1990 AHSME Problems/Problem 29

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Problem

A subset of the integers $1,2,\cdots,100$ has the property that none of its members is 3 times another. What is the largest number of members such a subset can have?

$\text{(A) } 50\quad \text{(B) } 66\quad \text{(C) } 67\quad \text{(D) } 76\quad \text{(E) } 78$

Solution 1

Notice that inclusion of the integers between 34 to 100 inclusive is allowed as long as no integer between 11 and 33 inclusive is within the set. This provides a total of 100 - 34 + 1 = 67 solutions.

Further analyzing the remaining integers between 1 and 10, we notice that we can include all the numbers except 3 (as including 3 would force us to remove both 9 and 1) to obtain the maximum number of 9 solutions.

Thus, 67 + 9 = 76 $\fbox{D}$

Solution 2

Write down in a column the elements $x$ which are indivisible by three, and then follow each one by $3x, 9x, 27x, \ldots$

\[\begin{array}{ccccc}1&3&9&27&81\\2&6&18&54\\4&12&36\\5&15&45\\7&21&63\\8&24&72\\10&30&90\\11&33&99\\13&39\\\vdots&\vdots\end{array}\] We can take at most $3$ elements from the first row, and at most $2$ elements from each of the next seven rows. After that we can take only $1$ from any following row. Thus the answer is $3+7\cdot 2\,+$ the number of integers between $13$ and $100$ inclusive which are indivisible by three.

There are $\tfrac13(99-12)=29$ multiples of three in that range, so there are $88-29=59$ non-multiples, and $3+14+59=76$, which is $\fbox{D}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 29
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