Difference between revisions of "2025 AMC 8 Problems/Problem 22"
(→Solution 1) |
|||
| (4 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
| − | ==Problem | + | ==Problem== |
A classroom has a row of 35 coat hooks. Paulina likes coats to be equally spaced, so that there is the same number of empty hooks before the first coat, after the last coat, and between every coat and the next one. Suppose there is at least 1 coat and at least 1 empty hook. How many different numbers of coats can satisfy Paulina's pattern? | A classroom has a row of 35 coat hooks. Paulina likes coats to be equally spaced, so that there is the same number of empty hooks before the first coat, after the last coat, and between every coat and the next one. Suppose there is at least 1 coat and at least 1 empty hook. How many different numbers of coats can satisfy Paulina's pattern? | ||
| Line 7: | Line 7: | ||
==Solution 1== | ==Solution 1== | ||
Suppose there are <math>c</math> coats on the rack. Notice that there are <math>c+1</math> "gaps" formed by these coats, each of which must have the same number of empty spaces (say, <math>k</math>). Then the values <math>k</math> and <math>c</math> must satisfy <math>c+k(c+1)=35 \implies kc+k+c=35</math>. We now use Simon's Favorite Factoring Trick as follows: <cmath>kc+k+c=35</cmath> <cmath>\implies kc+k+c+1=36</cmath> <cmath>\implies (k+1)(c+1)=36</cmath> Our only restrictions now are that <math>k>0 \implies k+1 > 1</math> and <math>c>0 \implies c+1>1</math>. Other than that, each factor pair of <math>36</math> produces a valid solution <math>(k,c)</math>, which in turn uniquely determines an arrangement. Since <math>36</math> has <math>9</math> factors, our answer is <math>9-2=\boxed{\textbf{(D)}~7}</math>. ~cxsmi | Suppose there are <math>c</math> coats on the rack. Notice that there are <math>c+1</math> "gaps" formed by these coats, each of which must have the same number of empty spaces (say, <math>k</math>). Then the values <math>k</math> and <math>c</math> must satisfy <math>c+k(c+1)=35 \implies kc+k+c=35</math>. We now use Simon's Favorite Factoring Trick as follows: <cmath>kc+k+c=35</cmath> <cmath>\implies kc+k+c+1=36</cmath> <cmath>\implies (k+1)(c+1)=36</cmath> Our only restrictions now are that <math>k>0 \implies k+1 > 1</math> and <math>c>0 \implies c+1>1</math>. Other than that, each factor pair of <math>36</math> produces a valid solution <math>(k,c)</math>, which in turn uniquely determines an arrangement. Since <math>36</math> has <math>9</math> factors, our answer is <math>9-2=\boxed{\textbf{(D)}~7}</math>. ~cxsmi | ||
| + | |||
| + | ==Solution 2== | ||
| + | Say Paulina placed <math>n</math> coats. That will divide the 35 hooks into <math>n+1</math> spaces and <math>35-n</math> empty hooks. Therefore, <cmath>n+1|35-n.</cmath>The values of <math>n</math> that satisfy this are <cmath>n\in{1,2,3,5,8,11,17}</cmath>The answer is <math>\boxed{\text{(D) }7}</math>. ~Tacos_are_yummy_1 | ||
| + | |||
| + | ==Vide Solution 1 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| + | |||
| + | ==Video Solution by Thinking Feet== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=21|num-a=23}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:14, 30 January 2025
Contents
Problem
A classroom has a row of 35 coat hooks. Paulina likes coats to be equally spaced, so that there is the same number of empty hooks before the first coat, after the last coat, and between every coat and the next one. Suppose there is at least 1 coat and at least 1 empty hook. How many different numbers of coats can satisfy Paulina's pattern?
Solution 1
Suppose there are 
coats on the rack. Notice that there are 
"gaps" formed by these coats, each of which must have the same number of empty spaces (say, 
). Then the values and must satisfy 
. We now use Simon's Favorite Factoring Trick as follows: ![\[kc+k+c=35\]](http://latex.artofproblemsolving.com/e/1/1/e118a88b36e69a43050d1d2e2e0b1aea81e8665b.png)
![\[\implies kc+k+c+1=36\]](http://latex.artofproblemsolving.com/9/f/b/9fba8c51fb077682dfce0e17ae35facf3e425981.png)
![\[\implies (k+1)(c+1)=36\]](http://latex.artofproblemsolving.com/5/9/b/59b3651ae7165e96a48a493a6ef31386f3ddcb49.png)
Our only restrictions now are that 
and 
. Other than that, each factor pair of 
produces a valid solution 
, which in turn uniquely determines an arrangement. Since has 
factors, our answer is 
. ~cxsmi
Solution 2
Say Paulina placed 
coats. That will divide the 35 hooks into 
spaces and 
empty hooks. Therefore, ![\[n+1|35-n.\]](http://latex.artofproblemsolving.com/b/2/d/b2de0be3d0bccee3a9584a1d1ff2dc40f2ddfd67.png)
The values of that satisfy this are ![\[n\in{1,2,3,5,8,11,17}\]](http://latex.artofproblemsolving.com/a/f/0/af0ff5f62fb36cbbc2e7f53a1d1752ef6d3c806c.png)
The answer is 
. ~Tacos_are_yummy_1
Vide Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Thinking Feet
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
