Difference between revisions of "2025 AMC 8 Problems/Problem 13"

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The 2025 AMC 8 is not held yet. Please do not post false problems.
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== Problem ==
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Each of the even numbers <math>2, 4, 6, \ldots, 50</math> is divided by <math>7</math>. The remainders are recorded. Which histogram displays the number of times each remainder occurs?
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<asy>
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/*By Reda_mandymath*/
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unitsize(15);
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void histogram(pair p, string _str, int[] n) {
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    /* p is shift transformation,
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    _str is choice string,
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    n[] is the array of number of remainders,
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    _pen is the pen style of block,
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    a is the width of block,
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    b is the width of gap
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    _scale is the font scale of labels*/
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    pen _pen;
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    real a = 0.8;
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    real b = 0.3;
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    real _scale = 0.8;
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    draw(shift(p) * ((0, 0) -- (9, 0) -- (9, 5) -- (0, 5) -- cycle));
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    label(scale(_scale) * rotate(90) * "Count", (-0.4, 2.5)+p);
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    label(scale(_scale) * "Remainder", (4.5, -1)+p);
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    for (int i = 0; i <= 6; ++i) {
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        if (n[i] == 3) {
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            _pen = mediumgray;
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        } else {
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            _pen = heavygray;
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        }
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        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);
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        label(scale(_scale) * string(i), shift(p) * (a*(i+1.5) + b*i, 0), S);
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        label(scale(_scale) * string(n[i]), shift(p) * (a*(i+1.5) + b*i, n[i]), N);
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    }
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    label(_str, shift(p) * (-0.4, 6));
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}
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histogram((0, 0), "$\textbf{(A)}$", new int[] {3, 4, 4, 3, 4, 3, 4});
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histogram((12, 0), "$\textbf{(B)}$", new int[] {3, 4, 4, 4, 3, 3, 4});
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histogram((24, 0), "$\textbf{(C)}$", new int[] {3, 4, 4, 4, 4, 3, 3});
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histogram((0, -8), "$\textbf{(D)}$", new int[] {4, 3, 4, 3, 4, 3, 4});
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histogram((12, -8), "$\textbf{(E)}$", new int[] {4, 4, 3, 4, 3, 4, 3});
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</asy>
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== Solution 1 ==
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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 <math>\boxed{\textbf{(A)}}</math>.
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~Sigmacuber
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== Solution 2 ==
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Writing down all the remainders gives us
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<cmath>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.</cmath>
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In this list, there are <math>3</math> numbers with remainder <math>0</math>, <math>4</math> numbers with remainder <math>1</math>, <math>4</math> numbers with remainder <math>2</math>, <math>3</math> numbers with remainder <math>3</math>, <math>4</math> numbers with remainder <math>4</math>, <math>3</math> numbers with remainder <math>5</math>, and <math>4</math> numbers with remainder <math>6</math>. Manually computation of every single term can be avoided by recognizing the pattern alternates from <math>0, 2, 4, 6</math> to <math>1, 3, 5</math> and there are <math>25</math> terms. The only histogram that matches this is <math>\boxed{\textbf{(A)}}</math>.
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~alwaysgonnagiveyouup
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== Solution 3 ==
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First, we find all of the multiples of <math>7</math> that are even, and are therefore, in this list. Knowing that <math>7</math> is odd, and that odd+odd=even, we can find all of the even multiples of <math>7</math> by simply finding all of the multiples of <math>14</math> that fit on this list. Doing this, we end up with
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<cmath>14, 28, 42.</cmath>
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Now, we can clearly see that there are only <math>3</math> numbers in this list with <math>0</math> as a remainder. This disproves <math>2</math> of our <math>5</math> answers immediately. Since our remaining answers are identical until we reach <math>3</math> as a remainder, we can skip right to there. Now, we need to find all even numbers that leave a remainder of <math>3</math> when divided by <math>7</math>. To do this, we add <math>3</math> to all ODD multiples of <math>7</math> because odd+odd=even. This gives us
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<cmath>10, 24, 38.</cmath>
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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, which say that there are <math>4</math> of these numbers. This reveals the answer to be <math>\boxed{\textbf{(A)}}</math>.
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~Kapurnicus
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~Minor edit by NYCnerd
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== Video Solution 1 ==
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https://youtu.be/VP7g-s8akMY?si=JkH2XHbYOhk5-PSc&t=1269
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~hsnacademy
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== Video Solution 2 by Thinking Feet ==
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https://youtu.be/PKMpTS6b988
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==Video Solution(Quick, fast, easy!)==
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https://youtu.be/fdG7EDW_7xk
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~MC
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== See Also ==
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{{AMC8 box|year=2025|num-b=12|num-a=14}}
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{{MAA Notice}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 18:43, 9 March 2025

Problem

Each of the even numbers $2, 4, 6, \ldots, 50$ is divided by $7$. 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]

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 $\boxed{\textbf{(A)}}$.

~Sigmacuber

Solution 2

Writing down all the remainders gives us

\[2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 0, 2, 4, 6, 1.\]

In this list, there are $3$ numbers with remainder $0$, $4$ numbers with remainder $1$, $4$ numbers with remainder $2$, $3$ numbers with remainder $3$, $4$ numbers with remainder $4$, $3$ numbers with remainder $5$, and $4$ numbers with remainder $6$. Manually computation of every single term can be avoided by recognizing the pattern alternates from $0, 2, 4, 6$ to $1, 3, 5$ and there are $25$ terms. The only histogram that matches this is $\boxed{\textbf{(A)}}$.

~alwaysgonnagiveyouup


Solution 3

First, we find all of the multiples of $7$ that are even, and are therefore, in this list. Knowing that $7$ is odd, and that odd+odd=even, we can find all of the even multiples of $7$ by simply finding all of the multiples of $14$ that fit on this list. Doing this, we end up with

\[14, 28, 42.\]

Now, we can clearly see that there are only $3$ numbers in this list with $0$ as a remainder. This disproves $2$ of our $5$ answers immediately. Since our remaining answers are identical until we reach $3$ as a remainder, we can skip right to there. Now, we need to find all even numbers that leave a remainder of $3$ when divided by $7$. To do this, we add $3$ to all ODD multiples of $7$ because odd+odd=even. This gives us

\[10, 24, 38.\]

Note that we don't add $3$ to $49$ because that exceeds $50$. This shows us that there are only $3$ numbers on this list that have a remainder of $3$ when divided by $7$, disproving histograms B and C, which say that there are $4$ of these numbers. This reveals the answer to be $\boxed{\textbf{(A)}}$.

~Kapurnicus ~Minor edit by NYCnerd

Video Solution 1

https://youtu.be/VP7g-s8akMY?si=JkH2XHbYOhk5-PSc&t=1269 ~hsnacademy

Video Solution 2 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

See Also

2025 AMC 8 (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png