All integral solutions for $abc=100$?

Question

I can't find the answer to this question anywhere: $abc=100$, find the number of integral values of $a$, $b$ and $c$. As you might have guessed, I don't know much advanced maths so please try to explain in simple terms. Thanks

Answer 1

There are no advanced maths here.

It takes little effort to guess that $a,b,c$ must be among the divisors of $100$, which are $1,2,4,5,10,20,25,50$ and $100$. You can guess them all mentally, or use a trick: once you have found the divisors until $10$, namely $1,2,4,5,10$, the remaining divisors are obtained by division: $100/5,100/4,100/1,100/1$.

Now take all combinations of two divisors and the third will follow (when possible).

$$a=1,b=1\to c=100$$ $$a=1,b=2\to c=50$$ $$a=2,b=5\to c=10$$ $$a=20,b=50\to c=-$$ $$\cdots$$

Answer 2

First of all, there are lots of different solutions.

As Mathematician 42 suggested in his answer, you should pick $a, b$ and $c$ from the set $\{\pm1, \pm2, \pm4, \pm5, \pm10, \pm 20, \pm25,\pm50, \pm100\}$

Let's find how many different solutions we have.

First notice, that $100=2^25^2$. Our divisors have to be in the form $\pm 2^k5^l$, where $k,l\in \{0,1,2\}$. Let's focus on the positive solutions.

Popular method of solving problems like this is stars and bars method.

In our case we have two types of stars - 2s and 5s.

For each group we have three places, where we can put two bars, eg. $|2|2, 5||5$ ($a=5, b=2, c=10$) or $22||,55||$ ($a=100, b=1, c=1)$.

Thus the number of different positive solutions is $$N=\binom{4}{2}\cdot \binom{4}{2} = 36$$

If we want to include also the negative solutions, we have to notice, that the result would be positive, if the number of negative factors is even, so we can have 0 or 2 negative factors. We can select them in $\binom{3}{0}+\binom{3}{2} = 4$ ways.

So in total there are $$M=4\cdot 36 = 144$$ different integer solutions.