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Difference between revisions of "1985 AHSME Problems/Problem 18"
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==Problem==
 
==Problem==
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Six bags of marbles contain <math> 18, 19, 21, 23, 25, </math> and <math> 34 </math> marbles, respectively. One bag contains chipped marbles only. The other <math> 5 </math> bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?
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Six bags of marbles contain <math> 18, 19, 21, 23, 25 </math> and <math> 34 </math> marbles, respectively. One bag contains chipped marbles only. The other <math> 5 </math> bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?
  
 
<math> \mathrm{(A)\ } 18 \qquad \mathrm{(B) \ }19 \qquad \mathrm{(C) \  } 21 \qquad \mathrm{(D) \  } 23 \qquad \mathrm{(E) \  }25 </math>
 
<math> \mathrm{(A)\ } 18 \qquad \mathrm{(B) \ }19 \qquad \mathrm{(C) \  } 21 \qquad \mathrm{(D) \  } 23 \qquad \mathrm{(E) \  }25 </math>
  
 
==Solution==
 
==Solution==
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Let the number of marbles George has be <math> x </math>, and so the number of marbles Jane has is <math> 2x </math>. Therefore, the total number of non-chipped marbles is <math> 3x\equiv0(\text{mod }3) </math>. However, the total number of marbles is <math> 18+19+21+23+25+34=140\equiv2(\text{mod }3) </math>. Therefore, we need a number of chipped marbles <math> \equiv2(\text{mod }3) </math> to get a number of non-chipped marbles <math> \equiv0(\text{mod }3) </math>.
+
Let the number of marbles George has be <math> x </math>, and so the number of marbles Jane has is <math> 2x </math>. Therefore, the total number of non-chipped marbles is <math> 3x\equiv0 \ (\text{mod }3) </math>. However, the total number of marbles is <math> 18+19+21+23+25+34=140\equiv2 \ (\text{mod }3) </math>. Therefore, we need a number of chipped marbles <math> \equiv2 \ (\text{mod }3) </math> to get a number of non-chipped marbles <math> \equiv0 \ (\text{mod }3) </math>.
  
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<math> 18\equiv0(\text{mod }3) </math>
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<math> 18\equiv0 \ (\text{mod }3) </math>
  
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<math> 19\equiv1(\text{mod }3) </math>
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<math> 19\equiv1 \ (\text{mod }3) </math>
  
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<math> 21\equiv0(\text{mod }3) </math>
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<math> 21\equiv0 \ (\text{mod }3) </math>
  
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<math> 23\equiv2(\text{mod }3) </math>
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<math> 23\equiv2 \ (\text{mod }3) </math>
  
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<math> 25\equiv1(\text{mod }3) </math>
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<math> 25\equiv1 \ (\text{mod }3) </math>
  
−
<math> 34\equiv1(\text{mod }3) </math>
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<math> 34\equiv1 \ (\text{mod }3) </math>
  
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Since <math> 23 </math> is the only one <math> \equiv2(\text{mod }3) </math>, it is the only possible number of chipped marbles, <math> \boxed{\text{D}} </math>.
+
Since <math> 23 </math> is the only one <math> \equiv2 \ (\text{mod }3) </math>, it is the only possible number of chipped marbles, <math> \boxed{\text{D}} </math>.
  
 
To check, we can see that if Jane takes the <math> 19, 25, </math> and <math> 34 </math> marble bags and George takes the <math> 18 </math> and <math> 21 </math> marble bags, this satisfies the conditions.
 
To check, we can see that if Jane takes the <math> 19, 25, </math> and <math> 34 </math> marble bags and George takes the <math> 18 </math> and <math> 21 </math> marble bags, this satisfies the conditions.

Revision as of 00:00, 3 April 2018

Problem

Six bags of marbles contain $18, 19, 21, 23, 25$ and $34$ marbles, respectively. One bag contains chipped marbles only. The other $5$ bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?

$\mathrm{(A)\ } 18 \qquad \mathrm{(B) \ }19 \qquad \mathrm{(C) \ } 21 \qquad \mathrm{(D) \ } 23 \qquad \mathrm{(E) \ }25$

Solution

Let the number of marbles George has be $x$, and so the number of marbles Jane has is $2x$. Therefore, the total number of non-chipped marbles is $3x\equiv0 \ (\text{mod }3)$. However, the total number of marbles is $18+19+21+23+25+34=140\equiv2 \ (\text{mod }3)$. Therefore, we need a number of chipped marbles $\equiv2 \ (\text{mod }3)$ to get a number of non-chipped marbles $\equiv0 \ (\text{mod }3)$.

$18\equiv0 \ (\text{mod }3)$

$19\equiv1 \ (\text{mod }3)$

$21\equiv0 \ (\text{mod }3)$

$23\equiv2 \ (\text{mod }3)$

$25\equiv1 \ (\text{mod }3)$

$34\equiv1 \ (\text{mod }3)$

Since $23$ is the only one $\equiv2 \ (\text{mod }3)$, it is the only possible number of chipped marbles, $\boxed{\text{D}}$.

To check, we can see that if Jane takes the $19, 25,$ and $34$ marble bags and George takes the $18$ and $21$ marble bags, this satisfies the conditions.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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