$35 \cdot 25$ Six bags contain $18, 19, 21, 23, 25$ and 34 marbles, respectively, one of which has all its marbles cracked, while the other five bags do not have such cracked marbles. Jenny took three of the bags, and George took two of the bags, leaving only the bag with the cracked marbles. If Jenny got exactly twice as many marbles as George, the number of marbles with cracks is
(A) 18.
(B) 19.
(C) 21.
(D) 23.
(E) 25.
(36th American High School Mathematics Examination, 1985)
［Solution］Assuming Jenny has twice as many balls as George, then the total number of balls they have should be a multiple of 3.
Also, $(18+19+21+23+25+34) \div 3=46 \cdots \cdots 2$,
thus the number of balls with a crack should be the one that leaves a remainder of 2 when divided by 3. Among the given numbers, only 23 has this property.
Therefore, the answer is $(D)$.