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If set N contains only consecutive positive integers, what [#permalink]
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06 Nov 2005, 10:32
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If set N contains only consecutive positive integers, what is the sum of the numbers in set N ? (1) Nineteen times the sum of the first number in the set and the last number in the set is 1729 (2) There are 38 numbers in the set.
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Last edited by Bunuel on 12 Oct 2013, 17:28, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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I will go with C
Stmt 1) Let f be the first number and l be the last number
19 ( f + l ) = 1729
> f + l = 91 (A)
Hence INSUFF
Stmt 2) is INSUFF.
Combining, we can calculate 'f' from (A) and Stmt 2  Total 38 numbers. Hence suff.
Pls let me know if I have missed anything.



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Re: ds  numbers N [#permalink]
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12 Oct 2013, 17:24
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christoph wrote: If set N contains only consecutive positive integers, what is the sum of the numbers in set N? (1) Nineteen times the sum of the first number in the set and the last number in the set is 1729 (2) There are 38 numbers in the set. I think i'll go with (C) on this one. Please allow me to explain. So they are asking for the sum of the set of consecutive integers. We therefore need 2 things: 1) # of terms = Last  First + 1 2) Average of terms = (First + Last) / 2 Statement 1: We are given that 19 (F+L) = 1729 Then we can find F+L = 91. Hence, we can find the average. Now can we find Last  First? I'm afraid this is not possible. Statement 2: We are given the number of terms but we know nothing about the average of the terms. Hence Insuff (1) and (2): Now we have everything to find the sum of the set. # of terms from Statement 2 Average from Statement 1. Voila Hence answer is (C) Kudos if it helps!



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Re: If set N contains only consecutive positive integers, what [#permalink]
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17 Dec 2013, 05:55
christoph wrote: If set N contains only consecutive positive integers, what is the sum of the numbers in set N ?
(1) Nineteen times the sum of the first number in the set and the last number in the set is 1729 (2) There are 38 numbers in the set. I was able to get the answer using A only , please tell me what I have done wrong sum of n consecutive terms =( n/2)( F+L), where n is the number of terms , f = first term and L is the last term now can 19 be expressed as n/2 ? of course we can , so we can write 38/2(f+L) = 1729 > 19(F+L)=1729 this is the exact expression for the sum of n consecutive integers , hence the number of terms is 38 and the sum is 1729 Please let me know if I have erred somewhere. statement 1 : 19(F+L)=1729 > (38/2) ( F+L)  > (n/2)( F+L) = 1729
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Re: If set N contains only consecutive positive integers, what [#permalink]
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17 Dec 2013, 07:35
stne wrote: christoph wrote: If set N contains only consecutive positive integers, what is the sum of the numbers in set N ?
(1) Nineteen times the sum of the first number in the set and the last number in the set is 1729 (2) There are 38 numbers in the set. I was able to get the answer using A only , please tell me what I have done wrong sum of n consecutive terms =( n/2)( F+L), where n is the number of terms , f = first term and L is the last term now can 19 be expressed as n/2 ? of course we can , so we can write 38/2(f+L) = 1729 > 19(F+L)=1729 this is the exact expression for the sum of n consecutive integers , hence the number of terms is 38 and the sum is 1729 Please let me know if I have erred somewhere. statement 1 : 19(F+L)=1729 > (38/2) ( F+L)  > (n/2)( F+L) = 1729 We need to find the sum of N consecutive integers. The sum = \(\frac{(first + last)}{2}*N\). From (1) we have that \(19*(first + last) = 1729\) > \((first + last) = 91\). So, we need to find the value of \(\frac{(first + last)}{2}*N = 91*\frac{N}{2}\). How can you find the value of this expression? You cannot! for example, N can be 2 (45, 46) and in this case the sum is 91, or N can be 90 (1, 2, ..., 90) and in this case the sum is 4095. Hope it helps.
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Re: If set N contains only consecutive positive integers, what [#permalink]
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18 Dec 2013, 04:58
Bunuel wrote: stne wrote: christoph wrote: If set N contains only consecutive positive integers, what is the sum of the numbers in set N ?
(1) Nineteen times the sum of the first number in the set and the last number in the set is 1729 (2) There are 38 numbers in the set. I was able to get the answer using A only , please tell me what I have done wrong sum of n consecutive terms =( n/2)( F+L), where n is the number of terms , f = first term and L is the last term now can 19 be expressed as n/2 ? of course we can , so we can write 38/2(f+L) = 1729 > 19(F+L)=1729 this is the exact expression for the sum of n consecutive integers , hence the number of terms is 38 and the sum is 1729 Please let me know if I have erred somewhere. statement 1 : 19(F+L)=1729 > (38/2) ( F+L)  > (n/2)( F+L) = 1729 We need to find the sum of N consecutive integers. The sum = \(\frac{(first + last)}{2}*N\). From (1) we have that \(19*(first + last) = 1729\) > \((first + last) = 91\). So, we need to find the value of \(\frac{(first + last)}{2}*N = 91*\frac{N}{2}\). How can you find the value of this expression? You cannot! for example, N can be 2 (45, 46) and in this case the sum is 91, or N can be 90 (1, 2, ..., 90) and in this case the sum is 4095. Hope it helps. Thank you for your reply. I could be wrong again, most probably am but I am just trying to see the light and find out what is wrong with my logic suppose I was told that (n/2)(F+L) = 1729 where n is the number of terms in a consecutive series and f is the first term and l is the last term , Then I could immediately say that the sum of these n consecutive terms is 1729, could I not? now aren't we give the exact same thing indirectly? We are told 19(F+L)= 1729 which is the same as saying (38/2)(F+L)= 1729 so looking at this and comparing with (n/2)(F+L)= 1729 why cannot I say n=38?
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Re: If set N contains only consecutive positive integers, what [#permalink]
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18 Dec 2013, 07:05
stne wrote: Bunuel wrote: stne wrote: I was able to get the answer using A only , please tell me what I have done wrong
sum of n consecutive terms =( n/2)( F+L), where n is the number of terms , f = first term and L is the last term now can 19 be expressed as n/2 ? of course we can , so we can write 38/2(f+L) = 1729 > 19(F+L)=1729 this is the exact expression for the sum of n consecutive integers , hence the number of terms is 38 and the sum is 1729
Please let me know if I have erred somewhere.
statement 1 : 19(F+L)=1729 > (38/2) ( F+L)  > (n/2)( F+L) = 1729
We need to find the sum of N consecutive integers. The sum = \(\frac{(first + last)}{2}*N\). From (1) we have that \(19*(first + last) = 1729\) > \((first + last) = 91\). So, we need to find the value of \(\frac{(first + last)}{2}*N = 91*\frac{N}{2}\). How can you find the value of this expression? You cannot! for example, N can be 2 (45, 46) and in this case the sum is 91, or N can be 90 (1, 2, ..., 90) and in this case the sum is 4095. Hope it helps. Thank you for your reply. I could be wrong again, most probably am but I am just trying to see the light and find out what is wrong with my logic suppose I was told that (n/2)(F+L) = 1729 where n is the number of terms in a consecutive series and f is the first term and l is the last term , Then I could immediately say that the sum of these n consecutive terms is 1729, could I not? now aren't we give the exact same thing indirectly? We are told 19(F+L)= 1729 which is the same as saying (38/2)(F+L)= 1729 so looking at this and comparing with (n/2)(F+L)= 1729 why cannot I say n=38? Suppose I tell you that the sum of the first and the last terms of n positive consecutive integers is 5. Can you find the sum of n consecutive integers? The answer is NO. The set can be {2, 3} or {1, 2, 3, 4}. In the first case the sum is 5 and in the second case the sum is 10.
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Re: If set N contains only consecutive positive integers, what [#permalink]
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18 Dec 2013, 08:02
I agree , sum of first and last term is not enough to give the number of consecutive terms in a sequence but we are given more than that, we are given 19 * ( F +T) this immediately forced me to tell that there are 38 consecutive terms in the sequence . because 19(F+T) can we written as 38/2(F+T) which can be compared to n/2(F+T) I know 19 can we written in other forms as well but our aim is to get it in the form n/2 where n is a integer. If I was given 20 *(F+T) and we were dealing with consecutive terms then could I not tell there are 40 terms in the sequence, by changing eqn. to 40/2(F+L)? So F+T = 1729/19 if we look at this form of the equation , then certainly this seems insufficient but after algebraic manipulation we get the equation in the form 19(F+T) = 1729 this form seems sufficient I have learnt that in algebraic manipulation the same equation in certain form can give the answer well as in other forms it may not give the answer. F+L = 1729/19 = insufficient but 19( F+T) = 1729 or (38/2) (F+L) = 1729 or (n/2)(F+L) is sufficient. Is there a possibility that this could be a case here? Is there an issue of algebraic manipulation Thank you for trying to make me understand.
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Re: If set N contains only consecutive positive integers, what [#permalink]
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18 Dec 2013, 08:07
stne wrote: I agree , sum of first and last term is not enough to give the number of consecutive terms in a sequence but we are given more than that, we are given 19 * ( F +T) this immediately forced me to tell that there are 38 consecutive terms in the sequence .
because 19(F+T) can we written as 38/2(F+T) which can be compared to n/2(F+T) I know 19 can we written in other forms as well but our aim is to get it in the form n/2 where n is a integer.
If I was given 20 *(F+T) and we were dealing with consecutive terms then could I not tell there are 40 terms in the sequence, by changing eqn. to 40/2(F+L)?
So F+T = 1729/19 if we look at this form of the equation , then certainly this seems insufficient but after algebraic manipulation we get the equation in the form 19(F+T) = 1729 this form seems sufficient
I have learnt that in algebraic manipulation the same equation in certain form can give the answer well as in other forms it may not give the answer.
F+L = 1729/19 = insufficient but 19( F+T) = 1729 or (38/2) (F+L) = 1729 or (n/2)(F+L) is sufficient.
Is there a possibility that this could be a case here? Is there an issue of algebraic manipulation
Thank you for trying to make me understand. Again from (38/2) (F+L) = 1729 you cannot tell that n=38.
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Re: If set N contains only consecutive positive integers, what [#permalink]
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18 Dec 2013, 11:27
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Stne,
for the set {2,3} and {1,2,3,4} , if i were given 2* (F+T) = 10, as per you logic i can conclude there are 4 elements in the set. But that is not true as {2,3} can also be a set that satisfies 2* (F+T) = 10. The sum of {2,3} and {1,2,3,4} are different.



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Re: If set N contains only consecutive positive integers, what [#permalink]
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19 Dec 2013, 11:04
anjancgc wrote: Stne,
for the set {2,3} and {1,2,3,4} , if i were given 2* (F+T) = 10, as per you logic i can conclude there are 4 elements in the set. But that is not true as {2,3} can also be a set that satisfies 2* (F+T) = 10. The sum of {2,3} and {1,2,3,4} are different. Makes sense, you have managed to convince me.That's a wonderful first post by the way. +1 to you and also to Bunuel for his consistent assistance.
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If set N contains only consecutive positive integers, what [#permalink]
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15 Jul 2016, 09:52
christoph wrote: If set N contains only consecutive positive integers, what is the sum of the numbers in set N ?
(1) Nineteen times the sum of the first number in the set and the last number in the set is 1729 (2) There are 38 numbers in the set. Statment 1) 19 (First+last)=1729 First + Last=1729/19=91 We don't know first and last digit InSufficient (2) There are 38 numbers in the set Numbers can be anything (2,4,6,8,...........\(38^{th}\)term) or (120,122,124,.......38^{th} term) Insufficient Merging both SUFFICIENT Here is the trick, most people don't know that If the total number of terms in a set containing even numbers is even then average of first and last term of set gives the average of the entire set. For example 2,4,6,8 standard way to find average= (2+4+6+8)/4=20/4= 5 The short cut method to find average is =(first+last)/2 ==> (2+8)/2= 5 This relationship hold true in case of even numbers having even terms So we now know that 19(f+l)=1729 f+l=91 (f+l) =91/2=40.5 thats tha average To find the sum ;multiply it with the number of terms 38 * 40.5 = 1539 C is the answer
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