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Consecutive integers -MGMAT

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by selango » Sun Jul 11, 2010 12:32 pm
x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

A. x = w

B. x > w

C. x/y is an integer

D. w/z is an integer

E. x/z is an integer

OA C
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by barcebal » Sun Jul 11, 2010 2:27 pm
This is a tough problem to solve algebraically. I just go through the answers, using plug and chug.

Consider each option:

A. x can equal w. Assume y is 2 (integers 2 and 3) which means that x=5. Based on that assumption for A to be possible z must equal 1. So is there one single consecutive integer that equals 5? Yes it is five.

B. x can be > w. Assume y is 2 (integers 2 and 3) which means that x=5. Based on that assumption for A to be possible z must equal 1. So is there one single consecutive integer (w) that is less than 5 (x)? Yes, there are multiple: 4, 3, 2, 1.

C. x/y cannot be an integer. Since y is double z, y must be an even number. This is key. For example if y were 3, z would be 1.5, which is not possible since z is a positive integer. So assume y is 2. In that case are there any two consecutive integers whose sum is divisible by 2. No. Two consecutive integers will always sum to an odd number (even+odd=odd). This means x must be an odd number and y must be an even number. No odd number can be divided by an even number and produce an integer. C is not possible, so C is the correct answer.

D. w/z can be an integer since z can be odd or even, which means that w can be odd or even. If z is odd, let's say 3, can we come up with three consecutive integers whose sum is divisible by 3? Yes. 1+2+3=6/3=2. In fact the sum of any three consecutive integers is divisible by 3.

E. x/z can be an integer. If z is 1, then it doesn't matter what x is, x/z=x, which is an integer.
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by singhpreet1 » Mon Jul 12, 2010 8:53 am
x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

A. x = w

B. x > w

C. x/y is an integer

D. w/z is an integer

E. x/z is an integer
barcebal wrote:This is a tough problem to solve algebraically. I just go through the answers, using plug and chug.

C. x/y cannot be an integer. Since y is double z, y must be an even number. This is key. For example if y were 3, z would be 1.5, which is not possible since z is a positive integer. So assume y is 2. In that case are there any two consecutive integers whose sum is divisible by 2. No. Two consecutive integers will always sum to an odd number (even+odd=odd). This means x must be an odd number and y must be an even number. No odd number can be divided by an even number and produce an integer. C is not possible, so C is the correct answer.


i think you are missing something here..if y=2z, then if y is 3, z will be 6 and not 1.5 also i fail to understand why y must be an even integer.

would highly appreciate additional responses.

Preet
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by barcebal » Mon Jul 12, 2010 9:02 am
i think you are missing something here..if y=2z, then if y is 3, z will be 6 and not 1.5 also i fail to understand why y must be an even integer.
z will not be 6 if y is 3. If y=3 then 3=(2)z and z=1.5. Another way to look at it is (1/2)*y=z or y/2=z

Since z must be an integer the only numbers divisible by 2 that create integers are even numbers so y must be even for y/2=z

Plug in any number of odds for y into the equation y=2z or y/2=z and you will see that you get a non-integer for z.

Hope this helps!
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by debmalya_dutta » Mon Jul 12, 2010 9:39 am
A. x = w
X= 1+2+3+4+5+6
Y= 55+56+57
So surely this is not necessary
B. x > w
based on above example, this is not necessary
C. x/y is an integer
say a is the first term..subsequent terms are a+1,a+2......,a+(Y-1)
X= Ya + (1+2+3+........+Y-1) = Ya + Y/2( 2+ Y-2)
X/Y= a + ½(Y) ...
We know that Y=2Z , this implies that Y for sure is even
So X/Y is an integer
D. w/z is an integer
say b is the first term..subsequent terms are b+1,b+2......,b+(Y-1)
W= Zb + (1+2+3+........+Z-1) = Zb + Z/2( 2+ Z-2)
W/Z= b + ½(Z) ... So whether X/Y is an integer depends on whether Z is even or not
Since Z can either be odd or even , W/Z may or may not be an integer
E. x/z is an integer
X/ Z from the above example is not an integer
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by barcebal » Mon Jul 12, 2010 11:49 am
debmalya_dutta wrote: C. x/y is an integer
say a is the first term..subsequent terms are a+1,a+2......,a+(Y-1)
X= Ya + (1+2+3+........+Y-1) = Ya + Y/2( 2+ Y-2)
X/Y= a + ½(Y) ...
We know that Y=2Z , this implies that Y for sure is even
So X/Y is an integer

If Y is even X/Y is ONLY even if X is even. Otherwise you have an odd/even cannot EVER be an integer.

So now the question is can X be even. The short answer is NO.

If x is even that means you will have the sum of an even number of consecutive integers.

Let's expound.

Say Y=2, this means that x=sum of 2 consecutive integers. Two consecutive integers means that you have an odd and an even. odd+even=odd. X is odd if y is even.

Say Y=4. Now we CAN get four consecutive integers to be even. odd+even+odd+even=even, but the 4 is too big in the denominator. 1+2+3+4=10/4 is not an integer.
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by ez2dj86 » Tue Jul 13, 2010 5:13 am
Since answer is provided, I will just delve on C using algebraic terms.

Barcebal has provided the reason why Y is EVEN.

So let's say X= a+a+1+a+2+....a+Y-1 given that 'a' can be any +ve integer.

then using arithmetic progression formula, the sum of those numbers will lead to

X=Y(a+a+Y-1)/2 --> X=Y(2a+Y-1)/2

So X/Y becomes
(2a+Y-1)/2

We know Y is EVEN number and whatever the 'a' is, it's always going to be even when it's multiplied by 2.
So EVEN+EVEN-1 is going to give us an ODD number.

So ODD/2 will NOT give integer value. This was my first post to BGMAT and hope this helps :)
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by debmalya_dutta » Tue Jul 13, 2010 6:14 am
@barcebal
We know from the facts mentioned in the statement that y = 2z and y and z are integers.
Y is even irrespective of what Z is ...try substituting values for z
if Z is even , say Z=2 , then Y=4
if Z is odd , say Z=3 , then Y=6

So, Y is always EVEN
Getting back to my answer
You see how the following equation was established
X/Y= a + ½(Y)
we know that a is integer . Since Y is even, Y/2 is an integer
Hence X/Y is an integer

barcebal wrote:
debmalya_dutta wrote: C. x/y is an integer
say a is the first term..subsequent terms are a+1,a+2......,a+(Y-1)
X= Ya + (1+2+3+........+Y-1) = Ya + Y/2( 2+ Y-2)
X/Y= a + ½(Y) ...
We know that Y=2Z , this implies that Y for sure is even
So X/Y is an integer

If Y is even X/Y is ONLY even if X is even. Otherwise you have an odd/even cannot EVER be an integer.

So now the question is can X be even. The short answer is NO.

If x is even that means you will have the sum of an even number of consecutive integers.

Let's expound.

Say Y=2, this means that x=sum of 2 consecutive integers. Two consecutive integers means that you have an odd and an even. odd+even=odd. X is odd if y is even.

Say Y=4. Now we CAN get four consecutive integers to be even. odd+even+odd+even=even, but the 4 is too big in the denominator. 1+2+3+4=10/4 is not an integer.
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by vijaynaik » Tue Jul 13, 2010 1:29 pm
@debmalya
----
C. x/y is an integer
say a is the first term..subsequent terms are a+1,a+2......,a+(Y-1)
X= Ya + (1+2+3+........+Y-1) = Ya + Y/2( 2+ Y-2)
X/Y= a + ½(Y) ...
----

Actually the above equation is X/Y = a + 1/2(Y-1) which is not an Integer.
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by ez2dj86 » Wed Jul 14, 2010 2:55 am
Hi debmalya,

You have made a mistake on your equation.
As I explained above,

X/Y is:
(2a+Y-1)/2
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by jube » Wed Jul 14, 2010 9:47 am
Hey, this one might be easy if we keep in mind that for an Arithmetic Series:
1) the median will be equal to the mean.
2) Sum/No. of Items = Mean

So if I have n consecutive integers (which is an AP with difference 1) then if n is odd I will have a median which is an integer. Therefore my mean i.e. sum/no. of items is also an integer.

However, if I have n consecutive integers and n is even then I will have a median which will NOT be an integer (since 2 consecutive integers will be an odd & even pair and their sum will always be odd). Therefore, my mean or Sum/No. of Items will also not be an integer.

The question tells us that y is an even no. Therefore any set with y no. of items will not have an integer value for median or an integer value for the mean i.e. x/y which is what C is.
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by vijaynaik » Wed Jul 14, 2010 12:30 pm
thanks for explaining it in much simpler way.
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