The number \(5\,41G\,507\,2H6 \) 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 GH only once, even if it results from multiple G, H pairs.)
If it is divisible by 72, we prime factorize 72 to find its distinct factors. They are 2, 3, 4. This means the entire number is divisible by 2, 3, 4
Knowing 541G5072H6 is divisible by them, we consider the divisibility rules.
- Divisible by 2 if the number is even
- Divisible by 3 if the sum of the digits of the number are divisible by 3
- Divisible by 4 if the last two digits of the number are divisible by 4.
This means that the value of H must be 1, 3, 5, 7, or 9 (if you want the entire number to be divisible by 4. Because H is the second to last digit)
So either it is
Now considering the divisibility rule of 3 (summing up the digits so we can accord to the divisibility rule)
31 + G
33 + G
35 + G
37 + G
39 + G
Now we guess and check the values of G to find what values make the entire number divisible by 3
If H is 7 or 1, the value of G is 2, 5, or 8.
If H is 3 or 9 the value of G is 0, 3, 6, 9
If H is 5, then the value of G is 4, or 7
Now we must list the distinct pairs of G * H
2, 5, 8, 14, 35, 56, 0, 9, 18, 27, 54, 81, 20
Then find the sum
329 is your answer
Someone please check this it was a long process I might've made a silly error