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

Latest revision as of 23:24, 8 September 2025

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?

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

Soluiton 1

This is a modular arithmetic problem.

Let the number of coins be $N$ . Then:

$N \equiv 4 \pmod{6}$

$N \equiv 3 \pmod{5}$

Solving this system, the smallest such $N$ is:

$N = 28$

Now, dividing 28 by 7:

$28 \div 7 = 4 \text{ with a remainder of } \boxed{0}$

This is the number of coins left when the box is divided equally among seven people.

Solution 2

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 3

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 1

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

Video Solution 2 by WhyMath

https://youtu.be/-GteVuETb14

@@ Line 1: / Line 1: @@
 ==Problem==
-A box contains gold coins. If the coins are equally divided among
+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
-among ¯ve 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?
-(A) 0 (B) 1 (C) 2 (D) 3 (E) 5
-==Solution==
+<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 </math>
-(A) The counting numbers that leave a remainder of 4 when divided by 6 are
+==Soluiton 1==
-; 10; 16; 22; 28; 34; : : :. The counting numbers that leave a remainder of 3 when
+This is a '''modular arithmetic''' problem.
-divided by 5 are 3; 8; 13; 18; 23; 28; 33; : : :. So 28 is the smallest possible number
-of coins that meets both conditions. Because 4 £ 7 = 28, there are no coins left
+Let the number of coins be <math>N</math>. Then:
+<cmath>
+N \equiv 4 \pmod{6}
+</cmath>
+<cmath>
+N \equiv 3 \pmod{5}
+</cmath>
+Solving this system, the smallest such <math>N</math> is:
+<cmath>
+N = 28
+</cmath>
+Now, dividing 28 by 7:
+<cmath>
+\div 7 = 4 \text{ with a remainder of } \boxed{0}
+</cmath>
+This is the number of coins left when the box is divided equally among seven people.
+==Solution 2==
+The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are
+<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
+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.
-OR
+==Solution 3==
 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
+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.
+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 1==
+https://www.youtube.com/watch?v=uMBev3FUoTs   ~David
+==Video Solution 2 by WhyMath==
+https://youtu.be/-GteVuETb14
+==See Also==
+{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}
+{{MAA Notice}}
+[[Category:Introductory Number Theory Problems]]

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