Question

If k is an even integer, which of the following must be an odd integer?

Answer 1

a. k + 2
b. 3k
c. 3k + 2
d. k(k - 1)
e. (k + 1) (k - 1)

Answer 2

how would i do this?

Answer 3

And even + an even is always an even

Answer 4

so c

Answer 5

That gets rid of one answer, you have to get rid of the rest

Answer 6

oh d

Answer 7

because an even times an even is always an even so u substract 1

Answer 8

it's E

Answer 9

One way to solve it is by assuming that k is 2, since 2 is even, and testing mit in all the equations

Answer 10

okay thank you

Answer 11

so all it has to do is end up odd right?

Answer 12

Yes

Answer 13

1 is odd and odd plus even is always odd

Answer 14

3 times an even number is also even

Answer 15

If odd is O and even is E, then
O + E = O
E + E = E
O + O = E
O x O = O
O x E = E
E x E = E

Answer 16

expand the last one
(k + 1)(k - 1) = k^2 - 1

even or odd? if k is even

Answer 17

me?

Answer 18

k^2 - 1 must be odd because k^2 will always be even

Answer 19

\[k ^{2}\] is even, but \[k ^{2}-1 \] is odd

Answer 20

the whole problem is odd right? because its zero if u solve it all

Answer 21

Need Help?
http://dotstorming.com/topic/570171aef45290eb0029bb89

Answer 22

\[2^{2}\] is 4