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2007 Alabama ARML TST Problems/Problem 4

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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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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