54 Arithmetic & Algebraic Problems
(2) If a man lives on the even side and all the numbers on one side of him
equal those on the other side, how many houses are there, and what is his
number?
We assume that there are more than fifty houses on each side of the street
and fewer than five hundred.
- CORRECTING AN ERROR
Hilda Wilson was given a certain number to multiply by 409, but she made
a blunder that is very common with children when learning the elements of
simple arithmetic: she placed the first figure of her product by 4 below the
second figure from the right instead of below the third. We have all done that
as youngsters when there has happened to be a 0 in the multiplier. The result
of Hilda's mistake was that her answer was wrong by 328,320, entirely
in consequence of that little slip. What was the multiplicand-the number she
was given to multiply by 409?
- THE SEVENTEEN HORSES
"I suppose you all know this old puzzle," said Jeffries. "A farmer left
seventeen horses to be divided among his three sons in the following propor-
tions: To the eldest, one-half; to the second, one-third; and to the youngest,
one-ninth. How should they be divided?"
"Yes; I think we all know that," said Robinson, "but it can't be done. The
answer always given is a fallacy."
"I suppose you mean," Prodgers suggested, "the answer where they borrow
another chap's horse to make eighteen for the purpose of the division. Then
the three sons take 9, 6, and 2 respectively, and return the borrowed horse to
the lender."
"Exactly," replied Robinson, "and each son receives more than his share."
"Stop!" cried Benson. "That can't be right. If each man received more
than his share the total must exceed seventeen horses, but 9, 6, and 2 add up
to 17 correctly."
"At first sight that certainly looks queer," Robinson admitted, "but the
explanation is that if each man received his true fractional snare, these frac-
tions would add up to less than seventeen. In fact, there would be a fraction
left undistributed. The thing can't really be done."