Difference between revisions of "2001 AMC 12 Problems/Problem 21"
Aopsuser101 (talk | contribs) (→Solution 2) |
(Basically, the problem displayed on the page was the incorrect problem from the AMC, and the solution didn't make sense in context of the incorrect problem. I have edited to replace the incorrect problem with the correct problem.) |
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== Problem == | == Problem == | ||
− | + | Four positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> have a product of <math>8!</math> and satisfy: | |
+ | |||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | a | + | ab + a + b & = 524 |
− | b &= | + | \\ |
− | d &= | + | bc + b + c & = 146 |
+ | \\ | ||
+ | cd + c + d & = 104 | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
+ | |||
+ | What is <math>a-d</math>? | ||
+ | |||
+ | <math> | ||
+ | \text{(A) }4 | ||
+ | \qquad | ||
+ | \text{(B) }6 | ||
+ | \qquad | ||
+ | \text{(C) }8 | ||
+ | \qquad | ||
+ | \text{(D) }10 | ||
+ | \qquad | ||
+ | \text{(E) }12 | ||
+ | </math> | ||
== Solution 1 == | == Solution 1 == |
Revision as of 13:10, 17 October 2019
Problem
Four positive integers , , , and have a product of and satisfy:
What is ?
Solution 1
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.