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
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 = \(\dfrac{1}{2} \times \left[ {(p + 1).(q + 1).(r + 1)...} \right]\) = \(\dfrac{1}{2} \times \left[ {(2 + 1).(1 + 1).(1 + 1).} \right] = 6\)
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 = \(\dfrac{1}{2} \times \left[ {(p + 1).(q + 1).(r + 1)... - 1} \right]\) = \(\dfrac{1}{2} \times \left[ {(4 + 1).(2 + 1) - 1} \right]\) = \(7\)
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 = \(\dfrac{1}{2} \times \left[ {(p + 1).(q + 1).(r + 1)... + 1} \right]\) = \(\dfrac{1}{2} \times \left[ {(4 + 1).(2 + 1) + 1} \right]\) = \(8\)
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
