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Reverse Digits

Problem

If the number 41 is added to its reverse, 41 + 14 = 55.

The total, 55, is divisible by 5. But not many numbers have this property. For example, 27 + 72 = 99, which is not divisible by 5.

How many numbers under 100 have this property?


Solution

Let the original number, n = 10a + b, so its reverse, m = 10b + a.

Therefore, n + m = 11a + 11b = 11(a + b).

Clearly, n + m is divisible by 5 iff a + b congruent 0 mod 5.

Listing combinations:
a = 0, b = 5  05,50
a = 1, b = 4 or 9  14, 41, 19, 91
a = 2, b = 3 or 8  23, 32, 28, 82
a = 3, b = 7  37, 73
a = 4, b = 6  46, 64
a = 5, b = 5  55
a = 6, b = 9  69, 96
a = 7, b = 8  78, 87

That is, 19 numbers in total.

What about under 1000?

Problem ID: 82 (Oct 2002)     Difficulty: 2 Star

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