Difference between revisions of "1985 AJHSME Problem 2"
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==Solution 3== | ==Solution 3== | ||
− | You see that <math>90+91+92+93+94+95+96+97+98+99</math> is equal to <math>0+1+3+4+5+6+7+8+9+900</math>. You can use the formula <math>\frac{n(n+1)}{2}</math> to get 45+900=945 | + | You see that <math>90+91+92+93+94+95+96+97+98+99</math> is equal to <math>0+1+3+4+5+6+7+8+9+900</math>. You can use the formula <math>\frac{n(n+1)}{2}</math> to get <math>45+900=\boxed{\textbf{(B)}\ 945}</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 08:39, 22 March 2025
Contents
[hide]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
.
Solution 3 (Cheap and Quick)
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
.
Video Solution
~savannahsolver