What is the smallest number x such that $180\times x$ is a perfect cube?

Question

Hint: if a number is a perfect cube, then each prime factor must appear in the factorization with an exponent divisible by 3. — Aaron, Aug 01 '14 at 23:14
I'm not sure, this is to gain some extra credit for my summer course math class and she gave us 10 questions that were taken from a math contest a few months ago from some very bright students. — Bev, Aug 01 '14 at 23:27

score 4 · Answer 1 · answered Aug 01 '14 at 23:16

4

Hint

We have

$$180=2^2\times 3^2\times 5$$ so we choose $x=2\times 3\times 5^2$. Why?

answered Aug 01 '14 at 23:16

score 2 · Answer 2 · answered Aug 02 '14 at 12:41

$$180 = 2^2 \cdot 3^2 \cdot 5,$$ So multiplying by $$x = 2\cdot 3\cdot 5^2 = 150$$ gives us $$180 x = \underbrace{(2^2\cdot 3^2 \cdot 5)}_{\large =\,180} \cdot \underbrace{(2 \cdot 3\cdot 5^2)}_{\large =\,x} = 2^3\cdot 3^3 \cdot 5^3 = (2\cdot 3\cdot 5)^3 = (30)^3 = 27000$$

as desired.

score 1 · Answer 3 · answered Aug 01 '14 at 23:15

1

$180=2^2 3^2 5$ so the smallest $x$ is $2 \cdot 3\cdot 5^2=150$

answered Aug 01 '14 at 23:15

Rene Schipperus

39,526

What is the smallest number x such that $180\times x$ is a perfect cube?

3 Answers3