7. Find the number of factors of 100.
a. 2
b. 5
c. 7
d. 9

Answer: D

Explanation:
We know that \(100\) = \({2^2} \times {5^2}\)
So number of factors of 100 = (2 +1 ).(2 +1) = 9.
Infact the factors are 1, 2, 4, 5, 10, 20, 25, 50, 100

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8. Find the sum of the factors of 72
a. 152
b. 195
c. 205
d. 216

Answer: B

Explanation:
\(72\) = $({2^3} \times {3^2}).$
Sum of all the factors of 72 = $\dfrac{{{2^{3 + 1}} - 1}}{{2 - 1}} \times \dfrac{{{3^{2 + 1}} - 1}}{{3 - 1}}$ $= 15 \times 13 = 195$

9. Find the number of ways of writing 140 as a product of two factors
a. 6
b. 9
c. 10
d. 12

Answer: A

Explanation:
The prime factorization of 140 = \({2^2} \times 5 \times 7\)
So number of ways of writing 140 as a product of two factors = \(\dfrac{1}{2} \times \left[ {(p + 1).(q + 1).(r + 1)...} \right]\) = \(\dfrac{1}{2} \times \left[ {(2 + 1).(1 + 1).(1 + 1).} \right] = 6\)

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10. Find the number of ways of writing 144 as a product of two 'distinct factors'
a. 5
b. 6
c. 7
d. 8

Answer: C

Explanation:
The prime factorization of 144 = ${2^4} \times {3^2}$
If both factors are different, then = \(\dfrac{1}{2} \times \left[ {(p + 1).(q + 1).(r + 1)... - 1} \right]\) = \(\dfrac{1}{2} \times \left[ {(4 + 1).(2 + 1) - 1} \right]\) = \(7\)

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11. Find the number of ways of writing 144 as a product of two factors'
a. 5
b. 6
c. 7
d. 8

Answer: D

Explanation:
The prime factorization of 144 = ${2^4} \times {3^2}$
If both factors nere different, then = \(\dfrac{1}{2} \times \left[ {(p + 1).(q + 1).(r + 1)... + 1} \right]\) = \(\dfrac{1}{2} \times \left[ {(4 + 1).(2 + 1) + 1} \right]\) = \(8\)

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12. Find the number of co-primes to 144 which are less than that of it
a. 2
b. 36
c. 48
d. 56

Answer: C

Explanation:
The prime factorization of 144 = ${2^4} \times {3^2}$
The number of co-primes which are less than that of 144 = \(144 \times \left( {1 - \dfrac{1}{2}} \right) \times \left( {1 - \dfrac{1}{3}} \right)\) = 48