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

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==Problem==
 
==Problem==
  
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?  
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>
smallest number of coins that meets these two conditions, how
 
many coins are left when equally divided among seven people?
 
(A) 0 (B) 1 (C) 2 (D) 3 (E) 5
 
  
 
==Solution==
 
==Solution==
 
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===Solution 1===
(A) The counting numbers that leave a remainder of 4 when divided by 6 are
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The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are
4; 10; 16; 22; 28; 34;...; etc. The counting numbers that leave a remainder of 3 when
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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
divided by 5 are 3; 8; 13; 18; 23; 28; 33;...; etc. So 28 is the smallest possible number
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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
of coins that meets both conditions. Because 4 £ 7 = 28, there are no 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.
  
OR
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===Solution 2===
  
 
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
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
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>.
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==Video Solution==
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==Video Solution==
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https://www.youtube.com/watch?v=dQw4w9WgXcQ
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-Happytwin
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https://www.youtube.com/watch?v=uMBev3FUoTs  ~David
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==See Also==
 +
{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Latest revision as of 08:23, 18 June 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

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
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

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