1985 AMC 12 Problem 18

Question

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?

18

19

21

23

25

Answer: D

Concepts:divisibility modular arithmetic partitions and compositions

Difficulty rating: 1790

Small hint:

Add all six bag sizes, then remove the chipped bag

Big hint:

Jane’s and George’s totals are in the ratio

2:1,

so their combined total is divisible by

3

Solution:

The six bags total

140.

If the chipped bag contains

c

marbles, then the other bags total three times George’s amount, so

140-c

is divisible by

3.

Of the choices, only

c=23

has this property. It is attainable: George can take

18+21=39,

while Jane takes

19+25+34=78.

Thus the correct answer is D.