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 $7993$ .

2007 Alabama ARML TST (Problems)
Preceded by: Problem 3	Followed by: Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2007 Alabama ARML TST Problems/Problem 4

Problem

Solution

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