Difference between revisions of "2008 AMC 8 Problems/Problem 22"
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==Solution== | ==Solution== | ||
| − | If <math>\frac{n}{3}</math> is a three digit whole number, <math>n</math> must be divisible by 3 and be <math>\ge 100*3=300</math>. If <math>3n</math> is three digits, n must be <math>\le \frac{999}{3}=333</math> So it must be divisible by three and between 300 and 333. There are <math>12</math> | + | If <math>\frac{n}{3}</math> is a three digit whole number, <math>n</math> must be divisible by 3 and be <math>\ge 100*3=300</math>. If <math>3n</math> is three digits, n must be <math>\le \frac{999}{3}=333</math> So it must be divisible by three and between 300 and 333. There are <math>\boxed{\textbf{(A)}\ 12}</math> such numbers, which you can find by direct counting. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=21|num-a=23}} | {{AMC8 box|year=2008|num-b=21|num-a=23}} | ||
Revision as of 02:59, 25 December 2012
Problem
For how many positive integer values of 
are both 
and 
three-digit whole numbers?
Solution
If is a three digit whole number, must be divisible by 3 and be 
. If is three digits, n must be 
So it must be divisible by three and between 300 and 333. There are 
such numbers, which you can find by direct counting.
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
