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Difference between revisions of "1985 AHSME Problems/Problem 18"

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==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>.
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Let George's bags contain a total of <math>x</math> marbles, so Jane's bag contains <math>2x</math> marbles. This means the total number of non-chipped marbles is <math>3x \equiv 0 \pmod{3}</math>, while the total number of marbles is <math>18+19+21+23+25+34 = 140 \equiv 2 \pmod{3}</math>, so the number of chipped marbles must also be congruent to <math>2-0 \equiv 2 \pmod{3}</math>.
  
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<math> 18\equiv0 \ (\text{mod }3) </math>
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The answer choices are congruent modulo 3 to <math>0</math>, <math>1</math>, <math>0</math>, <math>2</math>, and <math>1</math> respectively, so the only possible number of chipped marbles among these is <math>23</math>. Indeed, if Jane takes the bags containing <math>19</math>, <math>25</math>, and <math>34</math> marbles and George takes the remaining bags containing <math>18</math> and <math>21</math> marbles, then Jane will have a total of <math>19+25+34 = 78</math> marbles, which is twice as many as George's <math>18+21 = 39</math> marbles, as desired. Thus the answer is precisely <math>\boxed{\text{(D)} \ 23}</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> 23\equiv2 \ (\text{mod }3) </math>
 
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<math> 25\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>.
 
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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.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=17|num-a=19}}
 
{{AHSME box|year=1985|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:50, 19 March 2024

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 George's bags contain a total of $x$ marbles, so Jane's bag contains $2x$ marbles. This means the total number of non-chipped marbles is $3x \equiv 0 \pmod{3}$, while the total number of marbles is $18+19+21+23+25+34 = 140 \equiv 2 \pmod{3}$, so the number of chipped marbles must also be congruent to $2-0 \equiv 2 \pmod{3}$.

The answer choices are congruent modulo 3 to $0$, $1$, $0$, $2$, and $1$ respectively, so the only possible number of chipped marbles among these is $23$. Indeed, if Jane takes the bags containing $19$, $25$, and $34$ marbles and George takes the remaining bags containing $18$ and $21$ marbles, then Jane will have a total of $19+25+34 = 78$ marbles, which is twice as many as George's $18+21 = 39$ marbles, as desired. Thus the answer is precisely $\boxed{\text{(D)} \ 23}$.

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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