2007 Alabama ARML TST Problems/Problem 4

Problem

Find the smallest positive integer $N$ such that the product $19999N$ ends in the four digits 2007.

Solution

The problem states that $19999N \equiv 2007 \pmod{10000}$. Since $20000N \equiv 0\pmod{10000}$, $N = 20000N - 19999N \equiv 0 - 2007 \equiv 7993 \pmod{10000}$.

Thus, the smallest positive integer solution is $7993$.

See also

2007 Alabama ARML TST (Problems)
Preceded by:
Problem 3
Followed by:
Problem 5
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