问题: help
Find the smallest natural number with 6 as the last digit, such that if the final 6 is moved to the front of the number it is multiplied by 4.
解答:
Solution
We have 4(10n+6) = 6·10m + n, where n has m digits. So 13n + 8 = 2·10m. Hence n = 2n' and 13n' = 10m - 4. Dividing, we quickly find that the smallest n', m satisfying this are: n' = 7692, m = 5. Hence the answer is 153846.
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