The number 541G5072H is divisible by 72. If G and H each represent a single digit, what is the sum of all distinct possible values of the product GH? (Count each possible value of only once, even if it results from multiple G,H pairs.)

Guest Apr 21, 2021

#1**+1 **

72 = 8*9, so the number 541G5072H must be divisible by both 8 and 9.

The divisibility rule for 8 is that if the last 3 digits divide 8, then the whole number divides 8. Notice that if you plug in 0 for H, it divides 8 (since 720 = 8*9*10). The only other possibility for H is 8 since no other digits work (you could deduce this by logic).

The divisibility rule for 9 is that if the sum of the digits of that number divides 9, then the whole number divides 9. Notice that we don't need to test out the possibilities of G if H is equal to 0, because the product of GH will be equal to zero (you still could test them out if you wanted to, but it's irrelevant). That means we will only need to test out the possibilities of G if H is equal to 8.

That means that the sum \(5+4+1+G+5+0+7+2+8=32+G\) must divide 9. The only possibility for G is 4, since 32+4=36, which divides 9. That means that \(G=4\) and \(H=8\), and their product is \(\boxed{32}\), which is our answer.

textot Apr 22, 2021