<s>[INST] Finding the number of divisors of $N=2^7\times3^5\times5^3$ Find the number of divisors of $N=2^7\times3^5\times5^3$ which are of the form $4t+1$, where $t$ is a natural number.
I understood the fact, that $5$ can be written in the form of $4t+1$. Also, every even power of $3$ can be written in the form of $4t+1.$
for example:-
$4(2)+1=9$ which is an even power of $3$
$4(0)+1=1$ which is an even power of $3$: $3^0$
$4(20)+1=81$ which is also an even power of $3$
Why are we multiplying power of $5$ that is $4$ powers of $5$ ($5^0$, $5^1$, $5^2$, $5^3$) and three powers of 3($3^0$, $3^2$, $3^4$)
That is $4\times3=12$.
And then why are we subtracting $-1$ from $12$. ( Final answer is $11$)
Why not adding $4+3=7$
 [/INST] We multiply instead of adding because we're counting the number of ways to mix and match elements from a set of $4$ with elements from a set of $3$. Watch:
$5^03^0, 5^03^2,5^03^4$
That's $3$.
$5^13^0, 5^13^2,5^13^4$
That's another $3$.
$5^23^0, 5^23^2,5^23^4$
Another $3$.
$5^33^0, 5^33^2,5^33^4$
Another $3$.
That's $3$, a total of $4$ times. Thus, $4$ times $3$.
Finally, we subtract $1$ because one of these answers, $5^03^0$ equals $1$, which is $4t+1$ with $t=0$. We don't count $0$ as a natural number, so we have to exclude that one.
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