Number System Questions and Answers – Divisibility Rules of Prime Numbers – Set 2

This set of Aptitude Questions and Answers (MCQs) focuses on “Divisibility Rules of Prime Numbers – Set 2”.

1. What least number should be added to 3000 to obtain a number exactly divisible by 17?
a) 9
b) 12
c) 5
d) 7
View Answer

Answer: a
Explanation: On dividing 3000 by 17, we get 8 as remainder.
Number to be added = (17 – 8) = 9.

2. Find the least number that must by subtracted from 2000 to get a number exactly divisible by 31.
a) 6
b) 16
c) 21
d) 30
View Answer

Answer: b
Explanation: On dividing 2000 by 31, we get 16 as remainder.
Required number to be subtracted = 16.

3. m is a positive integer such that m2 + 12 is divisible by m. Find all the possible values of m.
a) 1, 2, 4, 6 and 12
b) 1, 2, 3, 6 and 12
c) 1, 2, 3, 4, 6 and 12
d) 1, 2, 3, 4, 8 and 12
View Answer

Answer: c
Explanation: \(\frac{m^2+12}{m}\)=m+\(\frac{12}{m} \)
12 must be completely divisible by m.
So, possible values of m are 1, 2, 3, 4, 6 and 12.
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4. Find the greatest 5-digit number which is exactly divisible by 47.
a) 99969
b) 99981
c) 99975
d) 99993
View Answer

Answer: a
Explanation: The greatest 5-digit number is 99999.
On dividing 99999 by 47, we get 30 as remainder.
Therefore, the greatest 5-digit number which is exactly divisible by 47 is 99999-30 = 99969.

5. A chewing gum produced 58335 packs of gum. The company produced the same number of packs of each flavor of gum. How many different flavors of gum could the company have produced?
a) 6
b) 8
c) 7
d) 3
View Answer

Answer: b
Explanation: 58335 is divisible by 1, 3, 5, 15 and 3889.
1 pack each of 58335 flavors.
3 packs each of 19445 flavors.
5 packs each of 11667 flavors.
15 packs each of 3889 flavors.
3889 packs each of 15 flavors.
11667 packs each of 5 flavors.
19445 packs each of 3 flavors.
58335 packs each of 1 flavor.
Therefore, the company could have produced 8 different flavors of gum.

6. Which of the following numbers divide 111111111?
a) 3 and 37 only
b) 3, 37 and 111 only
c) 3, 11, 37 and 111 only
d) 3, 11, 37, 111 and 1001
View Answer

Answer: b
Explanation: Sum of all digits = 9, which is divisible by 3. So, the given number is divisible by 3.
The given number when divided by 37 gives 3003003. So, the given number is divisible by 37.
The given number when divided by 111 gives 1001001. Clearly, it is divisible by 111.
Hence, the given number is divisible by each one of 3, 37, and 111.

7. Which of the following is not necessarily true for a number m(m+1)(2m+1), m being a natural number?
a) It is divisible by 3
b) It is always even
c) It is never divisible by 237
d) It is divisible by the sum of the square of first n natural number
View Answer

Answer: c
Explanation: m(m+1)(2m+1) is divisible by 3 for all natural values of m. For example, if m=4, 4*5*9 is divisible by 3.
m(m+1)(2m+1) is always even. For example, if m=4, 4*5*9=180 is even number.
It is divisible by the sum of the square of first n natural number. For example, if m=4, 4*5*9=180 is divisible by 12+22+32+42=30.
m(m+1)(2m+1) is divisible by 239 for m=78, 78*79*156 is divisible by 237. Hence, the statement m(m+1)(2m+1) is never divisible by 237 is not necessarily true.
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8. What is the value of n when 34041 and 32506 are divided by a 3-digit number n, leave the same remainder?
a) 307
b) 461
c) 298
d) 359
View Answer

Answer: a
Explanation: Let the common remainder be x.
So, (34041-x) and (32506-x) both are divisible by n.
Hence, their difference is also divisible by n.
34041-x-32506+x = 1535 which is divisible by 307.
Therefore, when 34041 and 32506 are divided by a 3-digit number 307, leave the same remainder.

9. Find the number of positive integer p in the range 50≥x≥10, such that the product (p-1)(p-2)(9-3)……3*2*1 is not divisible by p.
a) 9
b) 11
c) 13
d) 8
View Answer

Answer: b
Explanation: For the product (p-1)(p-2)(9-3)……3*2*1 is not divisible by p, the value of p must be a prime number. Prime numbers between 10 and 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. Therefore, there are 11 positive integer values of p in the range 50≥x≥10, such that the product (p-1)(p-2)(9-3)……3*2*1 is not divisible by p.
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10. If the sum of numbers (x49)2 and x3 is divisible by 7, then which of the following may be a least value of x?
a) 5
b) 3
c) 7
d) 6
View Answer

Answer: b
Explanation: For x=3, 3492+33 = 121828, is divisible by 7.
For x=7, 7492+73 = 561344, is divisible by 7.
Therefore, the least value of x is 3.

To practice all aptitude questions, please visit “1000+ Quantitative Aptitude Questions”, “1000+ Logical Reasoning Questions”, and “Data Interpretation Questions”.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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