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the ans is 32

Ans. D) 32

Sum of ‘m’ consecutive integers = 8

So the integers have to be -7,-6,-5,…0,…,7,8. Only this sequence of consecutive integers would give us a sum of 8. So, m=16.

In the same logic, sum of ‘n’ consecutive integers = m = 16. So this sequence should run from -15 through to 16. So, n=32.

You got it!

Hi,

is the answer (D) 32?

i followed this method.

since the sum of the consecutive integers is 8 , it means the integers will be small values, so starting with 1,2,3,4 the sum is 10 , even if we take 2,3,4 it is still 9.. by doing a few more cases we find a pattern and also that this sequence should involve neg. numbers and . since its consecutive the series will go like this .-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8. that will be totally 16 ntegers making m=16.

also in the options there are 2 sets of values 16,15 & 31,32..

by logic we know that 16 +ve consec integers cant add up to 16 , so this series should also involve neg numbers, and by some pattern recognition we realise the series should go like -15,-14,-13…….-1,0,1,2,3,4,……,14,15,16 which is 32 integers, the answer.

Thanks for the thorough explanation and the correct response 🙂

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

16 = sum of n

-15, -14,-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16

n= 32

That’s it!

The answer is 32

Well if the sum on m consecutive integers is 8 than we can simply interpret it this way

The integers are from -7 to 8 (inclusive) . To count the number of terms inclusive of -7 and 8 , we do, 8-(-7) +1 , which is 16. Hence m is 16.

Now sum of n consecutive integers is 16 . Same old same old.

Integers are now -15 to 16 (inclusive) and to find the number of terms for it.

16- (-15) +1 . Which is 32.

Hence (D) is the right answer.

And also if anyone wondering that cant there be another case where consecutive integers sums up to 16 .

Well If consecutive integers must sum 16 , the maximum number in the series can be 8. as 8 plus 9 is greater than 16 and .

And before that no series sums to 16. I have checked all the cases.

Yep, there is no case. If a number cannot be arrived at by adding consecutive positive integers, it is called an impolite number. Any power of ‘2’ is impolite. Knowing that, of course, is way way beyond the scope of the GRE :).

Yep, you got it!

D，as the sum of consecutive positive integers cannot be 8，so there must be several consecutive negative integers ，those numbers are -7，-6，-5，-4，-3，-2，-1，0，1，2，3，4，5，6，7，8，so m=16，we can get that the value of n is 32 in the same way

You got it!

d

Answer is 32.

Workout :

For the sum of m integers to be 8, the series should be something like -7,-6..0..7,8, which yields a sum of 8 and thus the value of m is 16 (number of elements in the resulted series).

As stated in the question, the sum of n integers will be equal to m, which is found to be 16 in the last step.

And,For the sum of n integers to be 16, the series should be something like -15,-14..0..14,15,16, which yields a sum of 16 and thus the value of n is 32 (number of elements in the resulted series).

And I found this problem a lil harder. I managed to get the answer using the formula that is used to find the sum of Arithmetic progression series. There might be any other easy way to reach the answer and if so, please help me with it 🙂

Happy to post a comment in Magoosh for the first time!!! 🙂

Hi Pown,

Thanks for posting :).

So you definitely got the correct answer. The logic is all the negatives will cancel out the positives except for the number that is equal to the sum. So if you want to find which consecutive integers sum to, say, 99, you would start at -98 + -97… all the way up to 99.

Hope that makes sense!

Hi Chris,

Happy seeing your reply. I got your definition. Thank you so much.

Well, I’m not sure if this is right but the only solution I could come up with was this:

To achieve the sum of 8, one would start summing up from -7, so -7+-6+-5……+8=8. This would make 16 integers. m=16.

Then we can follow the same procedure to achieve 16. -15+-14+-13+—–+16=16. This would make 32 integers so the answer might be 32.

Since no consecutive positive integers add up to 8 (1+2+3=6, 2+3+4=9, etc), we must involve negative integers. Thus, the sum of -7,-6,…,6,7,8=8, which is m=16 consecutive numbers. Subsequently, n=32, the sum of 32 consecutive number -15,-14,…,14,15,16=16. Therefore, n=32.

Oops, that was only the first half of my answer! So if the sum of m consecutive integers is 8, where m = 16, then setting x = 16 gives:

(-15 + … + 15) + 16 = 16, the number of integers is: 15 x 2 + 1 (for zero) = 31, then +1 (for the 16) = 32.

32 consecutive integers, or answer D.

It’s pretty clear that 15 or more positive integers cannot add up to 8. Therefore, negative integers must be involved. But where to start and where to end?

Begin with a simple range of integers that start and end with the negative and positive version of that integer*:

(-4 + -3 + … + 3 + 4) = 0

So basically summing up any range of consecutive integers starting and ending with the same absolute value (in this case, |4| = |-4| = 4) will always be zero.

If you add one more integer to one of the ends:

(-4 + … + 4) + 5 = 5

or

-5 + (-4 + … + 4) = -5

Interesting. So one way to have the sum equal a particular number (say x) is to have the range go from – (x-1) + … + (x-1) + x. In our actual problem, let’s see what happens when x = 8.

(-7 + … + 7) + 8 = 8

This would be 16 integers: 7 x 2 + 1 (for the zero) = 15, then + 1 (for the 8) = 16.

Is this the only way? For an 8, yes, because no consecutive positive integers will add up to a power of 2 (it’s called an impolite number, I know, how rude). But what if x = 5?

(-4 + … + 4) + 5 = 5, and the number of integers = 4 x 2 + 1 (for the zero) = 9, then +1 (for the 5) = 10.

But … 5 can also be represented as the sum of the consecutive positive integers 2 + 3! So another, and shorter, solution is:

(-1 + … + 1) + 2 + 3 = 5, number of integers is 1 x 2 + 1 (for zero), then +2 (for 2 & 3) = 5.

*Parentheses were added to the sequence of integers to show which integers add up to zero, to emphasize the integers that were added past that range.

I started off thinking in terms of number line and this helped to see a pattern. The difference between the sum’s of consecutive numbers increase by 1 so this meant to find m, I had to find sum’s till the difference was 8, that is, 28 and 36. So the range would -28 to 36 which accounts for 16 numbers, -7 to 8. Extending this logic, we find n for 16 which turns out to be the range -15 to 16 which accounts for 32 numbers. Hence answer D.

Hello Mr. Lele, 32 (D). I hope it’s right!

The answer should be D i think.

Firstly, for consecutive integers to add up to 8, the numbers must range from -7 to 8 inclusive, i.e a total of 16 numbers. Therefore, m = 16

Secondly, the only combination of consecutive integers that will add up to 16 is from -15 to 16 inclusive, i.e a total of 32 integers if I am not mistaken.

Therefore, the answer should be D. Please correct me if I am mistaken. Thanks!

Ans E

Possible values of m are1 and 16

And for each value of m possible values of n are (1,2) and (1,32 ) respectively

Hence data insufficient.

32(D)

This is my solution to the math twister question

to get the sum of 8 for m consecutive numbers, I listed the number from -7 to 7 + 8. The set of number -7, thru 7 negates each others out.

{-7, + -6, + -5, … + 0 + 1 + 2 + 3+ 7 + 8 } = 8

this list contains 16 consecutive numbers

I applied the same technique to the second set and listed the consecutive numbers which sums up to 16.

{ -15, -14, -13, + ……..+ 9 + 10 + 11 + … 15 + 16} = 16

This list contains 32 consecutive numbers

therefore n=32

Hi,

This sure is a difficult question. It took me more than 15 minutes to solve! I was trying to get to the answer using the formula for the sum of an arithmetic progression, which, turns out, is rather difficult. Then I just drew a number-line, from the number -8 to +8, and suddenly, I had an epiphany, and realised that if I add all the numbers from -7 till +8, the sum of those numbers will be 8. By the same principle, if I add all the numbers from -15 to +16, the sum of those numbers will be 16.

Therefore,

The sum of 16(=m) numbers is 8, and:

The sum of 32(=n) numbers is 16(=m)

The answer is (D)

Yay!