Number of Cakes At Each Price

Here is a problem by J.A.H. Hunter.

'I put your cakes in the kitchen,' said Sam. 'Three different sorts, 12 cents, 14 cents, and 17 cents. Just two dollars altogether.' 'That's fine,' his mother declared. 'How many did you buy?' The boy told the total number, and his mother went on reading. But some moments later she stopped, and then did some figuring on a scrap of paper. 'I still can't tell how many you got of each,' she said. 'Did you buy only one of one sort?' Sam answered her question. Only one word, but enough to clear up any doubts as to the details of his purchase. How many had he bought at each price?

Please give the unique solution and your argument it is unique. Give also the one word Sam answered his mother's question.

Peter Heichelheim

Possible Spoiler Below

To buy cakes at 12¢, 14¢, and 17¢ and have them total $2.00, there are five combinations:

If telling her the total didn't help her figure out which combination he had bought, then he couldn't have bought 15 or 13, since each of those have only one combination. He must have bought 14 cakes.

'Did you buy only one of one sort?'

If he had bought either of the first two combinations, he would have had to answer 'Yes' (assuming he is neither a Hellian nor a Limboian) and would have left her in doubt as to exactly what combination he had purchased. Therefore, if his answer resolved her doubts, he must have purchased the third combination.

He bought four at 12¢, six at 14¢, and four at 17¢.

SPOILER

If there are x cakes at 12 cents, y at 14 cents, and z at 17 cents, then the only integer solutions to 12*x+14*y+17*z = 200 are:

x y z x+y+z 1 11 2 14 3 2 8 13 4 6 4 14 7 1 6 14 8 5 2 15

Out of these 5 solutions 3 have the same value for z+y+z - 14. The other 2 are unique. The mother could not figure out the individual numbers, so the total number of cakes must be 14.

From the 3 solutions two have 1 for one of the cakes. Therefore Sam's answer was 'no,' and based on this his mother could tell the numbers of cakes - 4 at 12 cents, 6 at 14 cents, and 4 at 17 cents.

Both of you got J.A.H. Hunter's answer.

Peter

I was wondering