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Problem

How many ordered pairs $(a,b)$ of positive integers have a product of $10!$ and a least common multiple of $9!$?

I was told my answer of $16$ to this problem was wrong. I don't see how since we just have to distribute the max powers of $2,3,5,$ and $7$ and there are $16$ ways to do that. We don't need to divide by $2$ since $(3,4)$ is different from $(4,3)$, for example.

Puzzled417
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    There's only one way to distribute the $5's$. We have $ord_5(10!)=2$ but $ord_5(9!)=1$ so there has to be one factor of $5$ in each of $a,b$. – lulu Dec 15 '15 at 13:30
  • @lulu $\text{ord}_5(10!)$ and $\text{ord}_5(9!)$ don't exist, because $10!^k\not\equiv 1, 9!^k\not\equiv 1\pmod{5}$, because $\gcd(10!,5)=\gcd(9!,5)\neq 1$. – user236182 Dec 15 '15 at 13:36
  • Since $ab=10!$ and $lcm(a,b)=9!$, thus $gcd(a,b)=10$. So $a$ and $b$ must have both $2$ and $5$ as their prime factors. Hope this helps. – Poypoyan Dec 15 '15 at 13:39
  • @user236182 By $ord_p(n)$ I mean the maximum $a$ such that $p^a$ divides $n$. As: http://math.stackexchange.com/questions/116356/what-is-the-meaning-of-operatornameord-pn – lulu Dec 15 '15 at 13:39
  • @lulu I think it's better to denote it by $v_p(n)$ or $\upsilon_p(n)$ instead of $\text{ord}_p(n)$ (see Wikipedia or the Lifting The Exponent Lemma paper), because $\text{ord}_p(n)$ has the different meaning of being the least positive integer $k$ such that $n^k\equiv 1\pmod{p}$. I've always seen the former notation and never the latter. – user236182 Dec 15 '15 at 14:14

2 Answers2

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Your approach is fine. We write $$10!=2^8\,3^4\,5^2\,7\;\;\;9!=2^7\,3^4\,5\,7$$

Your conditions then imply that:

$2^7$ must divide one or the other of $a,b$ while $2$ divides the other ($2$ choices).

$3^4$ must divide one or the other of $a,b$ ($2$ choices).

$5$ must divide both of $a,b$ ($1$ choice).

$7$ must divide one or the other of $a,b$ ($2$ choices).

Hence there are $\fbox 8$ choices all in all.

lulu
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You can rephrase it as: how many ordered pairs $(a,b)$ of positive integers satisfy $ab=10!$ and $\gcd(a,b)=10$, because $ab=\gcd(a,b)\text{lcm}(a,b)$.

I.e. let $a=10a_1, b=10b_1$, and you're asking how many ordered pairs $(a_1,b_1)$ of positive integers satisfy $a_1b_1=\frac{10!}{100}=2^6\cdot 3^4\cdot 7$ and $\gcd(a_1,b_1)=1$, so all the cases are $$a_1\in\{1,2^6,3^4,7,2^6\cdot 3^4,3^4\cdot 7,2^6\cdot 7,2^6\cdot 3^4\cdot 7\},$$

so the answer is $8$ ordered pairs.

user236182
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