13. Find the sum of all the co-primes of 144
a. 3456
b. 3168
c. 2276
d. 2256
Answer: A
Explanation:
The sum of co-primes of the 144 = $\phi (N) \times \dfrac{N}{2}$
\(\phi \left( {144} \right) = 144 \times \left( {1 - \dfrac{1}{2}} \right) \times \left( {1 - \dfrac{1}{3}} \right) = 48\)
$= 48 \times \dfrac{{144}}{2}$ = 3456
14. Find the number of ways of writing 60 as a product of two co - primes
a. 1
b. 2
c. 3
d. 4
Answer: D
Explanation:
The prime factorization of 60 = ${2^2} \times 3 \times 5$
The number of ways of writing 60 as a product of two co - primes = \({2^{n - 1}}\) = ${2^{3 - 1}}$ = 4
Here \(n\) is the number of prime factors of \(60\).
15. Find the product of all the factors of 50
a. \({50}\)
b. \({50^2}\)
c. \({50^3}\)
d. \({50^4}\)
Answer: C
Explanation:
Prime factorization of 50 = $2 \times {5^2}$
Product of all the factors of 50 = \({N^{\left( {{\textstyle{\text{Factors of N} \over 2}}} \right)}}\) = ${50^{\dfrac{{(1 + 1).(2 + 1)}}{2}}} = {50^3}$
16. Find the number of odd factors of 5400.
a. 16
b. 3
c. 8
d. 12
Answer: D
Explanation:
$\qquad \begin{array}{|l}
\llap{2~~~~} 5400 \\[4px] \hline
\llap{2~~~~} 2700 \\[4px] \hline
\llap{2~~~~} 1350 \\[4px] \hline
\llap{5~~~~} 675 \\[4px] \hline
\llap{5~~~~} 135 \\[4px] \hline
\llap{3~~~~} 27 \\[4px] \hline
\llap{3~~~~} 9 \\[4px] \hline
3
\end{array}$
$\therefore$ \(5400\) = \({2^3} \times {3^3} \times {5^2}\)
With little observation, you can write $5400$ as $54 \times 100$ $= 2 \times 27 \times {2^2} \times {5^2}$ $= 2 \times {3^3} \times {2^2} \times {5^2}$$ = {2^3} \times {3^3} \times {5^2}$
To get odd factors, we should not choose any factor from ${2^3}$.
Remaining factors of ${3^3} \times {5^2}$ are $\left( {3 + 1} \right).\left( {2 + 1} \right)$
$\therefore$ Total odd factors = $4 \times 3 = 12$
17. Find the sum of odd factors of 5400.
a. 956
b. 1156
c. 1200
d. 1240
Answer: D
Explanation:
\(5400\) = \({2^3} \times {3^3} \times {5^2}\)
To find the sum of odd factors we should not choose any factors of $2^3$.
Sum of odd factors = $\left( {1 + 3 + {3^2} + {3^3}} \right)\left( {1 + 5 + {5^2}} \right)$
$\therefore$ Sum of odd factors = $40 \times 31 = 1240$
18. Find the number of even factors of 5400.
a. 30
b. 36
c. 42
d. 46
Answer: B
Explanation:
\(5400\) = \({2^3} \times {3^3} \times {5^2}\)
To get even factors, we should choose atleast one two multiple from ${2^3}$. i.e.,$2,\,{2^2},\,{2^3}$
$\therefore$ Number of even factors = $3 \times 4 \times 3 = 36$
