Difference between revisions of "1989 AIME Problems/Problem 1"
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===Solution 5=== | ===Solution 5=== | ||
− | Multiplying <math>(31)(30)(29)(28)</math> gives us <math>755160</math>. Adding <math>1</math> to this gives <math>755161</math>. Now we must choose a number squared that is equal to <math>755161</math>. Taking the square root of this gives <math>\ | + | Multiplying <math>(31)(30)(29)(28)</math> gives us <math>755160</math>. Adding <math>1</math> to this gives <math>755161</math>. Now we must choose a number squared that is equal to <math>755161</math>. Taking the square root of this gives <math>\boxed{869}</math> |
== See also == | == See also == |
Revision as of 02:50, 4 March 2018
Contents
Problem
Compute .
Solution
Solution 1
Notice and
. So now our expression is
. Setting 870 equal to
, we get
which then equals
. So since
,
, our answer is
.
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
.
Solution 5
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
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.