Elevenses
Consider five-digit numbers like 13579, where the digits form an arithmetic progression (either ascending, descending, or constant).
How many of these five-digit numbers are multiples of 11?
The answer is 0.
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2 solutions
Alternative solution: Let c be the middle digit, then a b c d e = 1 1 1 1 1 ⋅ c + ( 2 0 0 0 0 + 1 0 0 0 − 1 0 − 2 ) ⋅ x = 1 1 1 1 1 ⋅ c + 2 0 9 8 8 ⋅ x . Now 2 0 9 8 8 = 1 1 ⋅ 1 9 0 8 , so that this number is only a multiple of 11 if the first term is. But 1 1 1 1 1 is prime; therefore c must be a multiple of 11. That gives c = 0 , etc.
Let the number required be ( a − 2 d ) ( a − d ) a ( a + d ) ( a + 2 d ) for some integer d .
Using the divisibility rule for 11, we get 2 a + 1 1 k = 3 a for some non-negative integer k . However, given the conditions of a < 1 0 and k > 0 , this equation has no solution.
The number a b c d e is divisible by 11 if a − b + c − d + e is. For an arithmetic progression starting at x and increasing by y each step, this means x − ( x + y ) + ( x + 2 y ) − ( x + 3 y ) + ( x + 4 y ) = x + 2 y is a multiple of 11 . But this is precisely the middle term of the progression. The only digit that is a multiple of 11 is zero; but that means that some of the surrounding digits should be negative (which is impossible) or all zero. Therefore only 0 0 0 0 0 satisfies this equation, but zero is not considered a 5-digit number.
Therefore there are no such numbers, and the answer is 0 .