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Difference between revisions of "1961 AHSME Problems/Problem 4"

(Created page with 'Let the set consisting of the squares of the positive integers be called ''u''; thus ''u'' is the set 1,4,9... . If a certain operation on one or more members of the set always y…')
 
(Solution to Problem 4)
 
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Let the set consisting of the squares of the positive integers be called ''u''; thus ''u'' is the set 1,4,9... . If a certain operation on one or more members of the set always yields a member of the set, we say that the set is closed under that operation. Then ''u'' is closed under:
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== Problem ==
  
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(A) addition  (B) multiplication  (C) division 
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Let the set consisting of the squares of the positive integers be called <math>u</math>; thus <math>u</math> is the set <math>1, 4, 9, 16 \ldots</math>.
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(D) extraction of a positive integral root (E) none of these
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If a certain operation on one or more members of the set always yields a member of the set,
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we say that the set is closed under that operation. Then <math>u</math> is closed under:
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<math>\textbf{(A)}\ \text{Addition}\qquad
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\textbf{(B)}\ \text{Multiplication} \qquad
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\textbf{(C)}\ \text{Division} \qquad\\
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\textbf{(D)}\ \text{Extraction of a positive integral root}\qquad
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\textbf{(E)}\ \text{None of these}  </math>
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==Solution==
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Consider each option, case by case.
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For option A, note that <math>1 + 4 = 5</math>, so the set is not closed under addition because <math>5</math> is not a perfect square.
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For option B, note that <math>9 \cdot 4 = 36</math> and <math>4 \cdot 25 = 100</math>.  Letting one member of the set be <math>a^2</math> and another member be <math>b^2</math> (where <math>a</math> and <math>b</math> are positive integers), the product of the two members is <math>(ab)^2</math>.  Since <math>ab</math> is an integer and <math>(ab)^2</math> is a perfect square, the set is closed under multiplication.
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For option C, note that <math>9 \div 4 = 2.25</math>, so the set is not closed under division because <math>2.25</math> is not an integer.
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For option D, note that <math>\sqrt{4} = 2</math>, so the set is not closed under extraction of a positive integral root because <math>2</math> is not a perfect square.
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Thus, the answer is <math>\boxed{\textbf{(B)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1961|num-b=3|num-a=5}}
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{{MAA Notice}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 18:47, 17 May 2018

Problem

Let the set consisting of the squares of the positive integers be called $u$; thus $u$ is the set $1, 4, 9, 16 \ldots$. If a certain operation on one or more members of the set always yields a member of the set, we say that the set is closed under that operation. Then $u$ is closed under:

$\textbf{(A)}\ \text{Addition}\qquad \textbf{(B)}\ \text{Multiplication} \qquad \textbf{(C)}\ \text{Division} \qquad\\ \textbf{(D)}\ \text{Extraction of a positive integral root}\qquad \textbf{(E)}\ \text{None of these}$

Solution

Consider each option, case by case.

For option A, note that $1 + 4 = 5$, so the set is not closed under addition because $5$ is not a perfect square.

For option B, note that $9 \cdot 4 = 36$ and $4 \cdot 25 = 100$. Letting one member of the set be $a^2$ and another member be $b^2$ (where $a$ and $b$ are positive integers), the product of the two members is $(ab)^2$. Since $ab$ is an integer and $(ab)^2$ is a perfect square, the set is closed under multiplication.

For option C, note that $9 \div 4 = 2.25$, so the set is not closed under division because $2.25$ is not an integer.

For option D, note that $\sqrt{4} = 2$, so the set is not closed under extraction of a positive integral root because $2$ is not a perfect square.

Thus, the answer is $\boxed{\textbf{(B)}}$.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions


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