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Difference between revisions of "1986 AJHSME Problems/Problem 17"
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==Solution==
 
==Solution==
  
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{{Solution}}
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We can solve this problem using logic.
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Let's say that <math>n</math> is odd. If <math>n</math> is odd, then obviously <math>no</math> will be odd as well, since <math>o</math> is odd, and odd times odd is odd. Since <math>o</math> is odd, <math>o^2</math> will also be odd, because <math>o^2 = oo</math>, and odd times odd is odd. And adding two odd numbers makes an even number, so if <math>n</math> is odd, the entire expression is even.
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Let's say that <math>n</math> is even. If <math>n</math> is even, then <math>no</math> will be even as well, because odd times even is even. <math>o^2</math> will still be odd. That means that the entire expression will be odd, since odd + even = odd.
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Looking at the multiple choices, we see that our second case fits choice E exactly.
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E
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 19:24, 24 January 2009

Problem

Let $\text{o}$ be an odd whole number and let $\text{n}$ be any whole number. Which of the following statements about the whole number $(\text{o}^2+\text{no})$ is always true?

$\text{(A)}\ \text{it is always odd} \qquad \text{(B)}\ \text{it is always even}$

$\text{(C)}\ \text{it is even only if n is even} \qquad \text{(D)}\ \text{it is odd only if n is odd}$

$\text{(E)}\ \text{it is odd only if n is even}$

Solution

We can solve this problem using logic.

Let's say that $n$ is odd. If $n$ is odd, then obviously $no$ will be odd as well, since $o$ is odd, and odd times odd is odd. Since $o$ is odd, $o^2$ will also be odd, because $o^2 = oo$, and odd times odd is odd. And adding two odd numbers makes an even number, so if $n$ is odd, the entire expression is even.

Let's say that $n$ is even. If $n$ is even, then $no$ will be even as well, because odd times even is even. $o^2$ will still be odd. That means that the entire expression will be odd, since odd + even = odd.

Looking at the multiple choices, we see that our second case fits choice E exactly.

E

See Also

1986 AJHSME Problems

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