Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 18 Aug 2009
Posts: 242

If the sum of the consecutive integers from –42 to n
[#permalink]
Show Tags
Updated on: 27 Jul 2012, 02:06
Question Stats:
67% (02:31) correct 33% (02:37) wrong based on 928 sessions
HideShow timer Statistics
If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n? A. 47 B. 48 C. 49 D. 50 E. 51
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by gmattokyo on 08 Nov 2009, 02:57.
Last edited by Bunuel on 27 Jul 2012, 02:06, edited 1 time in total.
Edited the question and added the OA.




Math Expert
Joined: 02 Sep 2009
Posts: 61471

Re: Sum of consecutive integers
[#permalink]
Show Tags
08 Nov 2009, 04:44
gmattokyo wrote: If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?
A. 47 B. 48 C. 49 D. 50 E. 51 # of terms =\(42+1+n=(n+43)\) \(Sum=372=(n+43)*\frac{(n42)}{2}\) \(744=(n+43)*(n42)\) \(n=50\) OR 42 terms after zero and 42 terms below zero will total 0. So, our new question will be consecutive integers with first term 43 have sum 372, what is the last term: \(\frac{43+n}{2}*(n43+1)=372\) \((n+43)*(n42)=744\) \(n=50\) Answer: D (50)
_________________




Manager
Joined: 18 Aug 2009
Posts: 242

Re: Sum of consecutive integers
[#permalink]
Show Tags
08 Nov 2009, 05:17
Bunuel wrote: gmattokyo wrote: If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?
A. 47 B. 48 C. 49 D. 50 E. 51 # of terms =\(42+1+n=(n+43)\) \(Sum=372=(n+43)*\frac{(n42)}{2}\) \(744=(n+43)*(n42)\) \(n=50\) OR 42 terms after zero and 42 terms below zero will total 0. So, our new question will be consecutive integers with first term 43 have sum 372, what is the last term: \(\frac{43+n}{2}*(n43+1)=372\) \((n+43)*(n42)=744\) \(n=50\) Answer: D (50) Nice solution. Correct as always. I was actually thinking of having a ballpark and reach the answer from 43 to 50. Add from 43, 44, 45.. to reach 372, need at least 7 (43x7=301) and less than 9 (43x9=387). Did you use any intermediary formulas for this part? (n+43)*(n42)=744 n=50



Math Expert
Joined: 02 Sep 2009
Posts: 61471

Re: Sum of consecutive integers
[#permalink]
Show Tags
08 Nov 2009, 05:49
gmattokyo wrote: Nice solution. Correct as always. I was actually thinking of having a ballpark and reach the answer from 43 to 50. Add from 43, 44, 45.. to reach 372, need at least 7 (43x7=301) and less than 9 (43x9=387).
Did you use any intermediary formulas for this part? (n+43)*(n42)=744 n=50 Sum of consecutive integers or sum of terms in any evenly spaced set (AP): \(\frac{first term+last term}{2}*number of terms\) In original question: First term: 42 Last term: n Number of terms: 42+n+1=n+43 \(\frac{(n42)}{2}*(n+43)=372\) \((n42)*(n+43)=744\)
_________________



Intern
Joined: 21 Jul 2009
Posts: 10

Re: Sum of consecutive integers
[#permalink]
Show Tags
08 Nov 2009, 06:57
Bunuel wrote: gmattokyo wrote: Nice solution. Correct as always. I was actually thinking of having a ballpark and reach the answer from 43 to 50. Add from 43, 44, 45.. to reach 372, need at least 7 (43x7=301) and less than 9 (43x9=387).
Did you use any intermediary formulas for this part? (n+43)*(n42)=744 n=50 Sum of consecutive integers or sum of terms in any evenly spaced set (AP): \(\frac{first term+last term}{2}*number of terms\) In original question: First term: 42 Last term: n Number of terms: 42+n+1=n+43 \(\frac{(n42)}{2}*(n+43)=372\) \((n42)*(n+43)=744\) Question is how we can solve such complex algebraic equation in short time. I suggest POE method. 1. Choose option C, put this value in the equation (First selection is based on instinct ) => (49+43)(4942) = 744 => 92 * 6 = 744 => 553 = 744 Here we can conclude that n should be greater than 49 . Options A,B and C are eliminated 2. Chose option D, put this value in the equation =>(50+43)(5042) = 744 => 93 * 8 = 744 => 744 = 744 Bingo !!! => n is 50 Is there any other method available?



Manager
Joined: 18 Aug 2009
Posts: 242

Re: Sum of consecutive integers
[#permalink]
Show Tags
08 Nov 2009, 18:31
Thanks guyz for the various ways. OA is D pray the mind in open enough during the actual exam to be able to think out benchmarking and the other techniques!
Here's some more addition from Bunuel.. Look at answer choices:
A. 47 B. 48 C. 49 D. 50 E. 51
n42 n+43
It can not be A as 4742=5 and 744 is not a multiple of 5. It can not be B as 4842=6 and 43+51=91 their multiple will have 6 for the last digit
It can not be C as 4942=7 and 744 is not a multiple of 7. It can not be E as 5142=9 and 744 is not a multiple of 9.



Manager
Joined: 23 Jun 2009
Posts: 53

Re: Sum of consecutive integers
[#permalink]
Show Tags
13 Nov 2009, 17:48
Bunuel wrote: gmattokyo wrote: Nice solution. Correct as always. I was actually thinking of having a ballpark and reach the answer from 43 to 50. Add from 43, 44, 45.. to reach 372, need at least 7 (43x7=301) and less than 9 (43x9=387).
Did you use any intermediary formulas for this part? (n+43)*(n42)=744 n=50 Sum of consecutive integers or sum of terms in any evenly spaced set (AP): \(\frac{first term+last term}{2}*number of terms\) In original question: First term: 42 Last term: n Number of terms: 42+n+1=n+43 \(\frac{(n42)}{2}*(n+43)=372\) \((n42)*(n+43)=744\) Bunel: I understand how you are calculating # of terms given the fact that the sum is +ve. What if the sum were ve? In that case "n" can be smaller or greater than 42. How do we calculate # of terms in that case. Thank you in advance.



Manager
Joined: 24 Jul 2009
Posts: 64
Location: United States

Re: Sum of consecutive integers
[#permalink]
Show Tags
13 Nov 2009, 19:25
I forgot about AP..... Bunnel you are amazing.......... This is how i tried which is ridiculous as it took arnd 4 mins Step1: Obviously, we need not worry about numbers till 42 as they will negate with the negative numbers when addition is performed: [strike]42, 41........0........., 41, 42[/strike], 43...n => 43+44+...+n = 372 Step2: the series can be represented by: 43 + 0, 43+1, 43+2.....43+(x1)# = 372 # > Series starting with zero Step3: 43x + [(x1)x]/2 = 372 > Solve quadratic equation. Step4: x^2  x + 86x = 372 => x^2 + 85n 744 = 0 => x^2 + 93x 8x  744 = 0 => (x+93)(x8)=0 x=8. Step5: But mind you the series ends with x1. Hence 43 + (81) = 50.



Manager
Joined: 05 Jun 2009
Posts: 65

Re: Sum of consecutive integers
[#permalink]
Show Tags
13 Nov 2009, 19:32
42  0 = 0  42 thus 372 must equal 43  X
If you average out to 50 and than kinda thumb it you see that there are 8 #'s over 42 that = 372 42 + 8 = 50
D



Math Expert
Joined: 02 Sep 2009
Posts: 61471

Re: Sum of consecutive integers
[#permalink]
Show Tags
13 Nov 2009, 21:36
Casinoking wrote: Bunel: I understand how you are calculating # of terms given the fact that the sum is +ve. What if the sum were ve? In that case "n" can be smaller or greater than 42. How do we calculate # of terms in that case. Thank you in advance. For the number of terms in the set of consecutive integers, it doesn't matter whether the sum is positive or negative. If we are told that integers from a to b are inclusive then the # of terms would be: last term (biggest)first term(smallest)+1=ba+1. eg. set from 42 to 35 inclusive: 35(42)+1=8. OR set from 59 to 42 inclusive: 42(59)+1=18 OR set from 1 to 4 inclusive: 4(1)+1=6 When a and b are not inclusive, # of terms between a and b: ba1
_________________



Director
Joined: 23 Apr 2010
Posts: 502

Re: Sum of consecutive integers
[#permalink]
Show Tags
22 Dec 2010, 05:48
Bunuel, is there a quick way to solve the equation:
744 = (n+43)(n42)
?
I mean the question is very simple. But the equation looks very tedious to me.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10113
Location: Pune, India

Re: Sum of consecutive integers
[#permalink]
Show Tags
22 Dec 2010, 18:13
gmattokyo wrote: If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?
A. 47 B. 48 C. 49 D. 50 E. 51 As suggested above, the quickest way to solve this question would be to ballpark it. We know we are looking for 43 + 44 + 45...... = 372 Now 40*8 = 320.... 372 is more than 320 40*9 = 360.... but it is very close to 372. When we add the numbers, we add 48, 49 etc which have an excess of 8, 9 etc as compared to 40. Hence, we need the sum to be some difference away from 372. Therefore, we need to add only 8 numbers, not 9. Keep in mind, when going from 43 to n to get 8 numbers, n will be 50, not 51. 43 to 50 give us 8 numbers (50  43 + 1).
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 22 Apr 2011
Posts: 112
Schools: Mccombs business school, Mays business school, Rotman Business School,

Re: Sum of consecutive integers
[#permalink]
Show Tags
27 Jul 2012, 02:04
hello bunuel, i do understand the process. but i really want to know is there any short cut for 744=(n+43)(n42) to arrive n=50
thanks in advance



Director
Joined: 22 Mar 2011
Posts: 581
WE: Science (Education)

Re: Sum of consecutive integers
[#permalink]
Show Tags
27 Jul 2012, 03:02
VeritasPrepKarishma wrote: gmattokyo wrote: If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?
A. 47 B. 48 C. 49 D. 50 E. 51 As suggested above, the quickest way to solve this question would be to ballpark it. We know we are looking for 43 + 44 + 45...... = 372 Now 40*8 = 320.... 372 is more than 320 40*9 = 360.... but it is very close to 372. When we add the numbers, we add 48, 49 etc which have an excess of 8, 9 etc as compared to 40. Hence, we need the sum to be some difference away from 372. Therefore, we need to add only 8 numbers, not 9. Keep in mind, when going from 43 to n to get 8 numbers, n will be 50, not 51. 43 to 50 give us 8 numbers (50  43 + 1). We can also use some divisibility properties, because we have a sequence of evenly spaced integers. If the number of terms in the sequence is odd, the sum of the numbers is a multiple of the middle term. If the number of terms is even, the total sum is a multiple of the sum of the two middle terms. If we factorize 372 we get 2*2*3*31. By combining the factors we cannot get a divisor of 372 in the range of 4050, but 3*31 = 93 = 46+47. So, we have an even number of terms, with 46 and 47 the two middle terms, first term 43, then the last must be 50. Not shorter than your solution, but it's fun to play with divisibility rules, sometimes just for the sake of practice...
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Director
Joined: 22 Mar 2011
Posts: 581
WE: Science (Education)

Re: Sum of consecutive integers
[#permalink]
Show Tags
27 Jul 2012, 04:50
nonameee wrote: Bunuel, is there a quick way to solve the equation:
744 = (n+43)(n42)
?
I mean the question is very simple. But the equation looks very tedious to me. Denote n  42 = x, where x is a positive integer, and of course, n + 43 = x + 85. You have to solve x(x + 85) = 744. Look for the factorization of 744: 744 = 2 * 2 * 2 * 3 * 31 = 8 * 93 (8 * 90 = 720, so you should try factors around 8 and 90) Therefore, x = 8 and n = 50.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10113
Location: Pune, India

Re: Sum of consecutive integers
[#permalink]
Show Tags
27 Jul 2012, 21:40
alchemist009 wrote: hello bunuel, i do understand the process. but i really want to know is there any short cut for 744=(n+43)(n42) to arrive n=50
thanks in advance n is the number of terms so it must be positive. It must be greater than 42 to ensure that (n  42) is not negative. 744 needs to be written as a product of two numbers. Factorize 744. You get 744 = 8*93 = 8*3*31 We need one of the numbers to be greater than 85 to account for (n+43) where n is greater than 42. We see 8*93 is possible. If the (n+43) factor is 93 but n must be 50.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 11 Jan 2013
Posts: 14
Location: United States

Re: If the sum of the consecutive integers from –42 to n
[#permalink]
Show Tags
30 Jun 2013, 08:55
As explained above n must be larger than 42 to compensate for the negative numbers in the set.
That means that 372 is the sum of x number of consecutive integers >43 (i.e. 43 + 44 + ... + (43+x)), i.e. n = 43+x.
Solving for x: If we divide 372 by 43, we yield a remainder of 28.
x is the number of consecutive integers, whose sum is 28: 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. Thus x is 7 and the n is 43+x = 43+7=50.
The answer is D.



Intern
Joined: 04 May 2013
Posts: 38

Re: Sum of consecutive integers
[#permalink]
Show Tags
22 Jul 2013, 17:38
Bunuel wrote: gmattokyo wrote: If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?
A. 47 B. 48 C. 49 D. 50 E. 51 # of terms =\(42+1+n=(n+43)\) \(Sum=372=(n+43)*\frac{(n42)}{2}\) \(744=(n+43)*(n42)\) \(n=50\) OR 42 terms after zero and 42 terms below zero will total 0. So, our new question will be consecutive integers with first term 43 have sum 372, what is the last term: \(\frac{43+n}{2}*(n43+1)=372\) \((n+43)*(n42)=744\) \(n=50\) Answer: D (50) Can you please explain how is the number of terms 42 + n +1? Thanks



Math Expert
Joined: 02 Sep 2009
Posts: 61471

Re: Sum of consecutive integers
[#permalink]
Show Tags
22 Jul 2013, 22:40
jjack0310 wrote: Bunuel wrote: gmattokyo wrote: If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?
A. 47 B. 48 C. 49 D. 50 E. 51 # of terms =\(42+1+n=(n+43)\) \(Sum=372=(n+43)*\frac{(n42)}{2}\) \(744=(n+43)*(n42)\) \(n=50\) OR 42 terms after zero and 42 terms below zero will total 0. So, our new question will be consecutive integers with first term 43 have sum 372, what is the last term: \(\frac{43+n}{2}*(n43+1)=372\) \((n+43)*(n42)=744\) \(n=50\) Answer: D (50) Can you please explain how is the number of terms 42 + n +1? Thanks # of term from a to b, inclusive is ba+1. # of term from 42 to n, inclusive is n(42)+1=42+n+1.
_________________



Intern
Joined: 04 May 2013
Posts: 38

Re: Sum of consecutive integers
[#permalink]
Show Tags
23 Jul 2013, 14:36
Bunuel wrote: jjack0310 wrote: Bunuel wrote: If the sum of the consecutive integers from –42 to n inclusive is 372, what is the value of n?
A. 47 B. 48 C. 49 D. 50 E. 51
# of terms =\(42+1+n=(n+43)\)
\(Sum=372=(n+43)*\frac{(n42)}{2}\)
\(744=(n+43)*(n42)\)
\(n=50\)
OR
42 terms after zero and 42 terms below zero will total 0. So, our new question will be consecutive integers with first term 43 have sum 372, what is the last term:
\(\frac{43+n}{2}*(n43+1)=372\)
\((n+43)*(n42)=744\)
\(n=50\)
Answer: D (50) Can you please explain how is the number of terms 42 + n +1? Thanks # of term from a to b, inclusive is ba+1. # of term from 42 to n, inclusive is n(42)+1=42+n+1. OK I get it. And finally, (N+43)*(N42) = 744 How is N = 50? Do you actually solve for N = 50? Multiply the two terms and subtract 744 or a easier and simpler way? Thanks




Re: Sum of consecutive integers
[#permalink]
23 Jul 2013, 14:36



Go to page
1 2 3
Next
[ 44 posts ]



