This question turned out to be not so clear to me.
When we concatenate two numbers (X and Y) the mod 23 of the new number XY is 0.
The concatenation of (X-a) and (Y-b) , which is (X-a)(Y-b) also yields 0 regarding mod 23.
Then, is it true that the mod 23 of the concatenation of a and b, which is ab must also be 0?
Thank you.
First, I assume you meant to write Y-b rather than X-b, otherwise there is no meaningful way to relate the two assertions. If I am correct, then the answer is generally NO:
First, assuming base ten, the catenation of X and Y may be represented arithmetically by $X \times 10^k + Y$, where k is the number of digits of Y. Then the catenation of (X-a) and (Y - b) will be $(X-a) \times 10^k + (Y-b)$, which is $(X \times 10^k + Y) - (a \times 10^k + b)$ PROVIDED (Y-b) has k digits also. IF Y, b, and (Y-b) all have k digits, THEN the latter is equal to the catenation of X and Y minus the catenation of a and b, in which case modular arithmetic forces the catenation of a and b to be divisible by 23 also.
However, these three values need not have the same number of digits, which generally messes up the divisibility. For example, if b has l digits, l < k, then $a \times 10^k + b$ equals $(a \times (10^k - 10^l)) + (a \times 10^l + b)$, the second term being the catenation of a and b, which is divisible by 23 only if the first term is also, and we can easily choose values for X, Y, a, and b to make this be not so.
Additional: It occurs to me that this would be more convincing with actual numbers. So: Let X = Y = 23, a = 2, and b = 7. Then the catenation of X and Y is 2323, which is obviously divisible by 23. The catenation of (X-a) and (Y-b) is 2116, which is 2323 - 207, and 207 is 23 times 9, so 2116 is also divisible by 23, hence your prerequisites are satisfied. BUT: The catenation of a and b is NOT 207, it is 27, which is clearly not divisible by 23.