Question

the six digit number 75A2B8 is a multiple of 36. If the 6 digit number is as large as possible, find the value of A-B.​

Answer 1

Explanation:

For a number to be divisible by 36, it must be divisible by both 4 and 9. For a number to be divisible by 4, its last two digits must be divisible by 4. Therefore, B must be even, and the last two digits of the number must be divisible by 4.

For a number to be divisible by 9, the sum of its digits must be divisible by 9. Therefore, we can add up the digits of the number and check if their sum is divisible by 9.

75A2B8 = 7 + 5 + A + 2 + B + 8 = 22 + A + B

For the sum of the digits to be divisible by 9, A + B must be 4, 13, 22, or 31.

Since we want the largest possible number, we want A and B to be as close to each other as possible. Therefore, A + B must be 31. The only even digit that can be used for B is 8, so B = 8 and A = 23 - B = 15.

Therefore, the six-digit number is 751528, and A - B = 15 - 8 = 7.