It was really easy, and can be found
here, under the heading
Next Week's Challenge.
From a 19th century trade card advertising Bassetts Horehound Troches, a remedy for coughs and colds: A man buys 20 pencils for 20 cents and gets three kinds of pencils in return. Some of the pencils cost 4 cents each, some are two for a penny and the rest are four for a penny. How many pencils of each type does the man get?
The facts can be related as
(the sum of all three kinds of pencils is 20) and
(the sum of the costs for the pencils is 20 cents). When you do elimination on these two equations, resulting in
for the last equation, there's still a free variable! But if you step through all possible values for
, you'll find that the only one where
and
are both integers (under the reasonable assumption that he can't have a fraction of any pencil) is two. If
, then
and
.
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