Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 12"
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We see a pattern when we look at the numbers that do fulfull this property. The first number is <math>1</math>. Then <math>3, 8, 9, 24, 27, ....</math>. This follows a pattern. The first number being <math>1</math>, and the rest being the previous: <math>+2, +5, +1, +15, +3, +19, +3, +15, +1, +5, +2</math>. This sequence then repeats itself. We hence find that there are a total of <math>11*15 - 1</math> or <math>\boxed{164}</math> numbers that satisfy the inequality. | We see a pattern when we look at the numbers that do fulfull this property. The first number is <math>1</math>. Then <math>3, 8, 9, 24, 27, ....</math>. This follows a pattern. The first number being <math>1</math>, and the rest being the previous: <math>+2, +5, +1, +15, +3, +19, +3, +15, +1, +5, +2</math>. This sequence then repeats itself. We hence find that there are a total of <math>11*15 - 1</math> or <math>\boxed{164}</math> numbers that satisfy the inequality. | ||
| − | ==See | + | ==See Also== |
| + | {{Mock AIME box|year=Pre 2005|n=3|num-b=11|num-a=13}} | ||
Revision as of 10:37, 4 April 2012
Problem
Determine the number of integers 
such that 
and 
is divisible by 
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
We see a pattern when we look at the numbers that do fulfull this property. The first number is 
. Then 
. This follows a pattern. The first number being , and the rest being the previous: 
. This sequence then repeats itself. We hence find that there are a total of 
or 
numbers that satisfy the inequality.
See Also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
