Difference between revisions of "1989 AIME Problems/Problem 1"
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== Solution 4 (Symmetry with Generalization) == | == Solution 4 (Symmetry with Generalization) == | ||
− | More generally, we can | + | More generally, we can show that one more than the product of four consecutive integers must be a perfect square: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
(a+3)(a+2)(a+1)(a)+1 &= \left[(a+3)(a)\right]\left[(a+2)(a+1)\right]+1 \\ | (a+3)(a+2)(a+1)(a)+1 &= \left[(a+3)(a)\right]\left[(a+2)(a+1)\right]+1 \\ |
Revision as of 14:02, 28 June 2021
Contents
Problem
Compute .
Solution 1 (Symmetry)
Note that the four numbers to multiply are symmetric with the center at .
Multiply the symmetric pairs to get
and
.
.
Solution 2 (Symmetry)
Notice that . Then we can notice that
and that
. Therefore,
. This is because we have that
as per the equation
.
~qwertysri987
Solution 3 (Symmetry with Generalization)
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
.
Solution 4 (Symmetry with Generalization)
More generally, we can show that one more than the product of four consecutive integers must be a perfect square:
At
we have
~Novus677 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 5 (Prime Factorizations)
Multiplying gives us
. Adding
to this gives
. Now we must choose a number squared that is equal to
. Taking the square root of this gives
Solution 6 (Observations)
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.