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

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==Problem==
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== Problem ==
  
 
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?
 
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?
 +
<asy>
 +
/*By Reda_mandymath*/
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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*/
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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);
 +
        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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    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});
 +
</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 <math>\boxed{\textbf{(A)}}</math>.
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==Solution==
 
−
 
 
−
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{\text{(C)}}</math>.
 
  
 
~Sigmacuber
 
~Sigmacuber
  
−
==Solution 2==
+
== Solution 2 ==
  
 
Writing down all the remainders gives us  
 
Writing down all the remainders gives us  
  
−
<cmath>
+
<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>
−
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>
 
  
 
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>.
 
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>.
Line 23: Line 55:
 
~alwaysgonnagiveyouup
 
~alwaysgonnagiveyouup
  
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==Video Solution by Thinking Feet==
+
== Video Solution 1 ==
 +
https://youtu.be/VP7g-s8akMY?si=JkH2XHbYOhk5-PSc&t=1269
 +
~hsnacademy
 +
 
 +
== Video Solution 2 by Thinking Feet ==
 
https://youtu.be/PKMpTS6b988
 
https://youtu.be/PKMpTS6b988
  
−
==See Also==
+
== See Also ==
 +
 
 
{{AMC8 box|year=2025|num-b=12|num-a=14}}
 
{{AMC8 box|year=2025|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Number Theory Problems]]

Latest revision as of 09:05, 5 February 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

Video Solution 1

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

Video Solution 2 by Thinking Feet

https://youtu.be/PKMpTS6b988

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. AMC logo.png

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