Difference between revisions of "1987 AIME Problems/Problem 14"
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<math>= \frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}</math></div> | <math>= \frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}</math></div> | ||
| − | Almost all of the terms cancel out! We are left with <math>\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = 373</math>. | + | Almost all of the terms cancel out! We are left with <math>\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}</math>. |
== See also == | == See also == | ||
Revision as of 19:44, 18 March 2014
Problem
Compute
.
Solution
The Sophie Germain Identity states that 
can be factorized as 
. Each of the terms is in the form of 
. Using Sophie-Germain, we get that 
.
![$\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]}$](http://latex.artofproblemsolving.com/4/d/f/4df1c85169d79080a1e4ad7cd3cb96f9c11ec0c7.png)
Almost all of the terms cancel out! We are left with 
.
See also
| 1987 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
