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

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==Solution 3 (Casework)==
 
==Solution 3 (Casework)==
  
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We split this up into 2 cases.
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We split this up into two cases:
  
 
'''Case 1: integer + integer'''
 
'''Case 1: integer + integer'''
  
−
The whole numbers we have are <math>\frac{10}{5}</math> (or <math>2</math>), <math>\frac{12}{3}</math> (or <math>4</math>), and <math>\frac{14}{1}</math> (or <math>14</math>). There are <math>\dbinom{3}{2}</math> ways to choose different-numbered pairs and <math>3</math> ways to choose the same-numbered pairs. So, <math>3+3=6</math>.
+
The whole numbers we have are <math>\frac{10}{5}</math> (or <math>2</math>), <math>\frac{12}{3}</math> (or <math>4</math>), and <math>\frac{14}{1}</math> (or <math>14</math>). There are <math>\dbinom{3}{2}=3</math> ways to choose different-numbered pairs and <math>3</math> ways to choose the same-numbered pairs. So, <math>3+3=6</math>.
  
 
'''Case 2: fraction + fraction'''
 
'''Case 2: fraction + fraction'''
  
−
The fractions we have are <math>\frac{5}{10}</math> (or <math>\frac{1}{2}</math>), <math>\frac{9}{6}</math> (or <math>\frac{3}{2}</math>), and <math>\frac{13}{2}</math>. Similarly, there are <math>\dbinom{3}{2}</math> ways to choose different-numbered pairs and <math>3</math> ways to choose the same-numbered pairs. So, <math>3+3=6</math>.
+
The fractions we have are <math>\frac{5}{10}</math> (or <math>\frac{1}{2}</math>), <math>\frac{9}{6}</math> (or <math>\frac{3}{2}</math>), and <math>\frac{13}{2}</math>. Similarly, there are <math>\dbinom{3}{2}=3</math> ways to choose different-numbered pairs and <math>3</math> ways to choose the same-numbered pairs. So, <math>3+3=6</math>.
  
 
Thus, <math>6+6=12</math>.
 
Thus, <math>6+6=12</math>.
−
 
−
 
  
 
So now you would just go ahead and innocently choose <math>\textbf{(D) }12</math>, right? No! We overcounted <math>8</math>, as <math>\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8</math>. Therefore, the correct answer is actually <math>12-1=\boxed{\textbf{(C)}\  11}</math>.
 
So now you would just go ahead and innocently choose <math>\textbf{(D) }12</math>, right? No! We overcounted <math>8</math>, as <math>\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8</math>. Therefore, the correct answer is actually <math>12-1=\boxed{\textbf{(C)}\  11}</math>.

Latest revision as of 20:25, 8 August 2023

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

The special fractions are \[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\] We rewrite them in the simplest form: \[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once: \[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] For the set $\{2,4,14\},$ two elements (not necessarily different) can sum to $4,6,8,16,18,28.$

For the set $\left\{\frac{1}{2},1\frac{1}{2},6\frac{1}{2}\right\},$ two elements (not necessarily different) can sum to $1,2,3,7,8,13.$

For the set $\left\{\frac{1}{4},2\frac{3}{4}\right\},$ two elements (not necessarily different) can sum to $3.$

Together, there are $\boxed{\textbf{(C)}\ 11}$ distinct integers that can be written as the sum of two, not necessarily different, special fractions: \[1,2,3,4,6,7,8,13,16,18,28.\] ~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.

The quarters are $\frac{15}{4},\frac{15}{12},$ which are $3\frac34,1\frac14,$ respectively. We get $5$ from this group of numbers.

Note that $10$ and $5$ each appear twice. Therefore, the answer is $\boxed{\textbf{(C)}\ 11}.$

~lopkiloinm

Solution 3 (Casework)

We split this up into two cases:

Case 1: integer + integer

The whole numbers we have are $\frac{10}{5}$ (or $2$), $\frac{12}{3}$ (or $4$), and $\frac{14}{1}$ (or $14$). There are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$.

Case 2: fraction + fraction

The fractions we have are $\frac{5}{10}$ (or $\frac{1}{2}$), $\frac{9}{6}$ (or $\frac{3}{2}$), and $\frac{13}{2}$. Similarly, there are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$.

Thus, $6+6=12$.

So now you would just go ahead and innocently choose $\textbf{(D) }12$, right? No! We overcounted $8$, as $\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8$. Therefore, the correct answer is actually $12-1=\boxed{\textbf{(C)}\ 11}$.

~MrThinker

Video Solution (Under 3 min!)

https://youtu.be/ZAWcU-LHJoI

~Education, the Study of Everything

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA?t=810

~Interstigation

Video Solution by WhyMath

https://youtu.be/HAHoyD06O18

~savannahsolver

Video Solution by TheBeautyofMath

For AMC 10: https://youtu.be/RyN-fKNtd3A?t=364

For AMC 12: https://youtu.be/yaE5aAmeesc?t=776

~IceMatrix

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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