The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B – A is perfectly divisible by 7, then which of the following is necessarily true?

The problem is asking for a condition that is always true given that we have a three-digit number A, its reverse B, B is greater than A, and the difference B - A is divisible by 7.

Let's denote the digits of A as X, Y, Z (in the order hundreds, tens, units). So, A = 100X + 10Y + Z.

The reverse of A is B = 100Z + 10Y + X.

Given that B > A, we have 100Z + 10Y + X > 100X + 10Y + Z. Simplifying this, we get 99Z > 99X, or Z > X.

So, one condition that is necessarily true is that the unit's digit of A is greater than the hundred's digit of A.

Now, let's consider the condition that B - A is divisible by 7.

B - A = (100Z + 10Y + X) - (100X + 10Y + Z) = 99(Z - X).

For this to be divisible by 7, Z - X must be divisible by 7. But since Z and X are digits, the only possible values for Z - X that are divisible by 7 are 0 and 7.

However, we already established that Z > X, so Z - X cannot be 0. Therefore, Z - X must be 7.

So, another condition that is necessarily true is that the difference between the unit's digit and the hundred's digit of A is 7.

In conclusion, the two conditions that are necessarily true are:

1. The unit's digit of A is greater than the hundred's digit of A.
2. The difference between the unit's digit and the hundred's digit of A is 7.