%  Challenge questions, Friday, February 5, 2010
%  21-110: Problem Solving in Recreational Mathematics, spring 2010
%  Brian Kell <bkell@cmu.edu>

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\centerline{\titlefont 21-110: Problem Solving in Recreational Mathematics}
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\centerline{Challenge questions: Friday, February~5, 2010}

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\question In the course of adding two numbers, sometimes it is necessary to
carry a~1. Explain why it is never necessary to carry a~2.

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\question In the course of multiplying two numbers, sometimes it is necessary
to carry. Explain why it is never necessary to carry a~9.

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\question (Problem~1 from {\it One Hundred Problems in Elementary
Mathematics\/} by Hugo Steinhaus.)\enspace We construct a sequence of numbers
as follows: The first number is~2, the next is~3,
$$2\times3=6,$$
the third number of the sequence is~6,
$$3\times6=18,$$
the fourth number is~1, and the fifth is~8,
$$6\times1=6,\qquad1\times8=8,$$
the sixth number is~6, then follows~8, etc.

\def\uu{\mathrel{\lower1ex\hbox{$\smile$}}}

This is the sequence which we obtain:
$$2\uu3\uu6\uu1\uu8\uu6\uu8\ldots$$

The little arcs under the numbers denote the multiplication carried out, the
result of which follows the last digit of the sequence. For example, we ought
to multiply now 6~by~8 and write down the numbers of the result, namely~4,~8.
There will never be a shortage of numbers for multiplication, since the number
of arcs is increased by one with each multiplication and the result will yield
at least one and often two digits, so that there always appears at least one
new digit.

Prove that numbers 5,~7 and~9 never appear in this sequence.

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