This set of Aptitude Questions and Answers (MCQs) focuses on “Divisibility Rules of Prime Numbers”.
1. Find the value of X, if 1245X42 is divisible by 11.
a) 4
b) 5
c) 8
d) 1
View Answer
Explanation: We know that, for a number to be divisible by 11, the difference between the sum of digits in odd places and the sum of digits in even places should be either 0 or a number divisible by 11.
(1+4+X+2) – (2+5+4) = 0 or multiple of 11
X – 4 = 0
Therefore, X = 4.
2. Which of the following is/are true?
I. For a number to be divisible by 19, multiply last digit by 2 and add it to the rest of the number and the answer should be a multiple of 19.
II. For a number to be divisible by 29, multiply last digit by 3 and subtract it to the rest of the number and the answer should be a multiple of 29.
a) Only I
b) Only II
c) Both I & II
d) Neither I nor II
View Answer
Explanation: The statement I is true i.e., for a number to be divisible by 19, multiply last digit by 2 and add it to the rest of the number and the answer should be a multiple of 19.
The statement II is false because, for a number to be divisible by 29, multiply last digit by 3 and add it to the rest of the number and the answer should be a multiple of 29.
3. If 23XY70 is a number with all distinct digits and divisible by 11, find XY.
a) 63
b) 61
c) 23 or 43
d) 61 or 17
View Answer
Explanation: We know that, for a number to be divisible by 11, the difference between the sum of digits in odd places and the sum of digits in even places should be either 0 or a number divisible by 11.
(2+X+5) – (3+Y+9) = 0 or multiple of 11
X – Y – 5 = 0 or multiple of 11
If X=6 and Y=1, X – Y – 5 = 0 and hence divisible by 11.
If X=1 and Y=7, X – Y – 5 = -11, which is a multiple of 11 and hence divisible by 11.
XY=17 cannot be the answer, because 23XY70 is a number with all distinct digits and 7 is already present in the number.
Therefore, XY is 61.
4. Which of the following number is divisible by 19?
a) 703
b) 629
c) 697
d) 559
View Answer
Explanation: For a number to be divisible by 19, multiply last digit by 2 and add it to the rest of the number and the answer should be a multiple of 19.
70 + 2*3 = 76 which is divisible by 19. Hence, 703 is divisible by 19.
5. Which of the following number divides 7386071?
a) 4
b) 9
c) 3
d) 11
View Answer
Explanation: For a number to divisible by 4, last two digits of the number should be divisible by 4. But 71 is not divisible by 4. Hence the number is not divisible by 4.
For a number to be divisible by 3, the sum of the digits should be divisible by 3. But, 7+3+8+6+0+7+1=32 is not divisible by 3. Hence the number is not divisible by 3.
For a number to be divisible by 9, the sum of the digits should be divisible by 9. But, 7+3+8+6+0+7+1=32 is not divisible by 9. Hence the number is not divisible by 9.
For a number to be divisible by 11, the difference between the sum of digits in odd places and the sum of digits in even places should be either 0 or a number divisible by 11. (7+8+0+1) – (3+6+7) = 0, which is divisible by 11. Hence the number is divisible by 11.
6. If a positive integer n is divisible by 7, the remainder is 3, which of the following number yields remainder 1 when divided by 7?
a) n+3
b) n-5
c) n+6
d) n+5
View Answer
Explanation: For example, let n=35. 35 is divisible by 7. Hence 38 is divisible by 7 with 3 as remainder.
We get remainder 1 when divided by 7 for 29, 36 and 43 i.e., n-8, n-2 and n+5.
From the given options, n+5 is the correct answer.
7. Which of the following number is divisible by 7, 11 and 13?
a) 2002
b) 2004
c) 2007
d) 2011
View Answer
Explanation: For a number to be divisible by 7, multiply last digit by 2 and subtract it to the rest of the number and the answer should be either 0 or multiple of 7. 200 – 2*2 = 196 = 19 – 6*2 = 7, which is divisible by 7. Hence, 2002 is divisible by 7.
(Sum of digits in odd place) – (Sum of digits in even place) = 2-2 = 0. So, 2002 is divisible by 11.
For a number to be divisible by 13, multiply last digit by 4 and add it to the rest of the number and the answer should be a multiple of 13. 200 + 4*2 = 208 = 20 + 4*8 = 52 which is divisible by 13. Hence, 2002 is divisible by 13.
8. If p and q are integers divisible by 5, which of the following is not necessarily true?
a) p + q is divisible by 10
b) p – q is divisible by 5
c) p2 – q2 is divisible by 25
d) p*q is divisible by 25
View Answer
Explanation: Let p = 15 and q = 10, which are divisible by 5.
p – q = 15 – 10 = 5, is divisible by 5.
p2 – q2 = 152 – 102 = 125, is divisible by 25.
p*q = 15*10 = 150, is divisible by 25.
p + q = 15 + 10 = 25, is not divisible by 10.
Hence, p + q is divisible by 10, is not necessarily true.
9. If a number is divisible by both 7 and 13, then in which of the number case the number is divisible?
a) 91
b) 20
c) 4
d) 40
View Answer
Explanation: We know that 7 and 13 are co-prime.
If a number is divisible by both 7 and 13, then it should also, be divisible by 7*13 = 91.
10. Which of the following is divisible by 29?
a) 899
b) 889
c) 799
d) 769
View Answer
Explanation: For a number to be divisible by 29, multiply last digit by 3 and add it to the rest of the number and the answer should be a multiple of 29.
89 + 3*9 = 116 = 11 + 3*6 = 29, which is divisible by 29. Hence, 899 is divisible by 29.
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