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

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==Solution 1==
 
==Solution 1==
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<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. These are both <math>6</math> integers and we see that <math>10</math> overlaps, so the answer is <math>\boxed{\textbf{(C)}\ 11}</math>.
 
  
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~lopkiloinm
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~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
  
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==Solution 2 (Enumeration)==
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==Solution 2==
 +
Let <math>a=15-b,</math> so the special fraction is <cmath>\frac ab = \frac{15-b}{b} = \frac{15}{b}-1.</cmath>
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We can ignore the <math>-1</math> part and only focus on <math>\frac{15}{b}.</math>
  
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Consider all the cases where <math>a+b=15</math>, and construct the following table:
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The integers are <math>\frac{15}{1},\frac{15}{3},\frac{15}{5},</math> which are <math>15,5,3,</math> respectively. We get <math>30,20,18,10,8,6</math> from this group of numbers.
  
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<cmath>\begin{array}{|c|c|c|}
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The halves are <math>\frac{15}{2},\frac{15}{6},\frac{15}{10},</math> which are <math>7\frac12,2\frac12,1\frac12,</math> respectively. We get <math>15,10,9,5,4,3</math> from this group of numbers.
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\hline & & \\ [-2ex]
 
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\textbf{a} & \textbf{b} & \textbf{a/b} \\ [0.5ex]
 
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\hline \hline
 
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& & \\ [-2ex]
 
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\ \ \ \ 1 \ \ \ \ & \ \ \ \ 14 \ \ \ \ & \ \ \ \ 1/14 \ \ \ \ \\ [1ex]
 
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2 & 13 & 2/13 \\ [1ex]
 
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3 & 12 & 1/4 \\ [1ex]
 
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4 & 11 & 4/11 \\ [1ex]
 
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5 & 10 & 1/2 \\ [1ex]
 
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6 & 9 & 2/3 \\ [1ex]
 
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7 & 8 & 7/8 \\ [1ex]
 
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8 & 7 & 8/7 \\ [1ex]
 
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9 & 6 & 3/2 \\ [1ex]
 
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10 & 5 & 2 \\ [1ex]
 
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11 & 4 & 11/4 \\ [1ex]
 
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12 & 3 & 4 \\ [1ex]
 
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13 & 2 & 13/2 \\ [1ex]
 
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14 & 1 & 14 \\ [1ex]
 
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\hline
 
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\end{array}</cmath>
 
  
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Let <math>\frac{a}{b}=n</math>.
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Note that <math>10</math> appears in both groups. Therefore, the answer is <math>6+6-1=\boxed{\textbf{(C)}\  11}.</math>
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Now, we list all the possible integers obtained from an addition of two values of <math>n</math>:
 
  
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<cmath>\begin{array}{|c|c|c|c|}
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~lopkiloinm
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\hline
 
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& & &\\ [-2ex]
 
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\textbf{n1} & \textbf{n2} & \textbf{sum} & \textbf{condition} \\ [0.5ex]
 
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\hline \hline
 
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& & &\\ [-2ex]
 
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\ \ \ \ 2 \ \ \ \ & \ \ \ \ 2 \ \ \ \ & \ \ \ \ 4 \ \ \ \ & \ \\ [1ex]
 
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& 4 & 6 & \\ [1ex]
 
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& 14 & 16 & \\ [1ex]
 
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\hline
 
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& & &\\ [-2ex]
 
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4 & 4 & 8 & \\ [1ex]
 
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& 14 & 18 & \\ [1ex]
 
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\hline
 
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& & & \\ [-2ex]
 
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14 & 14 & 28 & \\ [1ex]
 
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\hline
 
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& & &\\ [-2ex]
 
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1/2 & 1/2 & 1 & \\ [1ex]
 
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& 3/2 & 2 & \\ [1ex]
 
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& 13/2 & 7 & \\ [1ex]
 
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\hline
 
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& & &\\ [-2ex]
 
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3/2 & 3/2 & 3 & \\ [1ex]
 
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& 13/2 & 8 & \ \ Rep.\ \ \\ [1ex]
 
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\hline
 
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& & &\\ [-2ex]
 
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13/2 & 13/2 & 13 & \\ [1ex]
 
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\hline
 
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& & &\\ [-2ex]
 
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1/4 & 11/4 & 3 & Rep. \\ [1ex]
 
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\hline
 
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\end{array}</cmath>
 
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Although 13 terms are found in total, two numbers appear twice respectively. Taken repetition into account, we have a total of <math>\boxed{\textbf{(C)}\ 11}</math> terms.
 
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~Wilhelm Z
 
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== Solution 3 ==
 
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All special fractions are: <math>\frac{1}{14}</math>, <math>\frac{2}{13}</math>, <math>\frac{3}{12}</math>, <math>\frac{4}{11}</math>, <math>\frac{5}{10}</math>, <math>\frac{6}{9}</math>, <math>\frac{7}{8}</math>, <math>\frac{8}{7}</math>, <math>\frac{9}{6}</math>, <math>\frac{10}{5}</math>, <math>\frac{11}{4}</math>, <math>\frac{12}{3}</math>, <math>\frac{13}{2}</math>, <math>\frac{14}{1}</math>.
 
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Hence, the following numbers are integers: <math>\frac{3}{12} + \frac{11}{4}</math>, <math>\frac{5}{10} + \frac{5}{10}</math>, <math>\frac{5}{10} + \frac{9}{6}</math>, <math>\frac{5}{10} + \frac{13}{2}</math>, <math>\frac{9}{6} + \frac{9}{6}</math>, <math>\frac{9}{6} + \frac{13}{2}</math>, <math>\frac{10}{5} + \frac{10}{5}</math>, <math>\frac{10}{5} + \frac{12}{3}</math>, <math>\frac{10}{5} + \frac{14}{1}</math>, <math>\frac{12}{3} + \frac{12}{3}</math>, <math>\frac{12}{3} + \frac{14}{1}</math>, <math>\frac{13}{2} + \frac{13}{2}</math>, <math>\frac{14}{1} + \frac{14}{1}</math>.
 
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This leads to the following distinct integers: 3, 1, 2, 7, 8, 4, 6, 16, 18, 13, 28.
 
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Therefore, the answer is <math>\boxed{\textbf{(C) }11}</math>.
 
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~Steven Chen (www.professorchenedu.com)
 
  
 
==See Also==
 
==See Also==

Revision as of 05:36, 5 January 2022

The following problem is from both the 2021 Fall AMC 10B #7 and 2021 Fall AMC 12B #5, so both problems redirect to this page.

Problem

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 1

~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)

Solution 2

Let $a=15-b,$ so the special fraction is \[\frac ab = \frac{15-b}{b} = \frac{15}{b}-1.\] We can ignore the $-1$ part and only focus on $\frac{15}{b}.$

The integers are $\frac{15}{1},\frac{15}{3},\frac{15}{5},$ which are $15,5,3,$ respectively. We get $30,20,18,10,8,6$ from this group of numbers.

The halves are $\frac{15}{2},\frac{15}{6},\frac{15}{10},$ which are $7\frac12,2\frac12,1\frac12,$ respectively. We get $15,10,9,5,4,3$ from this group of numbers.

Note that $10$ appears in both groups. Therefore, the answer is $6+6-1=\boxed{\textbf{(C)}\ 11}.$

~lopkiloinm

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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