2012 AIME II Problem 1

Attempt Problem 1 of the 2012 AIME II 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 2012 AIME II solutions, or check the answer key.

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

1.

Find the number of ordered pairs of positive integer solutions

(m, n)

to the equation

20m + 12n = 2012.

Answer

Answer: 34

Concepts:Diophantine Equation modular arithmetic

Difficulty rating: 1870

Small hint:

Divide the equation by

4

to get

5m + 3n = 503

Big hint:

Modulo

3

the equation forces

m \equiv 1 \pmod{3};

count the values

m = 3k + 1

that keep

n

positive

Solution:

Dividing by

4

gives

5m + 3n = 503.

Reducing modulo

3,

we need

2m \equiv 503 \equiv 2 \pmod{3},

so

m \equiv 1 \pmod{3}.

Write

m = 3k + 1

with

k \ge 0;

then

3n = 503 - 5(3k + 1)

= 498 - 15k,

so

n = 166 - 5k.

This is positive exactly when

5k \le 165,

that is

k \le 33.

So

k = 0, 1, \ldots, 33

all work, giving

34

ordered pairs.