Can you prove that a three digit number whose first and third digits add up to the value of the second digit must be divisible by eleven?

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Let the first digit be x and the third digit be y.

The three digit number has a value of 100x + 10(x+y) + y

= 110x + 11y

= 11(10x + y)

As this has a factor of 11 it must be divisible by 11

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