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Difference between revisions of "Mock AIME 1 2010 Problems/Problem 1"

(Created page with 'Since <math>n^2 = 7k+4</math>, <math>n^2</math> is congruent to <math>4</math> mod <math>7</math>. Taking the square root of both sides, we obtain <math>n</math> is congruent to…')
 
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Since <math>n^2 = 7k+4</math>, <math>n^2</math> is congruent to <math>4</math> mod <math>7</math>.  Taking the square root of both sides, we obtain <math>n</math> is congruent to <math>\pm 2</math> mod <math>7</math>. Solving, <math>n</math> is congruent to either <math>2</math> or <math>5</math> mod <math>7</math>.  The maximum of <math>a_k</math> is <math>14074</math>, of which the square root is between <math>118</math> and <math>119</math>.  So, <math>n</math> must be less than <math>118</math>.  The series of solutions for <math>n</math> is <math>2, 9, 16...114</math> and <math>5, 12, 19...117</math>.  However, <math>2</math> does not work because <math>2^2 = 4</math>, a value not possible for <math>a_k</math>. Thus, there are <math>16 + 17</math> different solutions for a total of <math>\boxed{33}</math> solutions.
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== Problem ==
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Let <math>a_k = 7k + 4</math>. Find the number of perfect squares among <math>\{a_1, a_2, \ldots, a_{2010}\}</math>.
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== Solution ==
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Since <math>n^2 = 7k+4</math>, <math>n^2</math> is congruent to <math>4\pmod 7</math>.  Taking the square root of both sides yields <math>n \equiv \pm 2 \pmod 7</math>. The maximum of <math>a_k</math> is <math>14,074</math>, whose square root lies between <math>118</math> and <math>119</math>.  So, <math>n \leq 118</math>.  The series of solutions for <math>n</math> is <math>2, 9, 16...114</math> and <math>5, 12, 19...117</math>.  However, <math>2</math> does not work because <math>2^2 = 4</math>, a value too small for <math>a_k</math>. Thus, there are <math>16 + 17 = \boxed{033}</math> solutions.
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== See Also ==
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{{Mock AIME box|year=2010|n=1|before=First Problem|num-a=2}}

Latest revision as of 08:46, 2 August 2024

Problem

Let $a_k = 7k + 4$. Find the number of perfect squares among $\{a_1, a_2, \ldots, a_{2010}\}$.

Solution

Since $n^2 = 7k+4$, $n^2$ is congruent to $4\pmod 7$. Taking the square root of both sides yields $n \equiv \pm 2 \pmod 7$. The maximum of $a_k$ is $14,074$, whose square root lies between $118$ and $119$. So, $n \leq 118$. The series of solutions for $n$ is $2, 9, 16...114$ and $5, 12, 19...117$. However, $2$ does not work because $2^2 = 4$, a value too small for $a_k$. Thus, there are $16 + 17 = \boxed{033}$ solutions.

See Also

Mock AIME 1 2010 (Problems, Source)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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