1989 AIME Problems/Problem 1: Difference between revisions
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== Solution == | == Solution == | ||
=== Solution 1=== | |||
Let's call our four [[consecutive]] integers <math>(n-1), n, (n+1), (n+2)</math>. Notice that <math>(n-1)(n)(n+1)(n+2)+1=(n^2+n)^2-2(n^2+n)+1 \Rightarrow (n^2+n-1)^2</math>. Thus, <math>\sqrt{(31)(30)(29)(28)+1} = (29^2+29-1) = \boxed{869}</math>. | Let's call our four [[consecutive]] integers <math>(n-1), n, (n+1), (n+2)</math>. Notice that <math>(n-1)(n)(n+1)(n+2)+1=(n^2+n)^2-2(n^2+n)+1 \Rightarrow (n^2+n-1)^2</math>. Thus, <math>\sqrt{(31)(30)(29)(28)+1} = (29^2+29-1) = \boxed{869}</math>. | ||
=== Solution 2=== | |||
Note that the four numbers to multiply are symmetric with the center at <math>29.5</math>. Multiply the symmetric pairs to get <math>31\cdot 28=868</math> and <math>30\cdot 29=870</math>. Now clearly <math>868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1 + 1 = 869^2</math>. | |||
== See also == | == See also == | ||
Revision as of 14:21, 27 December 2008
Problem
Compute 
.
Solution
Solution 1
Let's call our four consecutive integers 
. Notice that 
. Thus, 
.
Solution 2
Note that the four numbers to multiply are symmetric with the center at 
. Multiply the symmetric pairs to get 
and 
. Now clearly 
.
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
