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Number of Divisor of $2^{2}.3^{3}.5^{3}.7^{5}$ of the form $4n+1$ where n$\in \mathbb{N}$ is ........

My approach is to solve it using remainder theoran like putting $2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}$ put $a=0; b=0,2;c=0,1,2,3;d=0,2,4$ but not able to approach

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I agree with previous answer that a=0 for any divosor. So only need to work with combinations of b,c and d. I will put now in this form for simplfy : $$ F(b,c,d)=3^b 5^c 7^d $$ What we need to get is a divisor which is $1\mod 4$. What happen if we multiply a ($1 \mod 4$) by 3. Then we will get ($3 \mod 4$). If we multiply a ($3 \mod 4$) we get ($1 \mod 4$). Same will happen if we multiply by 7.From 1 to 3, and from 3 to 1. If we multiply by 5 we get same $\mod 4$ : If we multiply ($1 \mod 4$) by 5 we get ($1 \mod 4$) and if we multiply ($3 \mod 4$) by 5 we get ($3 \mod 4$). What mean this ? We can use any c but b and d have a condition : $$ b+d \equiv 0 \mod 2 $$ Maximal values of b and c are 3, but maximal value of d is 5. We also count 0, and in case b=0,c=0 and d=0 also would be count as a divisor (1). For c, then we have 4 possibilities (0,1,2 and 3). For b and d, we put now (b,d) in order to count all possibilities :
(0,0) (0,2) (0,4)
(1,1) (1,3) (1,5)
(2,0) (2,2) (2,4)
(3,1) (3,3) (3,5)
So in total we have 3*4=12 possibilities for b and d. So in total we have : $$ 4*12=48 $$ possibilities.(4 possibilities for c and 12 possibilities for b and d)
So we have 48 divisors. Daniel

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  • $\begingroup$ Answer is 47 not 48 $\endgroup$ – Samar Imam Zaidi Oct 18 '17 at 13:22
  • $\begingroup$ Why 47 ? You not count 1 as a divisor ? See en.wikipedia.org/wiki/Divisor. "1" is a divisor but not is a "non-trivial divisor" $\endgroup$ – Daniel Pol Oct 18 '17 at 21:06
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Guide:

Clearly $a=0$.

$$3 \equiv -1 \pmod 4$$

$$7 \equiv -1 \pmod 4$$

$$5 \equiv 1 \pmod 4$$

Hence we require $b+d$ to be an even number.

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Segregate the primes by class $\bmod 4$ since that is what we care about. We can't have any factors of $2$ because then the number would either be $4n$ or $4n+2$. Note that $3^2$ is of the form $4n+1$. That is where your problem lies.

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