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

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Writing down all of the numbers modulo <math>7</math>, we have <math>2, 4, 6, 1, 3, 5, 0, 2, \ldots, 4, 6, 1</math>. Notice how the the cycle <math>2, 4, 6, 1, 3, 5, 0</math> repeats itself 3 times (because <math>\lfloor{\frac{50}{14}}\rfloor=3</math>). Then, we have <math>44</math>, <math>46</math>, <math>48</math>, and <math>50</math> remaining, which are <math>2</math>, <math>4</math>, <math>6</math>, and <math>1</math> mod 7, respectively. After adding them to our total count, the remainder <math>0</math> occurs <math>3</math> times, <math>1</math> occurs <math>4</math> times, <math>2</math> occurs <math>4</math> times, <math>3</math> occurs <math>3</math> times, <math>4</math> occurs <math>4</math> times, <math>5</math> occurs <math>3</math> times, and <math>6</math> occurs <math>4</math> times, which corresponds to histogram <math>\boxed{\text{(A)}}</math>.
 
Writing down all of the numbers modulo <math>7</math>, we have <math>2, 4, 6, 1, 3, 5, 0, 2, \ldots, 4, 6, 1</math>. Notice how the the cycle <math>2, 4, 6, 1, 3, 5, 0</math> repeats itself 3 times (because <math>\lfloor{\frac{50}{14}}\rfloor=3</math>). Then, we have <math>44</math>, <math>46</math>, <math>48</math>, and <math>50</math> remaining, which are <math>2</math>, <math>4</math>, <math>6</math>, and <math>1</math> mod 7, respectively. After adding them to our total count, the remainder <math>0</math> occurs <math>3</math> times, <math>1</math> occurs <math>4</math> times, <math>2</math> occurs <math>4</math> times, <math>3</math> occurs <math>3</math> times, <math>4</math> occurs <math>4</math> times, <math>5</math> occurs <math>3</math> times, and <math>6</math> occurs <math>4</math> times, which corresponds to histogram <math>\boxed{\text{(A)}}</math>.
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 +
==Solution 2==
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 +
Writing down all the remainders gives us
 +
 +
<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>.
  
 
~mrtnvlknv
 
~mrtnvlknv

Revision as of 23:16, 29 January 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?


Solution

Writing down all of the numbers modulo $7$, we have $2, 4, 6, 1, 3, 5, 0, 2, \ldots, 4, 6, 1$. Notice how the the cycle $2, 4, 6, 1, 3, 5, 0$ repeats itself 3 times (because $\lfloor{\frac{50}{14}}\rfloor=3$). Then, we have $44$, $46$, $48$, and $50$ remaining, which are $2$, $4$, $6$, and $1$ mod 7, respectively. After adding them to our total count, the remainder $0$ occurs $3$ times, $1$ occurs $4$ times, $2$ occurs $4$ times, $3$ occurs $3$ times, $4$ occurs $4$ times, $5$ occurs $3$ times, and $6$ occurs $4$ times, which corresponds to histogram $\boxed{\text{(A)}}$.

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

~mrtnvlknv

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