Difference between revisions of "2001 AMC 12 Problems/Problem 21"
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ab + a + b & = 524 | ab + a + b & = 524 | ||
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cd + c + d & = 104 | cd + c + d & = 104 | ||
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Clearly <math>7^2</math> divides <math>fg</math>. On the other hand, <math>7^2</math> can not divide <math>f</math>, as it then would divide <math>ef</math>. Similarly, <math>7^2</math> can not divide <math>g</math>. Hence <math>7</math> divides both <math>f</math> and <math>g</math>. This leaves us with only two cases: <math>(f,g)=(7,21)</math> and <math>(f,g)=(21,7)</math>. | Clearly <math>7^2</math> divides <math>fg</math>. On the other hand, <math>7^2</math> can not divide <math>f</math>, as it then would divide <math>ef</math>. Similarly, <math>7^2</math> can not divide <math>g</math>. Hence <math>7</math> divides both <math>f</math> and <math>g</math>. This leaves us with only two cases: <math>(f,g)=(7,21)</math> and <math>(f,g)=(21,7)</math>. | ||
− | The first case solves to <math>(e,f,g,h)=(75,7,21,5)</math>, which gives us <math>(a,b,c,d)=(74,6,20,4)</math>, but then <math>abcd \not= 8!</math>. | + | The first case solves to <math>(e,f,g,h)=(75,7,21,5)</math>, which gives us <math>(a,b,c,d)=(74,6,20,4)</math>, but then <math>abcd \not= 8!</math>. We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by <math>7</math>. (Also, a - d equals <math>70</math> in this case, which is way too large to fit the answer choices.) |
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 == |
Latest revision as of 12:01, 18 September 2023
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.