2001 AMC 12 Problem 21

Attempt Problem 21 of the 2001 AMC 12 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 2001 AMC 12 solutions, or check the answer key.

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21.

Four positive integers

a,

b,

c,

and

d

have a product of

8!

and satisfy

ab + a + b = 524,

bc + b + c = 146,

cd + c + d = 104.

What is

a - d?

4

6

8

10

12

Answer: D

Concepts:Simon’s Favorite Factoring Trick divisibility system of equations

Difficulty rating: 1960

Small hint:

Add

1

to each equation so the left side factors:

(a + 1)(b + 1) = 525,

and so on

Big hint:

525 = 3 \cdot 5^2 \cdot 7

has a factor of

25

but

(b+1)(c+1) = 147

does not, so

a + 1

is a multiple of

25

Solution:

Adding

1

to each equation factors the left sides:

\begin{aligned} (a + 1)(b + 1) &= 525 = 3 \cdot 5^2 \cdot 7, \\ (b + 1)(c + 1) &= 147 = 3 \cdot 7^2, \\ (c + 1)(d + 1) &= 105 = 3 \cdot 5 \cdot 7. \end{aligned}

Since

525

has a factor of

25

while

147

is not divisible by

5,

the factor

a + 1

must carry the

25.

Among divisors of

525,

only

a + 1 = 25

makes

a = 24

divide

8! = 40320.

Then

b + 1 = 21,

c + 1 = 7,

and

d + 1 = 15,

giving

b = 20,

c = 6,

d = 14.

(Indeed

24 \cdot 20 \cdot 6 \cdot 14 = 40320 = 8!.

)

So

a - d = 24 - 14 = 10.

Thus, the correct answer is D.

Problem 21 in other years