Difference between revisions of "1985 AJHSME Problem 2"
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(→Solution 2) |
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Pair the numbers like so: | Pair the numbers like so: | ||
<cmath>(90+99)+(91+98)+(92+97)+(93+96)+(94+95)</cmath> | <cmath>(90+99)+(91+98)+(92+97)+(93+96)+(94+95)</cmath> | ||
− | The sum of each pair is <math>189</math> and there are <math>5</math> pairs, so the sum is <math>945</math> and the answer is <math>boxed{B)945}</math> | + | The sum of each pair is <math>189</math> and there are <math>5</math> pairs, so the sum is <math>945</math> and the answer is <math>\boxed{\textbf{(B)}\ 945}</math>. |
==Cheap Solution== | ==Cheap Solution== |
Revision as of 08:34, 22 March 2025
Problem
Solution 1
We can add as follows:
The answer is
Solution 2
Pair the numbers like so:
The sum of each pair is
and there are
pairs, so the sum is
and the answer is
.
Cheap Solution
We know that and
Quick estimation reveals that this sum is in between these two numbers, so the only answer available is
Solution 3
You see that is equal to
. You can use the formula
to get 45+900=945
Video Solution
~savannahsolver