Difference between revisions of "2020 AMC 12A Problems/Problem 13"
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The equation is then <math>N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}</math> which implies that <math>\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.</math> | The equation is then <math>N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}</math> which implies that <math>\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.</math> | ||
− | <math>a</math> has to be <math>2</math> since <math>\frac{25}{36}>\frac{ | + | <math>a</math> has to be <math>2</math> since <math>\frac{25}{36}>\frac{7}{12}</math>. <math>\frac{7}{12}</math> is the result when <math>a, b,</math> and <math>c</math> are <math>3, 2,</math> and <math>2</math> |
− | <math>b</math> being <math>3</math> will make the fraction <math>frac{2}{3}</math> which is close to <math>frac{25}{36}</math>. | + | <math>b</math> being <math>3</math> will make the fraction <math>\frac{2}{3}</math> which is close to <math>\frac{25}{36}</math>. |
− | Finally, with <math>c</math> being <math>6</math>, the fraction becomes <math>frac{25}{36}</math>. In this case <math>a, b,</math> and <math>c</math> work, which means that <math>b</math> must equal <math>\boxed{\textbf{(B) } 3.}</math>~lopkiloinm | + | Finally, with <math>c</math> being <math>6</math>, the fraction becomes <math>\frac{25}{36}</math>. In this case <math>a, b,</math> and <math>c</math> work, which means that <math>b</math> must equal <math>\boxed{\textbf{(B) } 3.}</math>~lopkiloinm |
Revision as of 15:12, 1 February 2020
Problem
There are integers and each greater than such that
Solution
can be simplified to
The equation is then which implies that
has to be since . is the result when and are and
being will make the fraction which is close to .
Finally, with being , the fraction becomes . In this case and work, which means that must equal ~lopkiloinm