Number Series

If S is a series of x consecutive odd integers and Z is the larget integer in S, what is the smallest integer in S.

Let the first number be 2n+1,
the series will be like
2n+1
2n+3
2n+5
:
:
Z

numbers 1,3,5,.....are in A.P. to calculate last number use [last number = first number + common difference (number of terms -1)]
L = 1 + 2(x-1) = 2x-1

so we get Z = 2n + 2x-1
Z+1 = (2n+1) +2x-1..............adding 1 on both sides
2n+1 = Z+1-2x+1 = Z-2x+2 = Z-2(x-1)

Answer C

Thank you for this explanation. Looks easier when it has been layed out logically. Was going round in circles and had to post something. thanks for all the responses thus far. I prefer to solve with algebra than plugging numbers where possible

Agreed .But plugging in numbers are pretty easy.

take ,5,7,9.....

HereZ=9,x=3

We know that our smallest number is 5.

Plug in the values of Z & X.

1.Z-2x
--->9- 2*3=3( Not 5, which is required)

2.Z-x+1
---> 9-3+1=7(Not the answer we want)

3.Z-2(x-1)
--->9-2(3-1)=5(Bingo! we got answer smallest integer)

ahinye wrote:If S is a series of x consecutive odd integers and Z is the largest integer in S, what is the smallest integer in S.

1) Z-2x
2)Z-x+1
3)Z-2(x-1)
4)x-6+Z
5)Z- (x/2)


How can I solve this algebraically?

It is a very simple question of AP
where a= S, Tn= Z and n c.d. = 2 (for odd numbers)
thus Z= S+(x-1)2 --> [spoiler]S= Z-2(x-1) Ans3[/spoiler]

gmatmachoman wrote:Agreed .But plugging in numbers are pretty easy.

take ,5,7,9.....

HereZ=9,x=3

We know that our smallest number is 5.

Plug in the values of Z & X.

1.Z-2x
--->9- 2*3=3( Not 5, which is required)

2.Z-x+1
---> 9-3+1=7(Not the answer we want)

3.Z-2(x-1)
--->9-2(3-1)=5(Bingo! we got answer smallest integer)

I don't think plugin is at all required here. Just go with simple concept of AP and u will get ans in seconds.. Though plug in really helps a lot many a times..

ahinye wrote:Thank you for this explanation. Looks easier when it has been layed out logically. Was going round in circles and had to post something. thanks for all the responses thus far. I prefer to solve with algebra than plugging numbers where possible

Well,I would say it should be decided upon seeing the question which technique to use:-In some cases that I have personally found,solving by the formal proper approach takes too much time and even the calculation process is very cumbersome.

But yeah,ofcourse,in case of even 10% doubt I would say adopt the formal approach.It is better not to change the strategy on the exam day.

It is better to start practising these techniques on a day-to-day basis so that the shortcut approaches becomes second nature to us.

In this question,going by the soln. that brent has posted,i guess it takes equal time by using both the techniques.
So any approach will do fine.

I think for many questions its a personal choice wether to use algebra or plug in numbers. For this problem, I would use algebra:
We have consecutive odd numbers "X", with largest number "Z"
Now, for evenly spaced sets,
The number of terms is calculated as:
X = (Z - L)/n + 1
n = common difference/spacing ( for Odd/even number n = 2)
L = SMALLEST ODD NUMBER OF THE SET,
Z = LARGEST ODD NUMBER OF THE SET
X = (Z - L)/2 + 1
Thus,
2X = Z - L + 2
Rearranging,
L = Z - 2X - 2
L = Z - 2(X - 1)