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

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==Problem==
 
==Problem==
  
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A box contains gold coins. If the coins are equally divided among
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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?  
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six people, four coins are left over. If the coins are equally divided
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among ¯ve people, three coins are left over. If the box holds the
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 </math>
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smallest number of coins that meets these two conditions, how
 
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many coins are left when equally divided among seven people?
 
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(A) 0 (B) 1 (C) 2 (D) 3 (E) 5
 
  
 
==Solution==
 
==Solution==
 +
===Solution 1===
 +
The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are
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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
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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 28 is divisible by 7, there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left
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when they are divided among seven people.
 +
 +
===Solution 2===
  
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(A) The counting numbers that leave a remainder of 4 when divided by 6 are
 
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4; 10; 16; 22; 28; 34; : : :. The counting numbers that leave a remainder of 3 when
 
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divided by 5 are 3; 8; 13; 18; 23; 28; 33; : : :. So 28 is the smallest possible number
 
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of coins that meets both conditions. Because 4 £ 7 = 28, there are no coins left
 
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when they are divided among seven people.
 
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OR
 
 
If there were two more coins in the box, the number of coins would be divisible
 
If there were two more coins in the box, the number of coins would be divisible
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by both 6 and 5. The smallest number that is divisible by 6 and 5 is 30, so the
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by both <math>6</math> and <math>5</math>. The smallest number that is divisible by <math>6</math> and <math>5</math> is <math>30</math>, so the
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smallest possible number of coins in the box is 28.
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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>.
 +
 
 +
==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==
 +
{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}
 +
{{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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