# Thread: I'm stuck and losing sleep!

1. ## I'm stuck and losing sleep!

Help if anyone can provide it! I'm stuck on a problem that I have the answer but don't know how they got it. Here it is:

A wire is cut into three equal parts. The resulting segments are then cut into 4, 6, and 8 parts respectively. If each of the resulting segments has an integer length, what is the minimum length of the wire?

The answer is 72. But the rationale for why was explained and I don't understand.

I do understand why each piece must be "able" to be cut into pieces of 8 and would therefore would be 24 inches (8 x 3) but how does this end up being 72? Or, how and why does this get muliplied by 3? I think it has do do with multiples but its not sinking in.

Thanks,

Rob

2. Originally Posted by GRE-Rob
Help if anyone can provide it! I'm stuck on a problem that I have the answer but don't know how they got it. Here it is:

A wire is cut into three equal parts. The resulting segments are then cut into 4, 6, and 8 parts respectively. If each of the resulting segments has an integer length, what is the minimum length of the wire?

The answer is 72. But the rationale for why was explained and I don't understand.

I do understand why each piece must be "able" to be cut into pieces of 8 and would therefore would be 24 inches (8 x 3) but how does this end up being 72? Or, how and why does this get muliplied by 3? I think it has do do with multiples but its not sinking in.

Thanks,

Rob
Let the wire be $\displaystyle x$ units of length long

Cutting it into equal parts gives $\displaystyle \frac{x}{3}$ for each piece.

Each of these thirds is then cut up into
$\displaystyle \frac{x}{3} \times \frac{1}{4} = \frac{x}{12}$

$\displaystyle \frac{x}{3} \times \frac{1}{6} = \frac{x}{18}$,

$\displaystyle \frac{x}{3} \times \frac{1}{8} = \frac{x}{24}$

For x to be an integer we need to find the LCM (lowest common multiple) of 12, 18 and 24 which happens to be 72 (see the spoiler for how to show this)

Spoiler:
To do this split into prime factors:
• $\displaystyle 12 = 2 \times 2 \times 3$
• $\displaystyle 18 = 2 \times 3 \times 3$
• $\displaystyle 24 = 2 \times 2 \times 2 \times 3$

Then multiply each unique prime factor: $\displaystyle 2 \times 2 \times 3 \times 2 \times 3 = 72$

The first 3 parts are from the prime factors of 12, the second 2 from the 24 and the last 3 from the 18.

Therefore x must be 72 or multiples thereof but as the question asks for the lowest value it must be 72

3. ## Thank you e^(i*pi)

Thank uo for your help. I sleep better now.