Question Sam Loyd's Cyclopedia of Puzzles Answer
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Mrs. Hubbard has invented a clever system for keeping tabs on her blackberry jam. She filled twenty-five jars and arranged the three sizes so as to have twenty quarts on each shelf. Can you guess her secret so as to tell how much one of the big jars contains?

Alfred Mercier says: “What we learn with pleasure we never forget,” which is a more elegant way of expressing Josh Billings’ trite saying: “There are better ways of knocking learning into a boy’s bead than with a wormwood club.” Tommy would like to learn how to extract unknown quantities from those jars of jam by reduction, elimination or even by the process of substitution of empty jars for full ones. The whole juvenile class would speedily reduce everything to the minimum quantities and clear off fractions if they were not awed by the mother’s radical terms.

Like good Mother Hubbard we will solve the problem by inspection, and prove the quantities in the different jars. Knowing that each shelf contains just 20 quarts, let us begin by cancelling off six little jars from the two lower shelves. The result proves that two big jars equal four medium ones, or one large one equals two medium size. Replace the jars and cancel the two large ones from the middle shelf and equalize the top shelf by removing the large one and two of the medium size. This shows that the one medium sized jar must hold as much as three little ones. Now multiply all the large jars by two and they are changed to mediums and multiply the number then representing all the mediums by three to reduce them to the smallest size, and when we add them all together we find that the entire amount could be contained in 54 of the small size, 18 of the medium, or 9 of the largest. As a large jar would contain one-ninth of 60 quarts, we see that it would hold just six and two-third quarts.

2. A Rebus

My first on my second often there lies,
My first and my last should be of a size;
My whole, like my second, is always the same.
So now, if you’re smart, just tell me my name.

Cipher Answer. ”6, 15, 15, 20, 19, 20, 15, 15, 12.


3. A Charade

My first is found in many mines;
And there my costly second shines.
As for my whole”what shall I say?
It seems intended to betray.
Then, oh! Beware, unthinking youth,
Adhere to honesty and truth.

Cipher Answer.”19, 20, 18, 1, 20, 5, 7, 5, 13.


4. Election Puzzle

Here is a simple but somewhat pretty problem which developed at a recent election where 5,219 votes were cast for four candidates. The victor exceeded his opponents by 22, 30 and 73 votes, and yet not one of them knew how to figure out the exact number of votes received by each.

Can you give a simple rule for giving the desired results?

In the election puzzle add the pluralities to the total vote and divide by the number of candidates. The quotient will be the vote of the successful one, from which the votes of the others can be ascertained by subtraction. The counts were 1,336, 1,314, 1,306 and 1,263.

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