Difference between revisions of "1998 AJHSME Problems/Problem 12"
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| Line 9: | Line 9: | ||
Taking the first product, we have | Taking the first product, we have | ||
| − | <math>(1-\frac{1}{2})=\frac{1}{2}</math> | + | <math>\left(1-\frac{1}{2}\right)=\frac{1}{2}</math> |
<math>\frac{1}{2}\times2=1</math> | <math>\frac{1}{2}\times2=1</math> | ||
| Line 15: | Line 15: | ||
Looking at the second, we get | Looking at the second, we get | ||
| − | <math>(1-\frac{1}{3})=\frac{2}{3}</math> | + | <math>\left(1-\frac{1}{3}\right)=\frac{2}{3}</math> |
<math>\frac{2}{3}\times3=2</math> | <math>\frac{2}{3}\times3=2</math> | ||
| Line 35: | Line 35: | ||
The sum of all numbers from <math>1</math> to <math>9</math> is | The sum of all numbers from <math>1</math> to <math>9</math> is | ||
| − | <math>\frac{ | + | <math>\frac{9\cdot10}{2}=45=\boxed{A}</math> |
| − | |||
== See also == | == See also == | ||
Latest revision as of 10:48, 20 December 2018
Problem
Solution
Taking the first product, we have
Looking at the second, we get
We seem to be going up by 
.
Just to check,
Now that we have discovered the pattern, we have to find the last term.
The sum of all numbers from to 
is
See also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
