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) | ||
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Followed by Problem 2 | |
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