Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (Bash))
(→‎Solution 3 (Casework))
 
(29 intermediate revisions by 9 users not shown)
Line 1: Line 1:
−
==Problem 5==
+
{{duplicate|[[2021 Fall AMC 10B Problems/Problem 7|2021 Fall AMC 10B #7]] and [[2021 Fall AMC 12B Problems/Problem 5|2021 Fall AMC 12B #5]]}}
 +
 
 +
==Problem==
 
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 is <math>15</math>. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
 
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 is <math>15</math>. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
  
Line 5: Line 7:
  
 
==Solution 1==
 
==Solution 1==
−
<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>.
+
The special fractions are <cmath>\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}.</cmath>
 +
We rewrite them in the simplest form: <cmath>\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.</cmath>
 +
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: <cmath>\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.</cmath>
 +
For the set <math>\{2,4,14\},</math> two elements (not necessarily different) can sum to <math>4,6,8,16,18,28.</math>
 +
 
 +
For the set <math>\left\{\frac{1}{2},1\frac{1}{2},6\frac{1}{2}\right\},</math> two elements (not necessarily different) can sum to <math>1,2,3,7,8,13.</math>
 +
 
 +
For the set <math>\left\{\frac{1}{4},2\frac{3}{4}\right\},</math> two elements (not necessarily different) can sum to <math>3.</math>
 +
 
 +
Together, there are <math>\boxed{\textbf{(C)}\  11}</math> distinct integers that can be written as the sum of two, not necessarily different, special fractions: <cmath>1,2,3,4,6,7,8,13,16,18,28.</cmath>
 +
~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
 +
 
 +
==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>
 +
We can ignore the <math>-1</math> part and only focus on <math>\frac{15}{b}.</math>
 +
 
 +
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.
 +
 
 +
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.
 +
 
 +
The quarters are <math>\frac{15}{4},\frac{15}{12},</math> which are <math>3\frac34,1\frac14,</math> respectively. We get <math>5</math> from this group of numbers.
 +
 
 +
Note that <math>10</math> and <math>5</math> each appear twice. Therefore, the answer is <math>\boxed{\textbf{(C)}\ 11}.</math>
  
 
~lopkiloinm
 
~lopkiloinm
  
−
==Solution 2 (Bash)==
+
==Solution 3 (Casework)==
 +
 
 +
We split this up into two cases:
 +
 
 +
'''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}=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'''
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
~MrThinker
 +
 
 +
==Video Solution (Under 3 min!)==
 +
https://youtu.be/ZAWcU-LHJoI
 +
 
 +
<i>~Education, the Study of Everything </i>
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/p9_RH4s-kBA?t=810
 +
 
 +
~Interstigation
  
−
Consider all the cases where <math>a+b=15</math>, and construct the following table:
+
==Video Solution by WhyMath==
 +
https://youtu.be/HAHoyD06O18
  
−
<cmath>\begin{array}{|c|c|c|}
+
~savannahsolver
−
\hline & & \\ [-2ex]
 
−
\textbf{a} & \textbf{b} & \textbf{a/b} \\ [0.5ex]
 
−
\hline \hline
 
−
& & \\ [-2ex]
 
−
\ \ \ \ 1 \ \ \ \ & \ \ \ \ 14 \ \ \ \ & \ \ \ \ 1/14 \ \ \ \ \\ [1ex]
 
−
2 & 13 & 2/13 \\ [1ex]
 
−
3 & 12 & 1/4 \\ [1ex]
 
−
4 & 11 & 4/11 \\ [1ex]
 
−
5 & 10 & 1/2 \\ [1ex]
 
−
6 & 9 & 2/3 \\ [1ex]
 
−
7 & 8 & 7/8 \\ [1ex]
 
−
8 & 7 & 8/7 \\ [1ex]
 
−
9 & 6 & 3/2 \\ [1ex]
 
−
10 & 5 & 2 \\ [1ex]
 
−
11 & 4 & 11/4 \\ [1ex]
 
−
12 & 3 & 4 \\ [1ex]
 
−
13 & 2 & 13/2 \\ [1ex]
 
−
14 & 1 & 14 \\ [1ex]
 
−
\end{array}</cmath>
 
  
−
Let <math>\frac{a}{b}=n</math>.
+
==Video Solution by TheBeautyofMath==
−
Now, we list all the possible integers obtained from an addition of two values of <math>n</math>:
+
For AMC 10: https://youtu.be/RyN-fKNtd3A?t=364
  
−
<cmath>\begin{array}{|c|c|c|c|}
+
For AMC 12: https://youtu.be/yaE5aAmeesc?t=776
−
\hline
 
−
& & &\\ [-2ex]
 
−
\textbf{n1} & \textbf{n2} & \textbf{sum} & \textbf{condition} \\ [0.5ex]
 
−
\hline \hline
 
−
& & &\\ [-2ex]
 
−
\ \ \ \ 2 \ \ \ \ & \ \ \ \ 2 \ \ \ \ & \ \ \ \ 4 \ \ \ \ & \ \\ [1ex]
 
−
& 4 & 6 & \\ [1ex]
 
−
& 14 & 16 & \\ [1ex]
 
−
\hline
 
−
& & &\\ [-2ex]
 
−
4 & 4 & 8 & \\ [1ex]
 
−
& 14 & 18 & \\ [1ex]
 
−
\hline
 
−
& & & \\ [-2ex]
 
−
14 & 14 & 28 & \\ [1ex]
 
−
\hline
 
−
& & &\\ [-2ex]
 
−
1/2 & 1/2 & 1 & \\ [1ex]
 
−
& 3/2 & 2 & \\ [1ex]
 
−
& 13/2 & 7 & \\ [1ex]
 
−
\hline
 
−
& & &\\ [-2ex]
 
−
3/2 & 3/2 & 3 & \\ [1ex]
 
−
& 13/2 & 8 & \ \ Rep.\ \ \\ [1ex]
 
−
\hline
 
−
& & &\\ [-2ex]
 
−
13/2 & 13/2 & 13 & \\ [1ex]
 
−
\hline
 
−
& & &\\ [-2ex]
 
−
1/4 & 11/4 & 3 & Rep. \\ [1ex]
 
−
\hline
 
−
\end{array}</cmath>
 
  
−
Although 13 terms are found in total, 8 and 3 have appeared twice respectively. Taken repetition into account, we have a total of <math>\boxed{\textbf{(C)}\ 11}</math> terms.
+
~IceMatrix
  
−
~Wilhelm Z
+
==See Also==
 +
{{AMC10 box|year=2021 Fall|ab=B|num-a=8|num-b=6}}
 +
{{AMC12 box|year=2021 Fall|ab=B|num-a=6|num-b=4}}
 +
{{MAA Notice}}

Latest revision as of 21:25, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_5&oldid=196664"