Number System Questions and Answers – Factors and Sum of Factors – Set 2

This set of Aptitude Questions and Answers (MCQs) focuses on “Factors and Sum of Factors – Set 2”.

1. Find the number of factors of 1512, excluding the factors which are perfect squares.
a) 18
b) 32
c) 28
d) 24
View Answer

Answer: c
Explanation: 1512 = 23*33*7 = 2*3*4*7*9.
The number of factors of 1512 are (3+1)(3+1)(1+1) = 32.
The factors which are perfect squares are (1+1)(1+1) = 4.
Therefore, the number of factors of 1512, excluding the factors which are perfect squares are 32-4 = 28.

2. Find the number of even factors of 5000.
a) 18
b) 15
c) 20
d) 12
View Answer

Answer: b
Explanation: 5000 = 23*54.
To find the number of even factors, we do not add 1 to the power of 2.
Therefore, the number of even factors of 5000 is (3)(4+1) = 15.

3. Find the number of odd factors of 1900.
a) 18
b) 13
c) 5
d) 6
View Answer

Answer: d
Explanation: 1900 = 22*52*19.
To find the number of odd factors we ignore the power of 2.
Therefore, the number of odd factors of 1900 is 1*(2+1)(1+1) = 6.
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4. Find the sum of all even factors of 600.
a) 1612
b) 1736
c) 1802
d) 1576
View Answer

Answer: b
Explanation: 600 = 23*3*52.
The sum of all factors is given by (20 + 21 + 22 + 23)(30 + 31)(50 + 51 + 52)
To find the sum of even factors, we ignore 20 from the above expression.
Hence, the sum of all even factors of 600 is (21 + 22 + 23)(30 + 31)(50 + 51 + 52) = 14*4*31 = 1736.

5. Find the sum of all odd factors of 2178.
a) 1729
b) 1639
c) 1549
d) 1891
View Answer

Answer: a
Explanation: 2178 = 2*32*112.
The sum of all factors is given by (20 + 21)(30 + 31+ 32)(110 + 111 + 112).
To find the sum of even factors, we ignore the entire set of 2, except 20 from the above expression.
Hence, the sum of all odd factors of 2178 is (20)(30 + 31 + 32)(110 + 111 + 112) = 1*13*133 = 1729.

6. Find the number of factors of 1200 which are multiple of 5.
a) 20
b) 30
c) 25
d) 18
View Answer

Answer: a
Explanation: 1200 = 24*3*52.
To find the number of factors which are multiple of 5, we do not add 1 to the power of 5.
Therefore, the number of factors of 1200 which are multiple of 5 is (1+5)(1+1)(2) = 20.

7. Find the sum of factors of 1400 which are multiple of 5.
a) 3400
b) 3200
c) 3800
d) 3600
View Answer

Answer: d
Explanation: 1400 = 23*52*7.
Sum of factors = (20 + 21 + 22 + 23)(50 + 51 + 52)(70 + 71).
To find the sum of factors which are multiple of 5, we ignore 50 from the above expression.
Hence, the sum of factors of 1400 which are multiple of 5 = (20 + 21 + 22 + 23)(51 + 52)(70 + 71) = 15*30*8 = 3600.
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8. Find the number of factors of 1750 which are multiple of 35.
a) 8
b) 6
c) 9
d) 10
View Answer

Answer: b
Explanation: 1750 = 2*53*7.
To find the number of factors which are multiple of 35, we do not add 1 to the power of 5 and 7.
Hence, the sum of factors of 1750 which are multiple of 35 is (1+1)(3)(1) = 6.

9. Find the sum of factors of 2800 which are multiple of 35.
a) 6560
b) 6550
c) 6510
d) 6520
View Answer

Answer: c
Explanation: 2800 = 24*52*7.
Sum of factors = (20 + 21 + 22 + 23 + 24)(50 + 51 + 52)(70 + 71).
To find the sum of factors which are multiple of 35, we ignore 50 and 70 from the above expression.
Hence, the sum of factors of 2800 which are multiple of 35 is (20 + 21 + 22 + 23 + 24)(51 + 52)(71) = 31*30*7 = 6510.
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10. Find the number of factors of 1440 which are divisible by 40.
a) 9
b) 12
c) 15
d) 18
View Answer

Answer: a
Explanation: 40 = 23*5 and 1440 = 25*32*5.
Therefore, the number of factors of 1440 which are divisible by 40 = 3*(2+1)*1 = 9.

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

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Manish Bhojasia - Founder & CTO at Sanfoundry
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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