Question: "Bart, Lisa, and Maggie have a bag with three balls in it. Each ball has a positive integer on it. Each of the three children randomly picks two balls from the bag, announces the product of the two numbers, and then replaces the balls in the bag. The numbers they announce are 36, 42, and 90. Unfortunately, Bart is not very good at arithmetic, and the number he announced was 2 different than the correct product, but the two girls correctly computed their products. What are the possible values of the numbers on the balls? (Note that we are not told which of the three numbers was announced by Bart.)"
"We assign variables to the three different balls; $a$, $b$, and $c$, in which $a>b>c$. Since $90$ is the greatest number, it is $ab$: Since $42$ is the second greatest number, it is $ac$: Since $36$ is the smallest number, it is $bc$. Since $90$ and $42$ both have a common factor, a, we can find the prime factorization and take the GCF:
$90 = 2*3^2*5$
$36 = 2^2 * 3^2$
$a$ must be $6$. We can find $b$ by considering $90$ and $36$: $b$ must be $18$; but since it must be less than $a$, this does not work. Is the problem messed up? We must remember that Bart is not very good at arithmetic; $42$ must be the number he messed up. We can consider $2$ choices: $40$, or $44$. $44$ does not work because $44$ is , and it is ac. $36$ is bc, and $11$ does not divide into $36$. Also, if $11$ were $a$, it would not divide into $90$ anyway. So we now know that the $3$ numbers are $90$, $40$, and $36$.
$40 = 2^3*5$
$a$ must be $10$. Also, we can find out that $b$ is $9$, because $ab$ is $90$, and $a$ is $10$. We can also apply this to $36$. $b$ is $9$, so $9c=36$, and $c=4$. So we have $a=10$, $b=9$, and $c=4$."
I don't really understand this solution. I'm okay with saying that the $a$ is 6 because of the GCF and that $b$ is 18 because of the GCF, but how does this solution imply that the problem number is $42$?