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,a, b,b, c,c, and dd have a product of 8!8! and satisfy

ab+a+b=524,ab + a + b = 524,bc+b+c=146,bc + b + c = 146,cd+c+d=104.cd + c + d = 104.

What is ad?a - d?

44

66

88

1010

1212

Answer: D
Concepts:Simon’s Favorite Factoring Trickdivisibilitysystem of equations
Difficulty rating: 1960
Solution:

Adding 11 to each equation factors the left sides: (a+1)(b+1)=525=3527,(b+1)(c+1)=147=372,(c+1)(d+1)=105=357. \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 525525 has a factor of 2525 while 147147 is not divisible by 5,5, the factor a+1a + 1 must carry the 25.25. Among divisors of 525,525, only a+1=25a + 1 = 25 makes a=24a = 24 divide 8!=40320.8! = 40320.

Then b+1=21,b + 1 = 21, c+1=7,c + 1 = 7, and d+1=15,d + 1 = 15, giving b=20,b = 20, c=6,c = 6, d=14.d = 14. (Indeed 2420614=40320=8!.24 \cdot 20 \cdot 6 \cdot 14 = 40320 = 8!.)

So ad=2414=10.a - d = 24 - 14 = 10.

Thus, the correct answer is D.

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