Difference between revisions of "2001 AMC 12 Problems/Problem 21"
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− | == Solution == | + | == Solution 1 == |
Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows: | Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows: | ||
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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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | As above, we can write the equations as follows: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | (a+1)(b+1) & = 525 | ||
+ | \\ | ||
+ | (b+1)(c+1) & = 147 | ||
+ | \\ | ||
+ | (c+1)(d+1) & = 105 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Looking at the first two equations, we know that <math>a+1</math> but not <math>b+1</math> is a multiple of 5, and looking at the last two equations, we know that <math>(d+1)</math> but not <math>(c+1)</math> must be a multiple of 5 (since if <math>b+1</math> or <math>c+1</math> was a multiple of 5, then <math>(b+1)(c+1)</math> would also be a multiple of 5). | ||
+ | |||
+ | Thus, <math>a \equiv d \equiv 0 \hspace{1 mm} \text{(mod 5)}</math>, and <math>a - d \equiv 0 \hspace{1 mm} \text{(mod 5)}</math>. The only answer choice where this is true is <math>\boxed{\text{D) 10}}</math>. | ||
== See Also == | == See Also == |
Revision as of 21:49, 10 July 2017
Contents
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 .
Solution 2
As above, we can write the equations as follows:
Looking at the first two equations, we know that but not is a multiple of 5, and looking at the last two equations, we know that but not must be a multiple of 5 (since if or was a multiple of 5, then would also be a multiple of 5).
Thus, , and . The only answer choice where this is true is .
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.