Difference between revisions of "1986 AJHSME Problems/Problem 17"

Latest revision as of 14:02, 4 April 2021

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

Solution 1

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.

$\boxed{\text{E}}$

Solution 2

We are given that $\text{o}\equiv 1\pmod{2}$ , so in mod $2$ we have $1^2+1(n) = n+1$ which is odd only if $\text{n}$ is even $\rightarrow \boxed{\text{E}}$

Solution 3 (easiest)

To make this problem simpler, we can assume a number to replace $\text{O}$ and $\text{N}$ . Let $\text{O}$ be $1$ and $\text{N}$ be $2$ . When we compute $(\text{o}^2+\text{no})$ , we get $1+2=3$ . We immediately rule out $\text{(B)}$ , $\text{(D)}$ , and $\text{(C)}$ . The only options left are $\text{(A)}$ and $\text{(E)}$ . This time let's assume $\text{O}$ is $2$ , and $\text{N}$ is $4$ . $(\text{o}^2+\text{no})$ comes out to be $4+8=12$ . $12$ isn't odd, so we cross out $\text{(A)}$ . Thus, the answer is $\boxed{\text{E}}$

~sakshamsethi

@@ Line 11: / Line 11: @@
 ==Solution==
-{{Solution}}
+===Solution 1===
+ We can solve this problem using logic.
+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.
+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.
+<math>\boxed{\text{E}}</math>
+===Solution 2===
+We are given that <math>\text{o}\equiv 1\pmod{2}</math>, so in mod <math>2</math> we have <cmath>1^2+1(n) = n+1</cmath> which is odd only if <math>\text{n}</math> is even <math>\rightarrow \boxed{\text{E}}</math>
+==Solution 3 (easiest)==
+To make this problem simpler, we can assume a number to replace <math>\text{O}</math> and <math>\text{N}</math>. Let <math>\text{O}</math> be <math>1</math> and <math>\text{N}</math> be <math>2</math>. When we compute <math>(\text{o}^2+\text{no})</math>, we get <math>1+2=3</math>. We immediately rule out
+<math>\text{(B)}</math>, <math>\text{(D)}</math>, and <math>\text{(C)}</math>. The only options left are <math>\text{(A)}</math> and <math>\text{(E)}</math>. This time let's
+assume <math>\text{O}</math> is <math>2</math>, and <math>\text{N}</math> is <math>4</math>. <math>(\text{o}^2+\text{no})</math> comes out to be <math>4+8=12</math>. <math>12</math> isn't odd, so we cross out <math>\text{(A)}</math>. Thus, the answer is <math>\boxed{\text{E}}</math>
+~sakshamsethi
 ==See Also==
-[[1986 AJHSME Problems]]
+{{AJHSME box|year=1986|num-b=16|num-a=18}}
+[[Category:Introductory Number Theory Problems]]
+{{MAA Notice}}

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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
All AJHSME/AMC 8 Problems and Solutions

Page

Toolbox

Search