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.)

xXxTenTacion Jul 19, 2019

#1**+1 **

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 541**G**5072**H**6 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

541G5072**1**6

541G5072**3**6

541G5072**5**6

541G5072**7**6

541G5072**9**6

...

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

CalculatorUser Jul 19, 2019