Difference between revisions of "1987 AJHSME Problems/Problem 3"
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<math>2(81+83+85+87+89+91+93+95+97+99)=</math> | <math>2(81+83+85+87+89+91+93+95+97+99)=</math> | ||
− | <math>\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{( | + | <math>\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(EF)}\ 1800</math> |
==Solution 1== | ==Solution 1== |
Revision as of 01:43, 20 September 2018
Contents
Problem
Solution 1
Find that
Which gives us
Solution 2
Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to
=
=
=
=
=
. Since we have
pairs, we multiply
by
to get
. But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get
, which is
.
See Also
1987 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.