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If a is the sum of x consecutive positive integers. b is the

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If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post Updated on: 06 Mar 2012, 02:50
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If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7

Originally posted by devinawilliam83 on 05 Mar 2012, 23:36.
Last edited by Bunuel on 06 Mar 2012, 02:50, edited 2 times in total.
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Re: sum of x consecutive positive integers  [#permalink]

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New post 06 Mar 2012, 02:48
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If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?
A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7

The sum n consecutive integers is give by: \(Sum=\frac{(2a_1+n-1)*n}{2}\) (check Number Theory chapter of Math Book for more: math-number-theory-88376.html);

Notice that:
If \(n=even=2*odd\), so when \(n\) (# of consecutive integers) is even but not a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*odd)}{2}=odd*odd=odd\);

If \(n=even=2*even\), so when \(n\) is a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*even)}{2}=odd*even=even\);

That's because a set of even number of consecutive integers has half even and half odd terms. The sum of even terms is obviously even. As for odd terms: their sum is even if their number is even (so total # of terms is multiple of 4) and their sum is odd if their number is odd (so total number of terms is even but not a multiple of 4);

So, the sum of 10 (not a multiple of 4) consecutive integers will be odd (the sum of 5 even and 5 odd integers) and the sum of 4 (multiple of 4) consecutive integers will be even (the sum of 2 even and 2 odd integers), so option D is not possible.

Answer: D.

Hope it's clear.
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 11 Apr 2014, 03:53
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devinawilliam83 wrote:
If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7



Sum of 2 consecutive integers = Odd
Sum of 4 consecutive integers = Even
Sum of 6 consecutive integers = Odd
Sum of 8 consecutive integer = Even and so on is the pattern.

If we look at answer option D. Sum of 10 consecutive integer = Odd, Sum of 4 consecutive integers = Even.

Odd cannot be equal to Even.

Hence the answer is D.
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 06 Mar 2012, 22:48
1
Hi,
I am struggling with the explanation.Here is what I had done. but was unable to eliminate answers
the premise was that for x and y to be equal both should be either even or odd
Please take a look and let me know
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 07 Mar 2012, 00:46
devinawilliam83 wrote:
Hi,
I am struggling with the explanation.Here is what I had done. but was unable to eliminate answers
the premise was that for x and y to be equal both should be either even or odd
Please take a look and let me know


I'm not sure what you are trying to say there with the diagram but the 4th row is not correct: if the # of terms is multiple of 4 then the sum is even, regardless of the first term.
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 27 Oct 2012, 20:25
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For consecutive terms , Sum = median * number of terms. So if m1 and m2 are two medians..

a= m1*x and b = 2*y

as per given condition. a=b => m1*x = m2*y if you put x=10 and y =4

10m1 = 4m2 => 5m1 = 2m2.....

When number of terms are even, the median is always a fraction. I.e 3.5 or 4.5 but when number of terms is odd then median is always an integer.
Now in above case. m1 and m2 both are fractions but m2's fraction can be nullified by 2 which is multiplied by m2. But for m1 it is not the case.

So all combinations are possible but if number of terms are multiple of 4, then the other can not be of the form 4n+1 or 4n+2 or 4n+3.

Hence D
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 02 Feb 2013, 10:05
Toooo good Bunuel...
Instead of learning formula i am more comfortable with the conceptual approach. Do you think learning formula is integral for GMAT.
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 03 Feb 2013, 02:21
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We know that if x=10, and say the first term is odd, then the sum(a) will be

5x(odd+even) = 5x(odd)= odd.

The first term can be even also, nonetheless, the sum will be 5x(even+odd)=5x(odd)= Still odd.

Now for y=4, whatever the first term be,(odd/even) the sum b = 2(odd+even) = 2x(odd)=even.

Thus as odd cant be equal to even, ans is D.
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Re: sum of x consecutive positive integers  [#permalink]

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New post 30 Nov 2013, 06:43
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Bunuel wrote:
If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?
A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7

The sum n consecutive integers is give by: \(Sum=\frac{(2a_1+n-1)*n}{2}\) (check Number Theory chapter of Math Book for more: math-number-theory-88376.html);

Notice that:
If \(n=even=2*odd\), so when \(n\) (# of consecutive integers) is even but not a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*odd)}{2}=odd*odd=odd\);

If \(n=even=2*even\), so when \(n\) is a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*even)}{2}=odd*even=even\);

That's because a set of even number of consecutive integers has half even and half odd terms. The sum of even terms is obviously even. As for odd terms: their sum is even if their number is even (so total # of terms is multiple of 4) and their sum is odd if their number is odd (so total number of terms is even but not a multiple of 4);

So, the sum of 10 (not a multiple of 4) consecutive integers will be odd (the sum of 5 even and 5 odd integers) and the sum of 4 (multiple of 4) consecutive integers will be even (the sum of 2 even and 2 odd integers), so option D is not possible.

Answer: D.

Hope it's clear.



Brunel ,

Sum = n/2( 2a +(n-1)d ).....i still not able to get ur sum formula . I guess for consecutive number you are taking d=1 ...
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Re: sum of x consecutive positive integers  [#permalink]

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New post 01 Dec 2013, 05:20
archit wrote:
Bunuel wrote:
If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?
A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7

The sum n consecutive integers is give by: \(Sum=\frac{(2a_1+n-1)*n}{2}\) (check Number Theory chapter of Math Book for more: math-number-theory-88376.html);

Notice that:
If \(n=even=2*odd\), so when \(n\) (# of consecutive integers) is even but not a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*odd)}{2}=odd*odd=odd\);

If \(n=even=2*even\), so when \(n\) is a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*even)}{2}=odd*even=even\);

That's because a set of even number of consecutive integers has half even and half odd terms. The sum of even terms is obviously even. As for odd terms: their sum is even if their number is even (so total # of terms is multiple of 4) and their sum is odd if their number is odd (so total number of terms is even but not a multiple of 4);

So, the sum of 10 (not a multiple of 4) consecutive integers will be odd (the sum of 5 even and 5 odd integers) and the sum of 4 (multiple of 4) consecutive integers will be even (the sum of 2 even and 2 odd integers), so option D is not possible.

Answer: D.

Hope it's clear.



Brunel ,

Sum = n/2( 2a +(n-1)d ).....i still not able to get ur sum formula . I guess for consecutive number you are taking d=1 ...


Yes, the common difference between the numbers in a set of consecutive numbers is 1.
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 01 Dec 2013, 07:09
devinawilliam83 wrote:
If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7



this is a very bad question ... no way a GMAT question... very time consuming ...
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 05 Jan 2015, 18:06
Excelente Private scores, that is a great approache which can allow us to face a gmat question like that!
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 05 Jan 2015, 20:03
Although time consuming, my approach is similar to others.

Sum of 2: 1+2 or 2+1 (odd+even or even+odd) = Odd
Sum of 3: 1+2+1 or 2+1+2 = Even/Odd
Sum of 4: 1+2+1+2 or 2+1+2+1 (Same both way) = Even
Etc. etc.

I got the question right over 2 minutes, hope this helps...

Etc. etc.
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 05 Jan 2015, 23:15
7
1
mdcash wrote:
devinawilliam83 wrote:
If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7



this is a very bad question ... no way a GMAT question... very time consuming ...


Actually, it isn't that bad. There is the logical approach of number properties that you can use but even if that doesn't work out in the limited time, go with brute force!

A. x = 2; y = 6

Sum of first 6 numbers is 6*7/2 = 21
Can the sum of two consecutive numbers be 21? Sure 10 and 11.
Out

B. x = 3; y = 6
Sum of first 6 numbers is 21. Can sum of three consecutive numbers be 21? Divide 21 by 3 to get 7. The three numbers will be 6, 7, 8.
Out

C. x = 7; y = 9
Sum of first 9 numbers is 9*10/2 = 45. Can sum of 7 consecutive numbers be 45? 45 is not divisible by 7 so this will not work. Try another method:
Sum1 = Sum2
7*Mean1 = 9*Mean2
If Mean1 = 9 and Mean2 = 7, it will satisfy. i.e. 6, 7, 8, 9, 10, 11, 12 and 3, 4, 5, 6, 7, 8, 9, 10, 11
Out

D. x = 10; y = 4
10*Mean1 = 4*Mean2
Mean1/Mean2 = 2/5
Both Mean1 and Mean2 must be fractions "something.5" (Even consecutive numbers)
Also, Mean1 = 2x and Mean2 = 5x
Hard. Hold it.

E. x = 10; y = 7
10*Mean1 = 7*Mean2
Mean1/Mean2 = 7/10
Mean1 must be a fraction "something.5" (10 consecutive numbers) and Mean2 must be an integer (7 consecutive numbers)
Such as 3.5/5 but you don't have 10 positive integers around 3.5
So perhaps 10.5/15
The numbers are
6, 7, 8, 9, 10, 11, 12, 13, 14, 15
and
12, 13, 14, 15, 16, 17, 18
Out

Answer (D) by elimination.
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 16 Jun 2015, 22:17
devinawilliam83 wrote:
If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?

A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7


Ok, I did it simply in the head...here it is

x = 10 : (sum of 5 even no.+ sum of 5 odd no.)=(even+odd)=odd number over all.
y = 4 : (sum of 2 even no.+sum of 2 odd no.)=(even+even)=even number over all.

Hope this helps!
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 27 Jun 2015, 07:05
Bunuel wrote:
If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?
A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7

The sum n consecutive integers is give by: \(Sum=\frac{(2a_1+n-1)*n}{2}\) (check Number Theory chapter of Math Book for more: math-number-theory-88376.html);

Notice that:
If \(n=even=2*odd\), so when \(n\) (# of consecutive integers) is even but not a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*odd)}{2}=odd*odd=odd\);

If \(n=even=2*even\), so when \(n\) is a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*even)}{2}=odd*even=even\);

That's because a set of even number of consecutive integers has half even and half odd terms. The sum of even terms is obviously even. As for odd terms: their sum is even if their number is even (so total # of terms is multiple of 4) and their sum is odd if their number is odd (so total number of terms is even but not a multiple of 4);

So, the sum of 10 (not a multiple of 4) consecutive integers will be odd (the sum of 5 even and 5 odd integers) and the sum of 4 (multiple of 4) consecutive integers will be even (the sum of 2 even and 2 odd integers), so option D is not possible.

Answer: D.

Hope it's clear.



Hi Bunuel,
Could you please explain why we're only dealing with n=even here? Also I'm unable to understand the part "As for odd terms: their sum is even if their number is even (so total # of terms is multiple of 4) and their sum is odd if their number is odd (so total number of terms is even but not a multiple of 4);". Can you please clarify? Thanks
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 04 Sep 2015, 11:23
devinawilliam83 wrote:
Hi,
I am struggling with the explanation.Here is what I had done. but was unable to eliminate answers
the premise was that for x and y to be equal both should be either even or odd
Please take a look and let me know


your approach is correct but have made mistake in the case of n=4. If you have 4 terms in the consecutive sequence, sum will always be even

\(n/2 *[2*a + (n-1)d]\)

since n=4 & d = 1, the expression reduces to:

\(2 * [2*a + 3]\)

which is always even
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 11 Sep 2015, 06:44
Consider the following scenario :

x=7, y=9

6,7,8,9,10,11,12 are 7 consecutive terms with sum (a) =63

3,4,5,6,7,8,9,10,11 are 9 consecutive term with sum (b) = 63

Thus, a=b and option C is correct.
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Re: If a is the sum of x consecutive positive integers. b is the  [#permalink]

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New post 27 Jun 2018, 14:47
Hi All,

We're told that A is the sum of X CONSECUTIVE positive integers and B is the sum of Y CONSECUTIVE positive integers. We're asked for which of the following values of X and Y is it IMPOSSIBLE that A = B. This question is based on a subtle Number Property rule (and if you know the rule, then the question becomes a whole lot easier). You can certainly TEST THE ANSWERS to prove what's possible and what's not:

Answer A: If X = 2 and Y = 6 you could have

A = 10 + 11 = 21
B = 1 + 2 + 3 + 4 + 5 + 6 = 21

In this example, A EQUALS B, so this is not the answer we're looking for.

Answer B: If X = 3 and Y = 6 you could have

A = 6 + 7 + 8 = 21
B = 1 + 2 + 3 + 4 + 5 + 6 = 21

In this example, A EQUALS B, so this is not the answer we're looking for.

You could continue with this method to figure out which of the other 3 answers are possible and which is not.

With the following Number Property though, you'll be able to quickly solve this problem:

Take a look at answer D....

X = 10 and Y = 4

The sum of 10 consecutive positive integers will ALWAYS be ODD... since there are 5 ODD numbers (try it and you'll see).
The sum of 4 consecutive positive integers will ALWAYS be EVEN... since there 2 ODD numbers (again, try it and you'll see)

So, A can NEVER EQUAL B in this circumstance.

Final Answer:

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Re: If a is the sum of x consecutive positive integers. b is the &nbs [#permalink] 27 Jun 2018, 14:47
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