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Difference between revisions of "2006 AMC 8 Problems/Problem 23"

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<math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of <math>3</math> when
 
<math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of <math>3</math> when
 
divided by <math>5</math> are <math>3,8,13,18,23,28,33, \cdots</math> So <math>28</math> is the smallest possible number
 
divided by <math>5</math> are <math>3,8,13,18,23,28,33, \cdots</math> So <math>28</math> is the smallest possible number
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of coins that meets both conditions. Because <math>4\cdot 7 = 28</math>, there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left
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of coins that meets both conditions. Because 28 is divisible by 7, there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left
 
when they are divided among seven people.
 
when they are divided among seven people.
  
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smallest possible number of coins in the box is <math>28</math> and the remainder when divided by <math>7</math> is <math>\boxed{\textbf{(A)}\ 0}</math>.
 
smallest possible number of coins in the box is <math>28</math> and the remainder when divided by <math>7</math> is <math>\boxed{\textbf{(A)}\ 0}</math>.
  
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===Solution 3===
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==Video Solution==
  
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We can set up a system of modular congruencies:
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==Video Solution==
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<cmath>g\equiv 4 \pmod{6}</cmath>
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https://www.youtube.com/watch?v=dQw4w9WgXcQ
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<cmath>g\equiv 3 \pmod{5}</cmath>
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-Happytwin
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We can use the division algorithm to say <math>g=6n+4</math> <math>\Rightarrow</math> <math>6n\equiv 4 \pmod{5}</math> <math>\Rightarrow</math> <math>n\equiv 4 \pmod{5}</math>. If we plug the division algorithm in again, we get <math>n=5q+4</math>. This means that <math>g=30q+28</math>, which means that <math>g\equiv 28 \pmod{30}</math>. From this, we can see that <math>28</math> is our smallest possible integer satisfying <math>g\equiv 28 \pmod{30}</math>. <math>28</math> <math>\div</math> <math>7=4</math>, making our remainder <math>0</math>. This means that there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left over when equally divided amongst <math>7</math> people.
 
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~Champion1234
 
  
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==Video Solution==
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https://www.youtube.com/watch?v=uMBev3FUoTs  ~David
  
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https://youtu.be/g1PLxYVZE_U
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==Video Solution by WhyMath==
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-Happytwin
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https://youtu.be/-GteVuETb14
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}
 
{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:10, 8 November 2024

Problem

A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5$

Solution

Solution 1

The counting numbers that leave a remainder of $4$ when divided by $6$ are $4, 10, 16, 22, 28, 34, \cdots$ The counting numbers that leave a remainder of $3$ when divided by $5$ are $3,8,13,18,23,28,33, \cdots$ So $28$ is the smallest possible number of coins that meets both conditions. Because 28 is divisible by 7, there are $\boxed{\textbf{(A)}\ 0}$ coins left when they are divided among seven people.

Solution 2

If there were two more coins in the box, the number of coins would be divisible by both $6$ and $5$. The smallest number that is divisible by $6$ and $5$ is $30$, so the smallest possible number of coins in the box is $28$ and the remainder when divided by $7$ is $\boxed{\textbf{(A)}\ 0}$.

Video Solution

Video Solution

https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin

https://www.youtube.com/watch?v=uMBev3FUoTs ~David

Video Solution by WhyMath

https://youtu.be/-GteVuETb14

See Also

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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