A four-digit number has the sum of its digits in the odd positions equal to the sum of its digits in the even positions. If it is subtracted from the number formed by reversing it, the result will always be divisible by ______.91113Both (A) and (B)
Question
A four-digit number has the sum of its digits in the odd positions equal to the sum of its digits in the even positions. If it is subtracted from the number formed by reversing it, the result will always be divisible by ______.91113Both (A) and (B)
Solution
The result will always be divisible by 11.
Here's why:
Let's denote the four-digit number as ABCD, where A, B, C, and D are its digits from left to right. According to the problem, A + C = B + D.
When we reverse the number, we get DCBA.
The difference between ABCD and DCBA is (1000A + 100B + 10C + D) - (1000D + 100C + 10B + A) = 999A - 999D = 999(A - D).
Since 999 is divisible by 11, the difference, 999(A - D), is also divisible by 11, regardless of the values of A and D.
Therefore, the result will always be divisible by 11.
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