Difference between revisions of "2006 AMC 8 Problems/Problem 23"
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==Solution== | ==Solution== | ||
===Solution 1=== | ===Solution 1=== | ||
− | The counting numbers that leave a remainder of 4 when divided by 6 are | + | 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 3 when | + | <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 <math>3,8,13,18,23,28,33, \cdots</math> So 28 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 |
− | of coins that meets both conditions. Because | + | 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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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 <math>30</math>, 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 <math>28</math> and the remainder when divided by 7 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>. |
− | == | + | ==Video Solution== |
− | + | ==Video Solution== | |
− | + | https://www.youtube.com/watch?v=dQw4w9WgXcQ | |
− | + | -Happytwin | |
− | |||
− | ~ | + | https://www.youtube.com/watch?v=uMBev3FUoTs ~David |
==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 09:23, 18 June 2024
Contents
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?
Solution
Solution 1
The counting numbers that leave a remainder of when divided by
are
The counting numbers that leave a remainder of
when
divided by
are
So
is the smallest possible number
of coins that meets both conditions. Because 28 is divisible by 7, there are
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 and
. The smallest number that is divisible by
and
is
, so the
smallest possible number of coins in the box is
and the remainder when divided by
is
.
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 (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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