2001 AMC 12 Problems/Problem 21: Difference between revisions
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ab + a + b & = 524 | ab + a + b & = 524 | ||
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cd + c + d & = 104 | cd + c + d & = 104 | ||
\end{ | \end{array} | ||
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Latest revision as of 12:01, 18 September 2023
Problem
Four positive integers 
, 
, 
, and 
have a product of 
and satisfy:
![\[\begin{array}{rl} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{array}\]](http://latex.artofproblemsolving.com/8/5/d/85dd8837f8a759a1ba67e80dc3a7d35203110e2f.png)
What is 
?
Solution
Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:
Let 
. We get:
Clearly 
divides 
. On the other hand, can not divide 
, as it then would divide 
. Similarly, can not divide 
. Hence 
divides both and . This leaves us with only two cases: 
and 
.
The first case solves to 
, which gives us 
, but then 
. We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by . (Also, a - d equals 
in this case, which is way too large to fit the answer choices.)
The second case solves to 
, which gives us a valid quadruple 
, and we have 
.
Video Solution
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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