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Mock AIME 1 2006-2007/Problem 12

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Problem

Let k be a positive integer with first digit 4 such that after removing the first digit, you get another positive integer, m, that satisfies 14m+1=k. Find the number of possible values of m between 0 and 10^{2007}.

Solution

The digit-removal condition is equivalent to the statement k = 4\cdot10^n + m where 10^n > m and n \geq 1. Thus 14m + 1 = 4\cdot 10^n + m so 13m = 4\cdot 10^n - 1 and m = \frac{4 \cdot 10^n - 1}{13}. It's easy to see that this value of m is small enough, so all we need to check is that it is an integer. That happens if and only if 13 is a divisor of 4\cdot 10^n - 1, so 4\cdot 10^n \equiv 1 \pmod{13} and multiplying by 4^{-1} \equiv 10 \pmod{13} we have that 10^n \equiv 10 \pmod {13} Certainly n = 1 is a solution. All we need is the order of 10 \pmod {13}. Now 10^2 = 100 \equiv 9 \pmod{13} so 10^3 \equiv 90 \equiv -1 \pmod{13}, 10^6 \equiv 1 \pmod{13} and the order of 10 mod 13 is 6. Thus, we get one value of m each time n = 6j + 1. There are 335 such values of n which fall in the required range.


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