+0  
 
+1
320
2
avatar+549 

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.)

 Jul 19, 2019
 #1
avatar+2856 
+3

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 

 

541G507216

 

541G507236

 

541G507256

 

541G507276

 

541G507296

...

 

 

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

 Jul 19, 2019
edited by CalculatorUser  Jul 19, 2019
edited by CalculatorUser  Jul 19, 2019
 #2
avatar
+1

There are 3 pairs that satisfy the conditions given in the question and are as follows:
G =5   and H =1
G =1   and H =5
G =6    and H=9
So, the distinct product values are:
5 x 1 = 5
6 x 9 = 54
Sum of the 2 distinct products =5 + 54 = 59 

 Jul 20, 2019

17 Online Users

avatar