Question

Question coming.

Answer 1

I can't believe it!!! You really have a doubt?

Answer 2

Lol @pratu043 .

Answer 3

odd times odd times even

Answer 4

Assume w as 5 then try to solve..

Answer 5

@waterineyes , why 5?

Answer 6

there is a typo in the question and it is damn hard to read

Answer 7

Let the w = 2k+1
Then we need the product of (2k+2)(2k-1)
Notice that from the two numbers, the larger one is even and is greater than the odd number by 3.
Now all you have to do is factor the options, and check if the larger factor is even and is greater than the odd factor by 3 or not

Answer 8

Okay..

Assume w as 1 3 5 7 and 9 then try to solve..

Answer 9

@UnkleRhaukus . I didn't typed it. :D

Answer 10

maybe you should have , then we could read it

Answer 11

2k + 1 ?
How? @FoolAroundMath

@waterineyes , ok. just a sec.

Answer 12

any odd number can be written as an even number + 1 and since 2k is even (if k is a natural number) so consequently, 2k+1 is odd (k is a whole number)

Answer 13

wait the product is just of two terms , one odd one even

Answer 14

If w is +ve odd integer, which of the following gives a possible value of the product value of one 1 than w and 2 less than w.

Answer 15

@UnkleRhaukus ^

Answer 16

(w+1)(w-2)=

Answer 17

cannot be even

Answer 18

(w+1)(w-2) = sth
If = 0
w^2 - x -2 =0
w = 2, or w=-1 (rejected)

if = 10
w^2 - x -2 =10
w = 4 or w =-3 (rejected)

if = 18
w^2 - x -2 =18
w = 5 or w =-4 => YES!

If = 28
w^2 - x -2 =28
w = 6 or w = -5 (rejected)

If = 54
w^2 - x -2 =54
w = 8 or w =07 (rejected)

Answer 19

w = 5

Answer 20

How can be w = 2??

Answer 21

w= -7 instead of w=07 for the last one sorry..

Answer 22

w is an odd positive integer..

Answer 23

o dear

Answer 24

So, only 18 is the possible answer! @waterineyes

Answer 25

yes..
18 is the possible answer for w = 5..

Answer 26

* sth = something..

Answer 27

Perfect guys. Thanks all.
I still owe a medal to @waterineyes , @Callisto and @FoolAroundMath . (:

Answer 28

take the medal off me i made to many mistakes

Answer 29

@UnkleRhaukus , atleast you tried. :D

Answer 30

See, w is an odd integer so it can be 1 3 5 7 9 and so on..

For w = 1,

(1 + 1)(1 - 2) = Negative (Ignore)

For w = 3,

(3 + 1)(3 - 2) = 4 (Not in option)

For w = 5,

(5 + 1)(5 - 2) = 18 (Right)

For w = 7,

(7 + 1)(7 - 2) = 40 (Not in option)

For w = 9,

(9 + 1)(9 - 2) = 70 (Not in option)



So you are left with w = 5 and the product possible is : 18...

Answer 31

saif....2k+1=2001

Answer 32

providing a formula for all numbers in that way
w=2k+1
(w+1)(w-2)=w^2-w-2=(2k+1)^2-(2k+1)-2=2(2k^2+k-1)

k w 2(2k^2+k-1)

0 1 -2
1 3 4
2 5 18 <---
3 7 40
...

Answer 33

another way of solving this:

since w is odd you can also write it as 2k - 1 (I use minus 1 here instead of plus 1 to take advantage of what the question says "1 more than w..." which will then get rid of this -1).

so, we have:

w = 2k - 1 for k=1,2,...
(w+1)(w-2) = 2k(2k - 3) => product is even.

so lets divide this and all the given options by 2 to get k(2k-3) which means we need a number that has these two as factors:
k(2k - 3) = 0 [k=0 is not valid --- reject]
= 5 = 1 x 5 [k=1 => 2k-3=-1 --- reject]
= 9 = 3 x 3 [k=3 => 2k-3=3 --- possible] <<<--- answer
= 14 = 2 x 7 [k=2 => 2k-3=1 --- reject]
= 27 = 3 x 9 [k=3 => 2k-3=3 --- reject]

so solution is when k=3, which gives a product of 18 (2 x 9)