2007 AIME II Problem 2

Attempt Problem 2 of the 2007 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 2007 AIME II solutions, or check the answer key.

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2.

Find the number of ordered triples

(a, b, c)

where

a,

b,

and

c

are positive integers,

a

is a factor of

b,

a

is a factor of

c,

and

a + b + c = 100.

Answer

Answer: 200

Concepts:divisibility Diophantine Equation casework

Difficulty rating: 2070

Small hint:

Since

a

divides both

b

and

c,

it must divide

a + b + c = 100

Big hint:

Write

b = as

and

c = at:

then

s + t = \frac{100}{a} - 1,

which has

\frac{100}{a} - 2

positive solutions

Solution:

Since

a

divides

b

and

c,

it divides

a + b + c = 100.

Write

b = as

and

c = at

with

s, t \ge 1;

then

a(1 + s + t) = 100,

so

s + t = \frac{100}{a} - 1.

For positive

s

and

t

we need

\frac{100}{a} \ge 3,

so

a \in \{1, 2, 4, 5, 10, 20, 25\}.

For each such

a,

the equation

s + t = \frac{100}{a} - 1

has

\frac{100}{a} - 2

ordered positive solutions. Summing,

\begin{aligned} &\small (100 + 50 + 25 + 20 + 10 + 5 + 4) \\ &\quad {}- 2 \cdot 7 \\ &= 214 - 14 = 200. \end{aligned}