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

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.

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

Toooo good Bunuel...
Instead of learning formula i am more comfortable with the conceptual approach. Do you think learning formula is integral for GMAT.

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.

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.

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

Excelente Private scores, that is a great approache which can allow us to face a gmat question like that!

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.

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.

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!

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

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

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.

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:

GMAT assassins aren't born, they're made,
Rich

How did you eliminate option B with this method? Sum of 3 consecutive integers is even and 6 consecutive integers is odd?