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

Latest revision as of 00:24, 31 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

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

~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 by Thinking Feet

https://youtu.be/PKMpTS6b988

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.

@@ Line 7: / Line 7: @@
 ==Solution==
-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>.
+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{(A)}}</math>.
-~mrtnvlknv
+~Sigmacuber
 ==Solution 2==
@@ Line 22: / Line 22: @@
 ~alwaysgonnagiveyouup
+==Video Solution by Thinking Feet==
+https://youtu.be/PKMpTS6b988
+==See Also==
+{{AMC8 box|year=2025|num-b=12|num-a=14}}
+{{MAA Notice}}

Page

Toolbox

Search