Section 6.3
179
23. [BB] After finitely many steps in the long division process, we will be adjoining a "0" at each stage to
the remainder from the division at the previous stage. Since only finitely many remainders are possible
(each remainder is less than the divisor), two must be the same and from the first repetition, all steps
(and corresponding decimal places) will repeat.
24. Let box 1 consist of the numbers 1,2,4,8,
... ,128; let box 3 consist of 3,6,12, .
.. ,(64)3, box 5
consist of 5,10,20,
... ,32(5), and so on; box 199 consists of 199 alone. (In general, for odd i, box i
consists of all numbers of the form
2ki
which do not exceed 200. There are 100 boxes in all.)
If
the
101 chosen numbers are al,
a2,
... ,a1Ol and
ai
is assigned to the box which contains it (by the hint,
every number belongs to some box), by the PigeonHole Principle, some box contains two numbers.
Since, for any two numbers in a box, one is a multiple of the other, we have the desired result.
25. (a) [BB] Suppose no two people have the same age so that there are at least 51 different ages in the
room. Let box 1 correspond to integers 1 and 2, box 2 correspond to integers 3 and 4, .
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 Summer '10
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 Graph Theory, Division, Remainder, Empty set, Natural number, largest possible sum, PigeonHole Principle

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