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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