The sum of m consecutive integers is 8. If the sum of n consecutive in

Hi..
On why m should be 16 and nothing else..
SUM is 8, and the SUM of any number of consecutive integers will never be 2^x...
example-- if we look at consecutive positive integers it is 1,2,3,4,5,6... Beyond 4, sum of two numbers too goes beyond 8 and other ways to add are 1+2=3, 1+2+3=6, 2+3=5,3+4=7

so it would be 8 itself and that will be possible when numbers are -7,-6,-5,....0,1,2,....6,7,8 here sum will be 8
and number of integers from -7 to 8 are 7 negative integers , 1 zero and 8 positive integers = 7+1+8=16..
so m is 16
again 16 is 2^4, so it will also require 16 as an integer... numbers become -15,-14....0...14,15,16 = 15+1+16=32

D..

only one problem i see if m is 1 that is 1 consecutive integer only 8 then answer does not fit in.

is there a way to solve this using Arithmatic Progression sum?

if we have odd consecutive integers 3 and 5 than m=2 how do we proceed?

took time to understand this logic ..nice explanation

We are given that the sum of m consecutive integers is 8. We see that the m (assuming m > 1) consecutive integers can’t be all positive because we can’t obtain a sum of 8 from m only-positive consecutive integers.. For example, if m = 2, 3 + 4 = 7, and 4 + 5 = 9, and if m = 3, 1 + 2 + 3 = 6, and 2 + 3 + 4 = 9. Thus, some of the integers must be positive and some must be negative, and we see that one of them must, therefore, be 0.

Since the sum is 8, a positive number, there must be more positive integers than negative integers. However, since the negative integers will cancel out their positive counterparts and we have mentioned that there can’t be 2 (or more) consecutive positive integers that add up to 8, there must be exactly 1 more positive integer than the number of negative integers. Furthermore, that positive integer (that won’t cancel with any of its negative counterparts) must be 8. That is, the m consecutive integers must be:

-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8

We see that the sum of all the integers from -7 to 8, inclusive, is 8 since the sum of all the integers from -7 to 7, inclusive, is 0. We also see that there are 16 integers in the above list, so m = 16.

Next we are given that the sum of n consecutive integers is m, which we now know is 16. We can use a similar argument to the one above to conclude that the n (assuming n > 1) consecutive integers can’t be all positive. Furthermore, there must be 1 more positive integer than the number of negative integers and that extra positive integer (the one without a negative counterpart) must be 16. That is, the n consecutive integers must be:

-15, -14, …, -1, 0, 1, …, 14, 15, 16

We see that there are 32 integers in the above list since there are 15 negative integers, 1 zero, and 16 positive integers. Thus, n = 32.

Answer: D

Key word = sum of m consecutive integers is 8,

how can we get that sum ??
Sum of 16 integers from -7 to 8, can help us in that

This 16 = m

Now how can we get 16 ??

Sum of 32 integers from -15 to 16 can help us in that

Therefore n = 32

Rule of thumb: Sum of k consecutive nos will be multiple of k, for example, sum of 3 consecutive nos will be multiple of 3.

So, sum of m consecutive nos is 8, so m will be multiple of 8. m can be 16,32...

Now, n is the sum of m consecutive nos, so it will be a multiple of m, i.e, multiple of 16,32..

so if m=16, n=32.

What if m is 1 so the sum is 8 = 8.
=> n = 1

Answer: E

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