x is the sum of y consecutive integers. w is the sum of z

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

Case 1: x = w
if y = [2,3] z = [5]
5=5, TRUE

Case 2: x > w
if y = [2,3] z = [1]
5>1, TRUE

Case 3: x/y is an integer
if y = [2,3,4] x = 9
9/3 is an integer, TRUE

Case 4: w/z is an integer
if z = [2] w = 2
2/1 is an integer, TRUE

Case 5: x/z is an integer
if x = 9, z = [2]
9/1 is an integer, TRUE

Am I doing something wrong here?

C

a) x= w if x = (1+2+3+4+5+6) = 21 and w = (6+ 7+8)=21.
b) x>w if x= (1+2+3+4+5+6) = 21 and w = (4+5+6)= 15
c) x/y => x = (1+2+3+4+5+6) = 21 and y = 6 then x/y is NOT an integer
=> x = (1+2+3+4) and y = 4 then x/y is NOT an integer
d) w/z =>w = (6+ 7+8)=21 and z=3 then w/z = 3
e) x/z => x = (1+2+3+4+5+6) = 21 and z = 3 then x/z = 7

IMO (C).

y will always be even since y=2z

assume 6 numbers
n-2 n-1 n n+1 n+2 n+3
x=6n+3, y=6, z=3

x/z = 2n+1

x/y = n+1/2 ==> cannot be integer

OA is C

sum of even no is not always odd.

consider 1+2+3+4=10 which is even

For any set of consecutive integers with an even number of terms, the sum of the integers is never a multiple of the number of terms.

consider above example 10 is not multiple of 4.

But For any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms.

1+2+3=6 multiple of 3

since y = 2z... it is even...

x is the sum of y (even no) of integers....

Hence x/y would never be an integer....

Therefore C....

Note
For any set of consecutive integers with an even number of terms, the sum of the integers is never a multiple of the number of terms.
But For any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms.

The question can be restated as:
If x is the sum of y consecutive integers, and w is the sum of y/2 consecutive integers, and neither of y or z is 0, then which of the following can NEVER be true?

Note that since y>0 and z=y/2 and z>0 and both y and z are integers, y will always be an even integer.

(a) x=w. This is certainly possible. Take x=-2,-1,0,1,2,3 and w=0,1,2
(b) x>w. This is certainly possible. Take x=-2,-1,0,1,2,3 and w=-2,-1,0
(c) Correct answer. You can never make a case that satisfies this. Why?
Sum of the AP = (y/2) [2a+(y-1)*1]
If y = 2k as y is always even, Sum = x = k [2a+ (2k-1)]
If we divide this by y, i.e. 2k, we get x/y = k[2a+2k-1]/2k. This is an odd number divided by an even number, and so can never be an integer.
(d) This is certainly possible. Evaluates to [2a+(z-1)]/2, which can be true if z is odd.
(e) This is certainly possible. Take x=-2,-1,0,1 and then z=2, and x/z=-1, which is an integer.

This is how you can work it out:

x = sum of y con. integers
w = sum of z con. integers

Plug in values such that most of these are easily made true.
Since z is in the denominator, I will try to put z = 1 to get w/z and x/z as integers without any complications.

If z = 1, y = 2
w = any one integer
x = sum of 2 consecutive integers
w/z and x/z will be integers so d and e can be true.

x = w is obviously possible. Say w = 9 and x = 4+5
x > w is also possible say if w = 8 while x = 4+5
Hence both a and b can be true.

Answer must be c.

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

x = w
x > w
x/y is an integer
w/z is an integer
x/z is an integer


average1= x/y
average2=w/z

y=2z,

implies that for every value of z y is even.
ie.
z=even say 4
y=2*4 =6 (even)

z=odd say 3

y=2*3 =6 (even)

the average of even numbers is not integer , so the answer is C.

For these kind of questions, we need to parse through the options to find an option which can NEVER be true. There are three scenarios possible with any option:
1. We can easily find out that this is possible. Then, we can eliminate this option and move forward.
2. We can easily find out that the option is never true. Thus, we can stop here because this is the answer.
3. Nothing can be easily inferred. We need to plug in values and do some reasoning to figure out. These options can be skipped for the moment to come back later, if we don't find an answer among other options. We want to SAVE our time.

Here is my attempt at this:
A. x = w - just looking at it, i can't say if this is possible or not. To find so will take time.I skip it to come back later
B. x > w - Easily possible is x is the sum of first 2 integers and w is the sum of first 1 integer
C. x/y is an integer - come back later
D. w/z is an integer - easily possible if z=1, w=1
E. x/z is an integer[/quote] - easily possible if z=1, x=3

now, we are left with A and C options. I begin with option C.

x=sum of consecutive y (or 2z) integers = n + (n+1) + (n+2) +....+ (n+2z-1) = 2zn + (1+2+3+...+2z-1) = 2zn + 2z*(2z-1)/2 (using formula for sum of first n natural numbers)
=> x/y or x/2z = n + (2z-1)/2
we know n is an integer. The other component can never be an integer, since the numerator is always odd and denominator is 2. So, x/y cannot be an integer.

Thus, C is the answer.

Cheers,
CJ

thanks this is a brilliant explanation.

where could i get other number properties such as this?

2. Properties of Integers

• Theory
  Math: Number Theory
  Number Properties: Tips and hints
  Advanced Number Properties on GMAT (All 5 parts)
  Do you make these 3 mistakes in GMAT Even-Odd Questions?
  Everything about Factorials on the GMAT
  Solving Complex Problems Using Number Line
  
  • Questions
    Tough New Set: Number Properties Set
    Power of a Number in a Factorial Problems
    Number Properties Ability Quiz
    Special Numbers and Sequences Problems
    The E-GMAT Number Properties Knockout: 10 Great 700+ Ques for you!
    Trailing Zeros Problems
    DS Questions
    PS Questions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.

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