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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 
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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