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

Revision as of 18:31, 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 $\text{n}$ is odd. If $\text{n}$ is odd, then obviously $\text{no}$ will be odd as well, since $\text{o}$ is odd, and the product of two odd numbers is odd. Since $\text{o}$ is odd, $\text{o}^2$ will also be odd. And adding two odd numbers makes an even number, so if $\text{n}$ is odd, the entire expression is even.

Let's say that $\text{n}$ is even. If $\text{n}$ is even, then $\text{no}$ will be even as well, because the product of an odd and an even is even. $\text{o}^2$ will still be odd. That means that the entire expression will be odd, since the sum of an odd and an even is odd.

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

$\boxed{\text{E}}$

See Also

1986 AJHSME Problems

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