What are the values for X & Y in 72X23Y for it to be perfectly divisible by 88?Options X = 1 & Y = 5 X = 7 & Y = 5 X = 3 & Y = 2 X = 7 & Y = 2

The question is asking for the values of X and Y in the number 72X23Y that would make it perfectly divisible by 88.

To solve this, we can use the divisibility rule for 88. A number is divisible by 88 if it is divisible by both 8 and 11, since 88 is a product of these two prime numbers.

First, let's check the divisibility by 8. For a number to be divisible by 8, the last three digits must form a number that is divisible by 8. So, we need to find a value for Y that makes 23Y divisible by 8. The only options for Y that satisfy this condition are 2 and 6. However, 6 is not one of the options given, so Y must be 2.

Next, let's check the divisibility by 11. For a number to be divisible by 11, the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions should be 0 or a multiple of 11. So, we need to find a value for X that satisfies this condition.

If we substitute Y = 2 into the number, we get 72X232. The sum of the digits in the odd positions is 7 + X + 3 = X + 10. The sum of the digits in the even positions is 2 + 2 + 2 = 6. The difference between these two sums is X + 10 - 6 = X + 4.

For X + 4 to be a multiple of 11, X must be 7. So, X = 7 and Y = 2. Therefore, the correct option is X = 7 & Y = 2.