Question

The sum of divisors of the number 3600 is

• A. 5 x 3 x 3
• B. (2⁵ – 1) $\left( \frac{3^{3} - 1}{2} \right)\mspace{6mu}\left( \frac{5^{3} - 1}{4} \right)$
• C. (2⁴ – 1) $\left( \frac{3^{2} - 1}{2} \right)\mspace{6mu}\left( \frac{52 - 1}{4} \right)$
• D. None of these

Answer 1

Answer: (B)

(2⁵ – 1) $\left( \frac{3^{3} - 1}{2} \right)\mspace{6mu}\left( \frac{5^{3} - 1}{4} \right)$

Answer 2

Now 3600 = 2⁴ x 3² x 5². Hence any divisor of 3600 is of the form 2^a 3^b 5^c where a∈{0, 1,2 , 3, 4}, b ∈ {0, 1, 2} and c ∈ {0, 1, 2}. All the divisors are terms in the product

(1+2+2² + 2³ + 2⁴) (1+3+3²). (1+5+5²) which equals the sum of all the divisors.

∴ Required sum

= $\left( \frac{2^{5} - 1}{2 - 1} \right)x\left( \frac{3^{3} - 1}{3 - 1} \right)x\left( \frac{5^{3} - 1}{5 - 1} \right)$