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Manager  Joined: 16 Feb 2012
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Concentration: Finance, Economics
The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Question Stats: 62% (02:19) correct 38% (02:17) wrong based on 319 sessions

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The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Stiv wrote:
The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

Sum of Numbers = Average * Number of numbers
If numbers are from 1 to 27, average will be the middle number 14.
Sum of numbers = 14 * 27

We need to find the option where the sum is the same.

A. -27 to 54
-27 to -1 will cancel out numbers from 1 to 27. So the sum would be the sum of numbers from 28 to 54. These will be 27 numbers but with a much higher average. So ignore this option.

B. 0 to 28
Here the sum will be 28 more than our desired sum. Ignore this option.

C. 15 to 45
Here we have 31 numbers with an average much more than 14. So the sum will be much more than the desired sum. Ignore.

D. 38 to 46
Here we have 9 numbers with an average of 42 (the middle number).
Sum = 42 * 9 = 14 * 27 (matches)
This is the answer.

E. 48 to 54
Just to complete the calculations, let me show you how you can compare this option too.
Here we have 7 numbers with an average of 51.
Sum = 7 * 51 = 14 * 51/2 = 14 * 25.5. This is less than the desired sum.

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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Hi gmatser1,

I'm going to give you a hint so that you can try this question again....

This question can be solved by 'bunching' (and a bit of pattern matching). To start, we have to figure out the sum of the integers from 1 to 27, inclusive. You are correct that you can start by 'bunching' the terms...

1+27 = 28
2+26 = 28
3+25 = 28
Etc.

Since there are 27 terms, there will be 13 'pairs' and a 'middle' term that does not get paired; the final sum will be 13(28) + 14. At this point, you can do the arithmetic in a couple of different ways.

(13)(28) + 14 =
(26)(14) + 14 =
(26)(14) + 1(14) =
(27)(14) =
378

So, you can either look for an answer that totals 378 OR you can look for an answer that can be 'bunched' into (27)(14).

Using this same type of approach, how would you work through the answer choices?

GMAT assassins aren't born, they're made,
Rich
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Stiv wrote:
The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from
A -27 to 54
B 0 to 28
C 15 to 45
D 38 to 46
E 48 to 54

I give kudos to anybody who post quick and efficient way for solving the problem (< 1 min.)

Sum of the integers from 1 to 27 is the average of first and last terms multiplied by # of terms. Applying that to the question stem gives you
(28/2) * (27)
-(14) * (27)
-(3 * 3 * 3 * 2 * 7).

Choice A is basically the sum of 27 - 54 which will be too big
Choice B is tempting but after applying the formula you will get (14) * (28)
Choice C will be too big as well, there are 30 terms and each term is bigger than it's corresponding term in the original sequence

Choice D looks like it could work, so apply the formula.

=(84/2) * (46 - 38 + 1)
=(42) * (9)
=(3 * 3 * 3 * 2 * 7)

Choice D is the answer
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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HI gmatser1,

You mentioned in your last post how "mapping out" all of the pairs would have taken too long. I agree, but you don't really have to map them all out to see the pattern and the total. Here's how:

You know that the first few 'pairs' are....

1+27 = 28
2+26 = 28

So the 'smaller number' + the 'bigger number' will total 28. Using that logic:

1) What would be the 'bigger number' when the 'smaller number' was 13?
2) Using all of the 'smaller numbers' for reference, how many total 'pairs' would there be?
3) Once you know what 'pairs up' with 13, how hard is it to find any 'leftover' numbers?

As you continue to study, you're going to find that most GMAT questions can be answered in a variety of ways. If you find that "your way" of approaching the prompt takes too long, then it might be that you're approaching the prompt in "the long way."

GMAT assassins aren't born, they're made,
Rich
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Stiv wrote:
The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from
A -27 to 54
B 0 to 28
C 15 to 45
D 38 to 46
E 48 to 54

I give kudos to anybody who post quick and efficient way for solving the problem (< 1 min.)

Summation : 27/2 . {2 . 1 + 26 . 1} = 27 . 14

we can eliminate A because there were 27 numbers on the question and the last was 27. After this 27 numbers every number is far more than these 27 numbers. So addition of same number of numbers after 27 must not be equal .
B eliminated by sight.
C contains 31 numbers , each is bigger than 1 to 27 serially. so eliminated.

only last two worth calculation.

Now D, So 9/2 {2 . 38 + (9-1) . 1} = 9 . 42 = 9 . 3 . 14 = 27 . 14 .
it equals the summation of 1 to 27 Answer (D)
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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I need some basic help on this one. Can you not start with:

1 + 27 = 28
2 + 26 = 28
3 + 25 = 28
...
...

(28-1+1)/2 = 14 total divisions

28(14) =

Then find the prime factorization and then match that prime factorization with all the answer choices?
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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sum of consecutive positve integers =n(n+1)/2
=27*28/2
=27*14
=378.
so now we see options
in option A -27 to 54.
if we substitute -27 to 27 sum equal to 0.
thereafter 28to54 is greater than 378.
so option B is 0t0 28.
sum equal 1to 28 it is greater than 1to 27.

option C sum 15 to 45,
total number is 31. medain number is 30.
sum of consecutive no equal to median *total no of terms.
=30*31=910.

option D 38 to 46 total 9 numbers are there.
median no 42*9=378.
option E 48 to 54 total numbers 7 . median no 51*7=357.
It is less than sum.
so option D is correct.
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Posts: 61
Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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EMPOWERgmatRichC wrote:
Hi gmatser1,

I'm going to give you a hint so that you can try this question again....

This question can be solved by 'bunching' (and a bit of pattern matching). To start, we have to figure out the sum of the integers from 1 to 27, inclusive. You are correct that you can start by 'bunching' the terms...

1+27 = 28
2+26 = 28
3+25 = 28
Etc.

Since there are 27 terms, there will be 13 'pairs' and a 'middle' term that does not get paired; the final sum will be 13(28) + 14. At this point, you can do the arithmetic in a couple of different ways.

(13)(28) + 14 =
(26)(14) + 14 =
(26)(14) + 1(14) =
(27)(14) =
378

So, you can either look for an answer that totals 378 OR you can look for an answer that can be 'bunched' into (27)(14).

Using this same type of approach, how would you work through the answer choices?

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

Thanks! I forgot about that 13th "middle" pair. Could have found that out by mapping it out, but that would have taken way too long.

In looking at the answers, the same dynamic applies with D. 46-38= 8 total numbers / 2 = 4 pairs with 42 left in the middle

38+46 = 84
39+45 = 84
40+44= 84
41+43 = 84

42

4(84) + 42 = 378

I think my original statement about prime factorization would be hard using this method. Nevermind!
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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D it is ... good explanations above _________________
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Stiv wrote:
The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

Sum of integers from 1 to 27= 27*28/2= 27*14 (Units digit will be 8)

A. -27 to 54
Effectively it will be sum of integers from 28 to 54 (It will definitely be greater than the sum of integers from 1 to 27)

B. 0 to 28
Integer 28 will be added more to the sum of integers from 1 to 27

C. 15 to 45
There are total 31 integers, and by looking closely at the numbers, we realize that most of the numbers are greater than 27 and hence the sum will also be greater

D. 38 to 46

By adding units digit of given numbers (8+9+0+1+2+3+4+5+6), we get 8 as the unit digit of resulting number. This is the answer.

E. 48 to 54
We do not get 8 as unit digit in this case as we got in option D
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Stiv wrote:
The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

Hi,

the property which I will use is that SUM of consecutive integers is Number of imtegers * average or center number..
sum of the positive integers from 1 to 27 = 27 * \frac{1+27}{2} = 27*14...
so the sum has to be EVEN and multiple of 7 and 27..

lets see the choices-

A. -27 to 54 --------- number of integers = 27+54+1 = 82.. this means the SUM will be multiple of PRIME 41... NO

B. 0 to 28-----------number of integers = 28+1 = 29.. this means the SUM will be multiple of PRIME 29... NO

C. 15 to 45-----------number of integers = 45-15+1 = 31.. this means the SUM will be multiple of PRIME 31... NO

D. 38 to 46-----------number of integers = 46-38+1 = 9..so we see the average now =$$\frac{46+38}{2}= 42$$..this means the SUM will be =9*42 = 9*6*7 = 27*14...YES

E. 48 to 54-----------number of integers = 54-48+1 = 7.... so we see the average now =$$\frac{48+54}{2}= 51$$..this means the SUM will be =51*7 OR an ODD number and also not a multiple of 27...NO

D
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The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Stiv wrote:
The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

i did the direct way...and it took more than 2 minutes...
A -> from 1 to 27 is 27*28/2 = 378. since we have -27, it's -378. -27 is inclusive, so subtract from -378-27, and get -351. sum from 0 to 54 = 54*55/2 = 27*55 -> last digit is 5...since we would add with -351, the last digit would be 4...so not good.
B -> 0 to 28 -> 28*29/2 = 14*29 -> way too big.
C -> sum from 0 to 14 = 14*15/2 = 7*15 = XX5. 0 to 45 is 45*46 = 23*45 -> last digit is 5. since we would subtract XX5, the last digit would be 0 and not 8. out.
D -> from 0 to 37 = 37*38/2 = 37*19 = smth smth last digit =3. 46*47/2 = 23*47 -> last digit =1. hm...if XX1 - XX3 -> we'll get last digit 8..this might be the answer...
E -> from 0 to 47 = 47*48/2 = 47*29 -> last digit is 3 (smth XX3). 0 to 54 = 54*55/2 = 27*55 = XX5..last digit 5...from 5 subtract 3 and get 2..we need 8
looks like only D works!
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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The sum of consecutive integers is given by n(n+1)/2...so how to calculate as per this?

acc to me,a option says -27 to 54,in which -26 to 26 will eventually cancel and remaining sum for 27 to 53 inclusive with 27 terms comes out to be 27*28/2...27*14 which is what coming previously(in the question)
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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chetan2u wrote:
Stiv wrote:
The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

Hi,

the property which I will use is that SUM of consecutive integers is Number of imtegers * average or center number..
sum of the positive integers from 1 to 27 = 27 * \frac{1+27}{2} = 27*14...
so the sum has to be EVEN and multiple of 7 and 27..

lets see the choices-

A. -27 to 54 --------- number of integers = 27+54+1 = 82.. this means the SUM will be multiple of PRIME 41... NO

B. 0 to 28-----------number of integers = 28+1 = 29.. this means the SUM will be multiple of PRIME 29... NO

C. 15 to 45-----------number of integers = 45-15+1 = 31.. this means the SUM will be multiple of PRIME 31... NO

D. 38 to 46-----------number of integers = 46-38+1 = 9..so we see the average now =$$\frac{46+38}{2}= 42$$..this means the SUM will be =9*42 = 9*6*7 = 27*14...YES

E. 48 to 54-----------number of integers = 54-48+1 = 7.... so we see the average now =$$\frac{48+54}{2}= 51$$..this means the SUM will be =51*7 OR an ODD number and also not a multiple of 27...NO

D

should we always consider the range as inclusive,even when not specified?
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Re: The sum of the positive integers from 1 to 27 is equivalent  [#permalink]

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Stiv wrote:
The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

Let’s analyze each choice.

A. -27 to 54

The sum of the integers from -27 to 27 is 0. Thus we only need to add the integers from 28 to 54. However, we can see that, though there are 27 integers from 28 to 54 (as many as from 1 to 27), each integer from 28 to 54 is greater than any of the integers from 1 to 27, so the sums can’t be the same.

B. 0 to 28

The integers from 0 to 28 include those from 1 to 27, plus two more integers: 0 and 28. Though adding an extra 0 will not change the sum, adding an extra 28 will. So the sums can’t be the same.

C. 15 to 45

There are 31 integers from 15 to 45 and if we exclude the last 4 integers, each of these is greater than the corresponding ones from 1 to 27. So the sum of the integers from 15 to 45 will be greater than the sum of the integers from 1 to 27. So the sums can’t be same.

D. 38 to 46

Though each integer from 38 to 46 seems to be much larger than those from 1 to 27, there are fewer integers from 38 to 46 than from 1 to 27 (9 integers vs. 27 integers). So the sums could be equal. The only way we can know that for sure is to calculate each sum. To do that, we can use the formula: sum = average x quantity.

Sum of integers 1 to 27 = (1 + 27)/2 x 27 = 14 x 27 = 378

Sum of integers 38 to 46 = (38 + 46)/2 x 9 = 42 x 9 = 378.

We see that they do have an equal sum. Thus, the answer is D.

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