/sci/ - Science & Math

>>11601095
We induct on [math]n[/math].
If [math]n=0[/math], [math]N=1[/math], and we're done.
Assuming that the statement is valid for [math]k[/math], we prove it for [math]k+1[/math].
First, we split [math]N= \prod_{i=1}^{n+1} p_i^{a_i}=p_{n+1}^{a_{n+1}} \prod_{i=1}^n p_i^{a_i}[/math]. For any divisor of [math]N[/math], we can split it into the multiplication of a divisor of [math]p_{n+1}^{a_{n+1}}[/math] and another of [math]\prod_{i=1}^n p_i^{a_i}[/math] (because they're coprime). Also, multiplying any divisor of the first by a divisor of the latter gives a divisor of the multiplication. This bijection between the divisors of [math]N[/math] and the product of the two sets of divisors essentially concludes the proof.
>>11601122
Eh, the usual way you'd do it is just considering that each exponent of a divisor is an integer between [math]0[/math] and [math]a_i[/math] and then conclude by the number of possible words.