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2006 AMC 8 Problems/Problem 23

Revision as of 19:15, 26 November 2023 by Golfrobot (talk | contribs) (→‎Solution 1)

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 $\boxed{\textbf{(A)}\ 0}$.

Solution 3

We can set up a system of modular congruencies: \[g\equiv 4 \pmod{6}\] \[g\equiv 3 \pmod{5}\] We can use the division algorithm to say $g=6n+4$ $\Rightarrow$ $6n\equiv 4 \pmod{5}$ $\Rightarrow$ $n\equiv 4 \pmod{5}$. If we plug the division algorithm in again, we get $n=5q+4$. This means that $g=30q+28$, which means that $g\equiv 28 \pmod{30}$. From this, we can see that $28$ is our smallest possible integer satisfying $g\equiv 28 \pmod{30}$. $28$ $\div$ $7=4$, making our remainder $0$. This means that there are $\boxed{\textbf{(A)}\ 0}$ coins left over when equally divided amongst $7$ people.

~Champion1234

Video Solution

https://youtu.be/Gxfjwxl3Sbo Soo, DRMS, NM

Video Solution

https://youtu.be/g1PLxYVZE_U -Happytwin

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

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

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