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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"

(Created page with "==Problem 5== Call a fraction <math>\frac{a}{b}</math>, not necessarily in the simplest form, ''special'' if <math>a</math> and <math>b</math> are positive integers whose sum...")
 
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==Solution==
 
==Solution==
 
<math>a=15-b</math> so the fraction is <math>\frac{15-b}{b}</math> which is <math>\frac{15}{b}-1</math>. We can just ignore the <math>-1</math> part and only care about <math>\frac{15}{b}</math>. Now we just group <math>\frac{15}{1}, \frac{15}{3}, \frac{15}{5}</math> as the integers and <math>\frac{15}{2}, \frac{15}{6}, \frac{15}{10}</math> as the halves. We get <math>30, 20, 18, 10, 8, 6</math> from the integers group and <math>15, 10, 9, 5, 4, 3</math> from the halves group (easily calculated by halving the values from the integers group). These are both <math>6</math> integers and we see that <math>10</math> overlaps, so the answer is <math>\boxed{(D)12}</math>.
 
<math>a=15-b</math> so the fraction is <math>\frac{15-b}{b}</math> which is <math>\frac{15}{b}-1</math>. We can just ignore the <math>-1</math> part and only care about <math>\frac{15}{b}</math>. Now we just group <math>\frac{15}{1}, \frac{15}{3}, \frac{15}{5}</math> as the integers and <math>\frac{15}{2}, \frac{15}{6}, \frac{15}{10}</math> as the halves. We get <math>30, 20, 18, 10, 8, 6</math> from the integers group and <math>15, 10, 9, 5, 4, 3</math> from the halves group (easily calculated by halving the values from the integers group). These are both <math>6</math> integers and we see that <math>10</math> overlaps, so the answer is <math>\boxed{(D)12}</math>.
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~lopkiloinm

Revision as of 01:03, 24 November 2021

Problem 5

Call a fraction $\frac{a}{b}$, not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?

$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$

Solution

$a=15-b$ so the fraction is $\frac{15-b}{b}$ which is $\frac{15}{b}-1$. We can just ignore the $-1$ part and only care about $\frac{15}{b}$. Now we just group $\frac{15}{1}, \frac{15}{3}, \frac{15}{5}$ as the integers and $\frac{15}{2}, \frac{15}{6}, \frac{15}{10}$ as the halves. We get $30, 20, 18, 10, 8, 6$ from the integers group and $15, 10, 9, 5, 4, 3$ from the halves group (easily calculated by halving the values from the integers group). These are both $6$ integers and we see that $10$ overlaps, so the answer is $\boxed{(D)12}$.

~lopkiloinm

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