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$2^{101p}$ and $2^{101q}$ are the factors of $2020^{2020}$. If $s(n)$ denotes all the sum of positive factors of $n$, then find the maximum value of $$4\cdot s\left(2^{p}\right) \cdot s\left(2^q\right)$$.

The options:

A) $\left(2^{20} - 1\right)\left(5^{20} - 1\right)$

B) $\left(2^{21} - 1\right)\left(5^{21} - 1\right)$

C) $\left(2^{40} - 1\right)\left(5^{20} - 1\right)$

D) $\left(2^{41} - 1\right)\left(5^{21} - 1\right)$

My first idea to solve this problem is factoring the number as follows:

$$2020^{2020}=2^{4040}\cdot 5^{2020}\cdot 101^{2020}$$

Then, (I'm not sure if the factor is the same as the dividor) my second thought is stating that $2^{4040} =(2^{1010})^4$ which put in me doubt if that would help. But if that so, that means $q=0$ as well and $p=q=0$. In this result, I really doubt if I'm doing this right. What the correct step should I do?

user516076
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1 Answers1

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I think there's a mistake in your question. Given the answers there should be $5^{101q}$ and $s(5^q)$.
Ginen that, note that if $n = p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ then $$s(n) = [\frac{(p_1^{a_1+1}-1)}{p_1-1}\frac{(p_2^{a_2+1}-1)}{p_2-1}...\frac{(p_k^{a_k+1}-1)}{p_k-1}]$$ The maximum value for p is 40 (101$*$40 = 4040) and the max value for q is 20 (101$*$20=2020). Then the answer will be D from the above formula.

Gray_Rhino
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