divisible by both 4 and 9.
Divisibility rule for 4:
A number is divisible by
4 if the last two digits are divisible by 4.
52346a7b => The last two digits are 7b. Therefore,
7b must be divisible by 4.
The possible value of b
are 2, 6
Divisibility rule for 9:
The sum of the digits of
a number is divisible by 9 if and only if the number is divisible by 9.
In this case, the sum of
the digits of 52346a7b ab is 5 + 2 + 3 + 4 + 6 + 7 + a + b. Therefore, 27 + a +
b must be divisible by 9.
The next multiple of 9
to 27+a+b is 36 => a+b = 9
Combining the
conditions:
a+b = 9
Taking b= 2 , a = 7
Taking b=6 , a=3
The largest possible
value of a is 7.
GRE Quantitative
Reasoning Questions on Number System- Divisibility Rules by 45
28) Find the largest possible value of a+b if the number 964a5b is
divisible by 45, where a and b are positive integers.
a)15 b)18 c)21 d)11
Answer : D
Explanation :
A number is divisible by
45 if and only if it is divisible by both 5 and 9.
Divisibility by 5:
A number is divisible by
5 if its unit’s digit is 0 or 5. In the case of
964a5b, the unit digit
is b, and for the entire number to be divisible by 5, b must be either 0 or 5.
Divisibility by 9:
A number is divisible by
9 if the sum of its digits is divisible by 9. In the case of
964a5b, the sum of the
digits is 9+6+4+a+5+b.
For the entire number to
be divisible by 9, (9+6+4+a+5+b) must be divisible by 9.
Now, let's consider the
options for b based on divisibility by 5:
b=0, then
9+6+4+a+5+0=24+a must be divisible by 9. => a = 3
In this case: a+b = 3+0
=3
b=5, then 9+6+4+a+5+5=29+a
must be divisible by 9. => a = 6
In this case a+b = 6 + 5
= 11
The largest possible
value of a+b is 11
CAT Quantitative
Aptitude Questions – Test of Divisibility Problems
29) Find the smallest
positive integer x such that the number 817267x is exactly divisible by 48.
a)0 b)1 c)2 d)3
Answer: C
Explanation:
817267x must be divisible
by both 3 and 16 because it is divisible by 48..
Divisibility by 3:
The sum of the digits of
817267x must be divisible by 3.
Adding digits:
8+1+7+2+6+7+x=31+x.
To make 817267x divisible by 3, x must be chosen such that 31+x is
divisible by 3.
x = 2 or 5 or 8
Divisibility by 16:
A number is divisible by 16
, if its last digit number is divisible by 16
In this case, the last
four digits are 267x
Substituting x = 2 or 5
or 8
If x=2 => 2672 which
is divisible by 16
Therefore the smallest value of x is 2
SAT Problem Solving
Questions on Divisibility BY 96 :
30) What are the largest possible values of (a + b) such that the number 49b18273a is
divisible by 96, where a and b are positive integers?
a)6 b)8 c)12 d)14
Answer
: D
Explanation
:
49b18273a is divisible by 96
then must be divisible by both 3 and 32.
Divisibility
by 32:
For a number to be divisible
by 32, its last five digits must form a multiple of 32. In this case, the last
five digits are 1273a.
Substituting a=6, we find
that 12736 which is divisible by 32.
Divisibility
by 3:
The sum of the digits of
49b18275a must be divisible by 3.
4+9+b+1+8+2+7+5=34+b.
34+b is divisible by 3.
b=2 or 5 or 8
The largest value for b is 8
Therefore, the required largest possible value for (a+b) is
(6+8)=14
SSC Mathematics: Number Systems Questions on Divisibility
31) What is the value of ‘a’ such that the number 364838a8 is
divisible by 48?
a)6 b)8 c)7 d)5
Answer: B
Explanation :
A number is divisible by 48
, if the number is divisible by 3 and 16.
So 364838a8 must be
divisible by both 3 and 16.
Divisibility
by 3:
The sum of the digits of
364838a8 must be divisible by 3.
3+6+4+8+3+8+a+8=40+a.
To make the entire number
divisible by 3, ‘a’ must be chosen such that (40+a) is divisible by 3.
The possible values of a are
2, 5 and 8
Divisibility
by 16:
A number is divisible by 16
, if its last digit number is divisible by 16
In this case, the last four
digits are 38a8.
3808 and 3888 are divisible
by 16.
So a can be 0 or 8.
Combining the
conditions: The value of a is 8. When a is 8, the number is divisible by both 3
and 16.
