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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 7"
(Changed solution to be more organized and precise, same solution idea maintained.)
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==Solution 1==  
 
==Solution 1==  
  
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Listing out all special fractions, we get:
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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>
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{<math>\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}</math>}
 
  
−
Simplifying and grouping based on their denominators gives
+
We take the fractional parts to obtain:
 +
<cmath>\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</cmath>
  
−
{<math>14, 4, 2</math>}, {<math> \frac{1}{2}, \frac{3}{2}, \frac{13}{2}</math>}, {<math>\frac{1}{4}, \frac{11}{4}</math>}
+
Looking at any of these fractions <math>a,</math> if there does not exist a corresponding fraction equivalent to <math>1-a</math> in this set, then <math>a</math> cannot possibly form an integer and we may disregard it. In this manner, we may disregard the following fractions:
  
−
Note all other special fractions have denominators that no other special fraction has, and therefore cannot be added with another special fraction to produce an integer. Furthermore, integers can only be produced by adding in these groupings because the denominators are equal so they could simplify to a denominator of 1 after being added.
+
<cmath>\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</cmath>
−
By simple inspection, the integers that can be expressed as the sum of two special fractions are:
 
  
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{28, 18, 16, 8, 6, 4, 1, 2, 7, 3, 13}
+
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.
 +
<cmath>(1)~~\frac{1}{4}, \frac{11}{4} \qquad (2)~~\frac{1}{2}, \frac{3}{2}, \frac{13}{2} \qquad (3)~~ 2,4,14</cmath>
  
−
And there are 11 of them, or <math>\boxed{C}</math>
+
Now, we simply try all combinations of two fractions within each subgroup to obtain
 +
<math>(1)</math> <cmath>\frac{1}{4}+\frac{11}{4}=3</cmath>
 +
<math>(2)</math> <cmath>\frac{1}{2}+\frac{1}{2}=1</cmath> <cmath>\frac{1}{2}+\frac{3}{2}=2</cmath> <cmath>\frac{1}{2}+\frac{13}{2}=7</cmath> <cmath>\frac{3}{2}+\frac{3}{2}+3</cmath> <cmath>\frac{3}{2}+\frac{13}{2}=8</cmath> <cmath>\frac{13}{2}+\frac{13}{2}=13</cmath>
 +
<math>(3)</math> <cmath>2+2=4</cmath> <cmath>2+4=6</cmath> <cmath>2+14=16</cmath> <cmath>4+4=8</cmath> <cmath>4+14=18</cmath> <cmath>14+14=28.</cmath>
  
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~KingRavi
+
Our final list of integer that may be formed consists of <cmath>1,2,3,4,6,7,8,13,16,18,28 \implies \boxed{\textbf{(C)}\ 11}.</cmath>
 +
 
 +
~samrocksnature
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 18:04, 27 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

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

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)

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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