Difference between revisions of "1986 AJHSME Problems/Problem 17"
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===Solution 1=== | ===Solution 1=== | ||
− | can solve this problem using logic. | + | 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 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. |
Latest revision as of 15:02, 4 April 2021
Problem
Let be an odd whole number and let
be any whole number. Which of the following statements about the whole number
is always true?
Solution
Solution 1
We can solve this problem using logic.
Let's say that is odd. If
is odd, then obviously
will be odd as well, since
is odd, and the product of two odd numbers is odd. Since
is odd,
will also be odd. And adding two odd numbers makes an even number, so if
is odd, the entire expression is even.
Let's say that is even. If
is even, then
will be even as well, because the product of an odd and an even is even.
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.
Solution 2
We are given that , so in mod
we have
which is odd only if
is even
Solution 3 (easiest)
To make this problem simpler, we can assume a number to replace and
. Let
be
and
be
. When we compute
, we get
. We immediately rule out
,
, and
. The only options left are
and
. This time let's
assume
is
, and
is
.
comes out to be
.
isn't odd, so we cross out
. Thus, the answer is
~sakshamsethi
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.