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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 7"

(→‎Solution 1)
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  12 \qquad\textbf{(E)}\ 13</math>
 
  12 \qquad\textbf{(E)}\ 13</math>
  
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==Solution 3 ==  
+
== Solution 1 (Simple) ==
 +
 
 +
We list out all possible special fractions. Because <math>a</math> and <math>b</math> are positive integers and must sum to 15, these 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>
 +
 
 +
Simplifying the ones that can be simplified, we obtain
 +
 
 +
<math>\frac{1}{14}</math>, <math>\frac{2}{13}</math>, <math>\frac{1}{4}</math>, <math>\frac{4}{11}</math>, <math>\frac{1}{2}</math>, <math>\frac{2}{3}</math>, <math>\frac{7}{8}</math>, <math>\frac{8}{7}</math>, <math>\frac{3}{2}</math>, <math>2</math>, <math>\frac{11}{4}</math>, <math>4</math>, <math>\frac{13}{2}</math>, <math>14</math>
 +
 
 +
It is impossible to obtain an integer from the sum of two fractions that have different denominators, and we can toss out the fractions that have denominators that appear only once because they cannot add to any other fraction to get an integer. Grouping the remaining fractions based on their denominator, we have:
 +
 
 +
<math>2</math>, <math>4</math>, <math>14</math>
 +
 
 +
<math>\frac{1}{2}</math>, <math>\frac{3}{2}</math>, <math>\frac{13}{2}</math>
 +
 
 +
<math>\frac{1}{4}</math>, <math>\frac{11}{4}</math>
 +
 
 +
Now we can methodically go through each group and find all integers that are possible;
 +
 
 +
<math>4, 6, 16, 8, 18, 28, 1, 2, 7, 3, 13</math>.
 +
 
 +
There are 11 of these integers so the answer is <math>\boxed{C}</math>
 +
 
 +
~KingRavi
 +
 
 +
 
 +
 
 +
== Solution 2 ==
 +
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>.
 +
 
 +
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>.
 +
 
 +
This leads to the following distinct integers: 3, 1, 2, 7, 8, 4, 6, 16, 18, 13, 28.
 +
 
 +
Therefore, the answer is <math>\boxed{\textbf{(C) }11}</math>.
 +
 
 +
~Steven Chen (www.professorchenedu.com)
 +
 
 +
==Solution 3 (Repeat of Solution 1) ==  
  
 
The possible special fractions are: <cmath>\frac{1}{14}, \frac{2}{13}, \frac{1}{4}, \frac{4}{11}, \frac{1}{2}, \frac{2}{3}, \frac{7}{8}, \frac{8}{7}, \frac{3}{2}, 2, \frac{11}{4}, 4, \frac{13}{2}, 14</cmath>
 
The possible special fractions are: <cmath>\frac{1}{14}, \frac{2}{13}, \frac{1}{4}, \frac{4}{11}, \frac{1}{2}, \frac{2}{3}, \frac{7}{8}, \frac{8}{7}, \frac{3}{2}, 2, \frac{11}{4}, 4, \frac{13}{2}, 14</cmath>
Line 29: Line 68:
  
 
~samrocksnature
 
~samrocksnature
−
 
−
== Solution 2 ==
 
−
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>.
 
−
 
−
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>.
 
−
 
−
This leads to the following distinct integers: 3, 1, 2, 7, 8, 4, 6, 16, 18, 13, 28.
 
−
 
−
Therefore, the answer is <math>\boxed{\textbf{(C) }11}</math>.
 
−
 
−
~Steven Chen (www.professorchenedu.com)
 
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==

Revision as of 18:39, 28 November 2021

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 (Simple)

We list out all possible special fractions. Because $a$ and $b$ are positive integers and must sum to 15, these are:

$\frac{1}{14}$, $\frac{2}{13}$, $\frac{3}{12}$, $\frac{4}{11}$, $\frac{5}{10}$, $\frac{6}{9}$, $\frac{7}{8}$, $\frac{8}{7}$, $\frac{9}{6}$, $\frac{10}{5}$, $\frac{11}{4}$, $\frac{12}{3}$, $\frac{13}{2}$, $\frac{14}{1}$

Simplifying the ones that can be simplified, we obtain

$\frac{1}{14}$, $\frac{2}{13}$, $\frac{1}{4}$, $\frac{4}{11}$, $\frac{1}{2}$, $\frac{2}{3}$, $\frac{7}{8}$, $\frac{8}{7}$, $\frac{3}{2}$, $2$, $\frac{11}{4}$, $4$, $\frac{13}{2}$, $14$

It is impossible to obtain an integer from the sum of two fractions that have different denominators, and we can toss out the fractions that have denominators that appear only once because they cannot add to any other fraction to get an integer. Grouping the remaining fractions based on their denominator, we have:

$2$, $4$, $14$

$\frac{1}{2}$, $\frac{3}{2}$, $\frac{13}{2}$

$\frac{1}{4}$, $\frac{11}{4}$

Now we can methodically go through each group and find all integers that are possible;

$4, 6, 16, 8, 18, 28, 1, 2, 7, 3, 13$.

There are 11 of these integers so the answer is $\boxed{C}$

~KingRavi


Solution 2

All special fractions are: $\frac{1}{14}$, $\frac{2}{13}$, $\frac{3}{12}$, $\frac{4}{11}$, $\frac{5}{10}$, $\frac{6}{9}$, $\frac{7}{8}$, $\frac{8}{7}$, $\frac{9}{6}$, $\frac{10}{5}$, $\frac{11}{4}$, $\frac{12}{3}$, $\frac{13}{2}$, $\frac{14}{1}$.

Hence, the following numbers are integers: $\frac{3}{12} + \frac{11}{4}$, $\frac{5}{10} + \frac{5}{10}$, $\frac{5}{10} + \frac{9}{6}$, $\frac{5}{10} + \frac{13}{2}$, $\frac{9}{6} + \frac{9}{6}$, $\frac{9}{6} + \frac{13}{2}$, $\frac{10}{5} + \frac{10}{5}$, $\frac{10}{5} + \frac{12}{3}$, $\frac{10}{5} + \frac{14}{1}$, $\frac{12}{3} + \frac{12}{3}$, $\frac{12}{3} + \frac{14}{1}$, $\frac{13}{2} + \frac{13}{2}$, $\frac{14}{1} + \frac{14}{1}$.

This leads to the following distinct integers: 3, 1, 2, 7, 8, 4, 6, 16, 18, 13, 28.

Therefore, the answer is $\boxed{\textbf{(C) }11}$.

~Steven Chen (www.professorchenedu.com)

Solution 3 (Repeat of Solution 1)

The possible special fractions are: \[\frac{1}{14}, \frac{2}{13}, \frac{1}{4}, \frac{4}{11}, \frac{1}{2}, \frac{2}{3}, \frac{7}{8}, \frac{8}{7}, \frac{3}{2}, 2, \frac{11}{4}, 4, \frac{13}{2}, 14\]

We take the fractional parts to obtain: \[\frac{1}{14}, \frac{2}{13}, \frac{1}{4}, \frac{4}{11}, \frac{1}{2}, \frac{2}{3}, \frac{7}{8}, \frac{1}{7}, \frac{1}{2}, 0, \frac{3}{4}, 0, \frac{1}{2}, 0\]

Looking at any of these fractions $a,$ if there does not exist a corresponding fraction equivalent to $1-a$ in this set, then $a$ cannot possibly form an integer and we may disregard it. In this manner, we may disregard the following fractions:

\[\cancel{\frac{1}{14}}, \cancel{\frac{2}{13}}, \frac{1}{4}, \cancel{\frac{4}{11}}, \frac{1}{2}, \cancel{\frac{2}{3}}, \cancel{\frac{7}{8}}, \cancel{\frac{1}{7}}, \frac{1}{2}, 0, \frac{3}{4}, 0, \frac{1}{2}, 0\]

We now convert the remaining fractions into their value before taking their fractional part, and divide them into subgroups in which they can form integer with the other fractions in that group. \[(1)~~\frac{1}{4}, \frac{11}{4} \qquad (2)~~\frac{1}{2}, \frac{3}{2}, \frac{13}{2} \qquad (3)~~ 2,4,14\]

Now, we simply try all combinations of two fractions within each subgroup to obtain

$(1)$ \[\frac{1}{4}+\frac{11}{4}=3\] $(2)$ \[\frac{1}{2}+\frac{1}{2}=1\] \[\frac{1}{2}+\frac{3}{2}=2\] \[\frac{1}{2}+\frac{13}{2}=7\] \[\frac{3}{2}+\frac{3}{2}+3\] \[\frac{3}{2}+\frac{13}{2}=8\] \[\frac{13}{2}+\frac{13}{2}=13\] $(3)$ \[2+2=4\] \[2+4=6\] \[2+14=16\] \[4+4=8\] \[4+14=18\] \[14+14=28.\]

Our final list of integer that may be formed consists of \[1,2,3,4,6,7,8,13,16,18,28 \implies \boxed{\textbf{(C)}\ 11}.\]

~samrocksnature

Video Solution by Interstigation

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

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

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