CHMMC 2015 Individual Problem 0.1.

Question

I am practicing for the CHMMC, and a lot of the problems are very hard for me...

Question $0.1.$ states:

The following number is the product of the divisors of $n$: $$2^63^3$$What is $n$? I did some searching on google and found this, but I don't completely understand it.

Can anyone give some needed formulas, hints, or solutions to help me solve this problem?

First, prove $n$ must be of the form $2^r3^s$ for some positive integers $r,s$. Then work out the product of the divisors of $2^r3^s$. — Gerry Myerson, Nov 16 '16 at 05:02

score 2 · Accepted Answer · edited Nov 16 '16 at 06:09

2

Prime factors of $n$ must be $2$ and $3$.

Factors of $n=2^r3^s$ are of the form of $2^{p}3^{q}$ where $0 \leq p \leq r$, $0 \leq q \leq s$

Hence the product is

$$\large{2^63^3=\prod_{p=0}^r \prod_{q=0}^s2^p3^q=\prod_{p=0}^r 2^{p(s+1)}3^{s(s+1)/2}=3^{\frac{(r+1)(s)(s+1)}{2}}2^\frac{(s+1)(r)(r+1)}{2}}$$ Comparing the power of $2$: $$(r)(s+1)(r+1)=2(6)=12$$ Comparing the pwoer of $3$: $$(s)(s+1)(r+1)=2(3)=6$$

$$r=2s$$

$$s(s+1)(2s+1)=6$$

$s=1, r=2$

Hence $n=2^23^1=12$

edited Nov 16 '16 at 06:09

suomynonA

7,303

answered Nov 16 '16 at 05:24

Siong Thye Goh

153,832

Can you explain the notation that looks like the roman numeral II? – suomynonA Nov 16 '16 at 06:10
That is the notation for products. $\prod_{i=0}^n a_i= a_0 \times a_1 \times \ldots a_n$ – Siong Thye Goh Nov 16 '16 at 06:11
Oh, so like summation notation but for products instead of sums? – suomynonA Nov 16 '16 at 06:12
yup, the corresponding notation for summation would be $\sum$. $\prod$ is for product. – Siong Thye Goh Nov 16 '16 at 06:13

CHMMC 2015 Individual Problem 0.1.

1 Answers1