# Confusing solution

• Module 5 Day 7 Your Turn Part 3 Mini-Question Solution

Let the GCD of a and b be g; then we can express a as g × A and b as g × B, where A and B don't have any factors in common. Then the LCM of a and b is g × A × B, and the LCM divided by the GCD would be A × B. Since this is equal to b, that means that g × B = A × B, which means that g = A. Then a = A^2, and b is a multiple of the square root of a.
Bold part is confusing

• @neatlobster (I'm gonna copy over the question first):

$$\text{Suppose that for integers } a \text{ and } b, \frac{\text{lcm}(a,b)}{\gcd(a,b)} =b$$
$$\text{Which of the following must be true?}$$

$$a=b^2 \\ \gcd(a,b)=1 \\ \text{One of the numbers is } b \text{ times the other} \\ a=1 \\ a \text{ is a square, and } b \text{ is a multiple of }\sqrt{a}$$

Following the explanation, we have the following:
$$\gcd(a,b)=g$$
$$a = g \cdot A$$
$$b = g \cdot B$$
$$\text{lcm}(a,b)=g \cdot A \cdot B$$

Remember, the problem gave us that $$\frac{\text{lcm}(a,b)}{\gcd(a,b)}=b$$. Using our substitutions for the lcm & gcd, we have that

$$\frac{g \cdot A \cdot B}{g} = b$$

Now, here's the clever substitution: recall that $$b = g \cdot B$$. So,
$$\frac{g \cdot A \cdot B}{g} = b \implies \frac{b \cdot A }{g} = b \implies \frac{A}{g} = 1 \implies A=g$$

Since $$a = g \cdot A , a = A^2$$, so $$a$$ is a perfect square.

And, since $$a=A^2 \implies \sqrt{a} = A = g$$

$$b = g \cdot B \implies b= \sqrt{a} \cdot B$$ , so $$b$$ is a multiple of $$\sqrt{a}$$.
Hence, the answer is the fifth one.

I know this is a lot, let me know if you have any more questions. I'd be glad to help!

P.S. Here's a hint: try using $$\gcd(a,b) \cdot \text{lcm}(a,b) = a \cdot b$$. It's a very useful fact!

• Thanks so much!

• @neatlobster @quacker88 Thank you!!
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