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The number of divisors of $2^2.3^3.5^3.7^5$ of the form (4n+1).

Since $4n +1$ is odd the divisors must be of the form $3^a.5^b.7^c$ Since $a,b,c$ can have $4,4,6$ values respectively, answer should be $4.4.6$
This answer says that answer is 48 but this question came in my test in which the correct answer was 47.
Please explain without the use of modular arithmetic.

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    $\begingroup$ Modular arithmetic is the useful way to think about this. You can unpack the same argument out of the language of modular arithmetic, and write a long story about remainders-when-divided-by-four, but it'd just be modular arithmetic in (out of?) disguise. The difference between 48 and 47 is perhaps down to the choice of whether or not $n = 0$, so $4n + 1 = 1$, counts as a solution. $\endgroup$ – Mees de Vries Feb 3 '18 at 15:50
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Your exponent $b$ can be any of $4$ options, but that's not entirely true for $a$ and $c$. In fact, if $a+c$ is odd, we have a problem. (Note, for example, that $3^17^0=3$ fails to be of the proper form.) Thus, for each choice of $a$, there are only $3$ viable choices for $c$, namely, those with the same parity as $a$.

I don't see any reason why the answer should be $47$ instead of $48$, though, unless we're excluding the number $1$ as a divisor? That would seem strange.

Edit

As noted below in a comment: If we exclude $0$ as a natural number, as some authors do, then the number $1$ is not of the correct form, so we would exclude the case $a=b=c=0$.

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  • $\begingroup$ If it is originally phrased as "of the form $4n + 1$ for $n \in \mathbb N$", and this is a source where $\mathbb{N} = \{1,2,3,\ldots\}$, it makes sense. $\endgroup$ – Mees de Vries Feb 3 '18 at 15:51
  • $\begingroup$ I still don't get it. Why the condition is only on a and c, and not on b. $\endgroup$ – Voneone Feb 3 '18 at 16:12
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    $\begingroup$ We don’t need a condition on $b$ because $5$, and all of its powers, are all of the form $4n+1$ already. On the other hand, $3$ and $7$ are not if the proper form, unless their combined power is even (e.g., $9,21,49$). $\endgroup$ – G Tony Jacobs Feb 3 '18 at 17:06

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