1985 AHSME Problems/Problem 18

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 $\boxed{\text{(D)} \ 23}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1985 AHSME Problems/Problem 18

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