i need to find the number of divisors to at-least divide one of the following numbers
$10^{40}$$,$ $20^{30}$ $,$ $40^{20}$ after factorising we get that $10^{40}$=$2^{40}$$*$ $5^{40}$ $,$ $20^{30}$=$2^{60}$$*$ $5^{30}$ and $40^{20}$ $=$ $2^{60}$$*$ $5^{20}$ i tried using the inclusion exclusion principle ( i used $|U- $A$_{1} $∪$ $A$_{2}$∪$ $A$_{3}|$ =S$_{0}$-S$_{1}$+S$_{2}$-S$_{3}$ ) so we get that S$_{0}$ = $41*41+61*31+61*21=4853$ which is all the options , but i get the same thing for S$_{1}$. for S$_{2}$ = A$_{1}$ $∩$ A$_{2}$$...$ i take the smallest out of the powers so i get
$41*31+41*21+61*21=3413$ , when i get to S$_{3}$ = A$_{1}$ $∩$ A$_{2}$ $∩$ A$_{3}$ i take the smallest out of them all $41*21=861$ my final answer is $4853-4853+3413-861=2552$ which is wrong , the answer should be 2301 but i cannot get to it
what am i doing wrong? am i using the inclusion exclusion principle wrong as well?
Edit: i noticed that if i do it without the S$_{1}$ i get the right answer but why? $4853-3413+861=2301$
thanks to any helpers
As you indicated, there are $41\times41$ factors of $10^{40}=2^{40}5^{40}$.
There are $20\times31$ factors of $20^{30}=2^{60}5^{30}$ that are not factors of $10^{40}=2^{40}5^{40}$;
they are of the form $2^n5^m$, where $41\le n\le 60$ and $0\le m\le 30$.
All factors of $40^{20} $ are factors of $20^{30}$.
Thus, there are $41\times41+20\times31$ factors of the three numbers.