Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 12"

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==Solution==
 
==Solution==
{{solution}}
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We see a pattern when we look at the numbers that do fulfill 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.
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==See Also==
 
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{{Mock AIME box|year=Pre 2005|n=3|num-b=11|num-a=13}}
==See also==
 

Latest revision as of 23:59, 24 April 2013

Problem

Determine the number of integers $n$ such that $1 \le n \le 1000$ and $n^{12} - 1$ is divisible by $73$.

Solution

We see a pattern when we look at the numbers that do fulfill this property. The first number is $1$. Then $3, 8, 9, 24, 27, ....$. This follows a pattern. The first number being $1$, and the rest being the previous: $+2, +5, +1, +15, +3, +19, +3, +15, +1, +5, +2$. This sequence then repeats itself. We hence find that there are a total of $11*15 - 1$ or $\boxed{164}$ 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
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