Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"
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− | ==Problem | + | {{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? | ||
<math>\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13</math> | <math>\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | <math>a=15-b</math> so the fraction is < | + | 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 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 | ||
+ | |||
+ | ==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== | ||
+ | {{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 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.
Contents
[hide]Problem
Call a fraction , not necessarily in the simplest form, special if and are positive integers whose sum is . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
Solution 1
The special fractions are We rewrite them in the simplest form: 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: For the set two elements (not necessarily different) can sum to
For the set two elements (not necessarily different) can sum to
For the set two elements (not necessarily different) can sum to
Together, there are distinct integers that can be written as the sum of two, not necessarily different, special fractions: ~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
Solution 2
Let so the special fraction is We can ignore the part and only focus on
The integers are which are respectively. We get from this group of numbers.
The halves are which are respectively. We get from this group of numbers.
The quarters are which are respectively. We get from this group of numbers.
Note that and each appear twice. Therefore, the answer is
~lopkiloinm
Solution 3 (Casework)
We split this up into two cases:
Case 1: integer + integer
The whole numbers we have are (or ), (or ), and (or ). There are ways to choose different-numbered pairs and ways to choose the same-numbered pairs. So, .
Case 2: fraction + fraction
The fractions we have are (or ), (or ), and . Similarly, there are ways to choose different-numbered pairs and ways to choose the same-numbered pairs. So, .
Thus, .
So now you would just go ahead and innocently choose , right? No! We overcounted , as . Therefore, the correct answer is actually .
~MrThinker
Video Solution (Under 3 min!)
~Education, the Study of Everything
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=810
~Interstigation
Video Solution by WhyMath
~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.