1995 AIME Problem 2

Attempt Problem 2 of the 1995 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AIME solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Find the last three digits of the product of the positive roots of

\sqrt{1995}\,x^{\log_{1995}x}=x^2.

Answer

Set

y=\log_{1995}x,

x=1995^y

Compare exponents of

1995

, then use the sum of the two values of

y

Put

y=\log_{1995}x,

x=1995^y.

The equation becomes

1995^{\frac{1}{2}+y^2}=1995^{2y},

and hence

(y-1)^2=\frac{1}{2}.

The two values of

y

have sum

2,

so the product of the corresponding positive roots is

1995^2.

Since

1995\equiv-5\pmod {1000},

its last three digits are

025.

The requested AIME answer is

25.

Problem 2 in other years