1989 AIME Problems/Problem 1
Contents
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 
.
.
Solution 3
The last digit under the radical is 
, so the square root must either end in or 
, since 
means 
. Additionally, the number must be near 
, narrowing the reasonable choices to 
and 
.
Continuing the logic, the next-to-last digit under the radical is the same as the last digit of 
, which is 
. Quick computation shows that 
ends in 
, while 
ends in 
. Thus, the answer is 
.
Solution 4
Similar to Solution 1 above, call the consecutive integers 
to make use of symmetry. Note that 
itself is not an integer - in this case, 
. The expression becomes 
. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives 
. The inside is a perfect square trinomial, since 
. It's equal to 
, which simplifies to 
. You can plug in the value of from there, or further simplify to 
, which is easier to compute. Either way, plugging in 
gives .
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 | ||
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