Difference between revisions of "2025 AMC 8 Problems/Problem 13"
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Note that we don't add <math>3</math> to <math>49</math> because that exceeds <math>50</math>. This shows us that there are only <math>3</math> numbers on this list that have a remainder of <math>3</math> when divided by <math>7</math>, disproving histograms B and C. This reveals the answer to be <math>\boxed{\textbf{(A)}}</math>. | Note that we don't add <math>3</math> to <math>49</math> because that exceeds <math>50</math>. This shows us that there are only <math>3</math> numbers on this list that have a remainder of <math>3</math> when divided by <math>7</math>, disproving histograms B and C. This reveals the answer to be <math>\boxed{\textbf{(A)}}</math>. | ||
| − | ~ | + | ~Kapurnicus |
(If anyone sees what i'm saying here, can you please try to clean this up? Thanks!) | (If anyone sees what i'm saying here, can you please try to clean this up? Thanks!) | ||
Revision as of 19:32, 24 February 2025
Contents
Problem
Each of the even numbers 
is divided by 
. The remainders are recorded. Which histogram displays the number of times each remainder occurs?
![[asy] /*By Reda_mandymath*/ unitsize(15); void histogram(pair p, string _str, int[] n) { /* p is shift transformation, _str is choice string, n[] is the array of number of remainders, _pen is the pen style of block, a is the width of block, b is the width of gap _scale is the font scale of labels*/ pen _pen; real a = 0.8; real b = 0.3; real _scale = 0.8; draw(shift(p) * ((0, 0) -- (9, 0) -- (9, 5) -- (0, 5) -- cycle)); label(scale(_scale) * rotate(90) * "Count", (-0.4, 2.5)+p); label(scale(_scale) * "Remainder", (4.5, -1)+p); for (int i = 0; i <= 6; ++i) { if (n[i] == 3) { _pen = mediumgray; } else { _pen = heavygray; } fill(shift(p) * ((a*(i+1) + b*i, 0) -- (a*(i+1) + b*i, n[i]) -- (a*(i+2) + b*i, n[i]) -- (a*(i+2) + b*i, 0) -- cycle), _pen); label(scale(_scale) * string(i), shift(p) * (a*(i+1.5) + b*i, 0), S); label(scale(_scale) * string(n[i]), shift(p) * (a*(i+1.5) + b*i, n[i]), N); } label(_str, shift(p) * (-0.4, 6)); } histogram((0, 0), "$\textbf{(A)}$", new int[] {3, 4, 4, 3, 4, 3, 4}); histogram((12, 0), "$\textbf{(B)}$", new int[] {3, 4, 4, 4, 3, 3, 4}); histogram((24, 0), "$\textbf{(C)}$", new int[] {3, 4, 4, 4, 4, 3, 3}); histogram((0, -8), "$\textbf{(D)}$", new int[] {4, 3, 4, 3, 4, 3, 4}); histogram((12, -8), "$\textbf{(E)}$", new int[] {4, 4, 3, 4, 3, 4, 3}); [/asy]](http://latex.artofproblemsolving.com/4/c/b/4cba093bf55b34f4838e00be60618cfcab45fd8f.png)
Solution 1
Let's take the numbers 2 through 14 (evens). The remainders will be 2, 4, 6, 1, 3, 5, and 0. This sequence keeps repeating itself over and over. We can take floor(50/14) = 3, so after the number 42, every remainder has been achieved 3 times. However, since 44, 46, 48, and 50 are left, the remainders of those will be 2, 4, 6, and 1 respectively. The only histogram in which those 4 numbers are set at 4 is histogram 
.
~Sigmacuber
Solution 2
Writing down all the remainders gives us
![\[2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1.\]](http://latex.artofproblemsolving.com/e/a/d/ead2ad7bd4fcb3c595a09558d722006e01c9b5ee.png)
In this list, there are 
numbers with remainder 
, 
numbers with remainder 
, numbers with remainder 
, numbers with remainder , numbers with remainder , numbers with remainder 
, and numbers with remainder 
. Manually computation of every single term can be avoided by recognizing the pattern alternates from 
to 
and there are 
terms. The only histogram that matches this is .
~alwaysgonnagiveyouup
Solution 3
First, we find all of the multiples of that are even, and are therefore, in this list. Knowing that is odd, and that odd+odd=even, we can find all of the even multiples of by simply finding all of the multiples of 
that fit on this list. Doing this, we end up with
![\[14, 28, 42.\]](http://latex.artofproblemsolving.com/5/c/a/5ca57df06e61ea4110789d82166b96c4c4d37d49.png)
Now, we can clearly see that there are only numbers in this list with as a remainder. This disproves of our answers immediately. Since our remaining answers are identical until we reach as a remainder, we can skip to right there. Now, we need to find all even numbers that leave a remainder of when divided by . To do this, we add to all ODD multiples of because odd+odd=even. This gives us
![\[10, 24, 38.\]](http://latex.artofproblemsolving.com/1/0/c/10c9eebd783da7b9f643c8071ee95ba5bb486c28.png)
Note that we don't add to 
because that exceeds 
. This shows us that there are only numbers on this list that have a remainder of when divided by , disproving histograms B and C. This reveals the answer to be .
~Kapurnicus
(If anyone sees what i'm saying here, can you please try to clean this up? Thanks!)
Video Solution 1
https://youtu.be/VP7g-s8akMY?si=JkH2XHbYOhk5-PSc&t=1269 ~hsnacademy
Video Solution 2 by Thinking Feet
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
