2025 AMC 8 Problems/Problem 22
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[hide]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:
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,
The values of
that satisfy this are
The answer is
. ~Tacos_are_yummy_1. i agree.
Solution 3 (Critical Thinking)
Say the number of empty coats before the first coat are .
By that, is the number of empty hooks between all coats. There is
gaps in every situation because
cannot be 0.
To solve the problem, you minus from 35(The
empty hook at the beginning and
empty hook from the end).
Then minus 1(The first occupied hook). After that, divide the rest of the number of hook by
(The difference of hooks between coats). We do this because the first occupied hook is always after
hooks, we just see if the rest can be divided by the difference.
To make it clear:
If =
is a integer so this situation satisfies the conditions.
For =
We can skip to the third step since every time you add to
it two less then last situation.
This situation satisfies the conditions.
We keep doing this until the difference is greater than and we see that
situations does satisfy the conditions. The answer is
.
Note: This method is slow but is easy to check and understand.
~ LIBAIKO
~ Edited by aoum
Solution 4
Let the number of empty hooks before the first coat be denoted as .
This means that there are empty hooks between all the coats. In every situation, there are
gaps, since
cannot be zero.
To solve the problem, subtract from 35 (accounting for the
empty hooks at the beginning and
empty hooks at the end). Then subtract 1 (for the first occupied hook). After that, divide the remaining number of hooks by
(the number of hooks between coats). We do this because the first occupied hook is always after
hooks; we just need to check if the remaining hooks can be evenly divided by the number of gaps.
To clarify:
- If
:
Since 16 is an integer, this situation satisfies the conditions.
- If
:
We can skip the the third step, as each time you add 1 to , it decreases by 2 from the previous situation.
This situation also satisfies the conditions.
Continue this process until the difference exceeds . We find that 7 situations satisfy the conditions. Therefore, the answer is 7.
Note: This method is slow but easy to check and understand.
~aoum
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=q76L7lqj8IXEUGuL&t=3152 ~hsnacademy
Video Solution by Thinking Feet
Video Solution by Dr. David
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 |
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