Difference between revisions of "2001 AMC 12 Problems/Problem 21"
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The second case solves to <math>(e,f,g,h)=(25,21,7,15)</math>, which gives us a valid quadruple <math>(a,b,c,d)=(24,20,6,14)</math>, and we have <math>a-d=24-14 =\boxed{10}</math>. | The second case solves to <math>(e,f,g,h)=(25,21,7,15)</math>, which gives us a valid quadruple <math>(a,b,c,d)=(24,20,6,14)</math>, and we have <math>a-d=24-14 =\boxed{10}</math>. | ||
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/zlzievV5e-U | ||
== See Also == | == See Also == | ||
Revision as of 21:21, 18 April 2022
Contents
Problem
Four positive integers 
, 
, 
, and 
have a product of 
and satisfy:
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
