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Re: If n is a positive odd number, and the sum of all the even numbers bet [#permalink]
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14 Jul 2015, 01:29
D is the answer. I solve it by using choices. If you take D=159 and use AP formula's will give the sum of 2+4+6...+158 equal to multiplication of 79*80. Thanks



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Re: If n is a positive odd number, and the sum of all the even numbers bet [#permalink]
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14 Jul 2015, 03:09
Bunuel wrote: If n is a positive odd number, and the sum of all the even numbers between 1 and n is equal to the product of 79 and 80, then what is the value of n?
A. 125 B. 133 C. 151 D. 159 E. 177
Kudos for a correct solution. Solution  Given that, sum of all even numbers between 1 and n = 79*80 The sequence will be of the form {2,4,6......n1}. Total no. of terms = (n1)/2. Middle term = (initial+final)/2 = (2+n1)/2 = (n1)/2 + 1 Sum of the sequence = Total no. of terms * middle term = (n1)/2 * (n1)/2 + 1 = 79*80 (n1)/2 and (n1)/2 + 1 are consecutive integers. (n1)/2 = 79 > n=159. ANS D
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Re: If n is a positive odd number, and the sum of all the even numbers bet [#permalink]
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14 Jul 2015, 03:21
Bunuel wrote: If n is a positive odd number, and the sum of all the even numbers between 1 and n is equal to the product of 79 and 80, then what is the value of n?
A. 125 B. 133 C. 151 D. 159 E. 177
Kudos for a correct solution. Bunuel: I think I have seen this question here on GC already But anyways... Property: Sum of Even Integers from 2 to 2a = a*(a+1)so Sum of Even Integers from 2 to (n1) = [(n1)/2]*[(n1)/2 +1] = [(n1)/2]*[(n+1)/2] = 79*80 i.e. (n1)(n+1) = 158*160 i.e. n1 = 158 (For positive values of n) i.e. n = 159 Answer: Option D
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Re: The sum of the even numbers between 1 and n is 79*80, where [#permalink]
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18 Jul 2015, 05:52
Bunuel wrote: If n is a positive odd number, and the sum of all the even numbers between 1 and n is equal to the product of 79 and 80, then what is the value of n?
A. 125 B. 133 C. 151 D. 159 E. 177
Kudos for a correct solution. Sum of N cons. even number = N(N+1) where N = (First Term + Last Term)/2  1 we know that the first term is 2 and the last term is n1 (2+n1)/2  1 = 79 => n+1 = 160 => n=159 Option D



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Re: The sum of the even numbers between 1 and n is 79*80, where [#permalink]
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03 Aug 2015, 06:59
Bunuel wrote: nonameee wrote: Can I ask someone to take a look at this one and provide a short explanation? Thank you. The sum of the even numbers between 1 and n is 79*80, where k is an odd number, then n=? (A) 79 (B) 80 (C) 81 (D) 157 (E) 159 \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\) (check this: totallybasic94862.html and this: mathnumbertheory88376.html). Now, as as \(n\) is odd then the last even number between 1 and \(n\) will be \(n1\) and the first even number will be 2. So, there are \(\frac{(n1)2}{2}+1=\frac{n1}{2}\) even numbers (multiples of 2) between 1 and \(n\). Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms > \(\frac{2+(n1)}{2}*\frac{n1}{2}=79*80\) > \((n1)(n+1)=158*160\) > \(n=159\) Answer E. Hi Bunuel ! First of all thanks a lot for the GMAT math book that you have shared with us..it really is a very valuable resource.. Regarding this method...instead of using the formula of no. of multiples cant we use this logic (would like to know if the concept is correct) Sum of consecutive integers = [(first term + last term)/2] * no. of even terms like you rightly mentioned first term would be 2 and last term will be n1 . For the no. of terms , cant we consider this concept : if from 1 to n, (n is the last and an even integer) , then the no. of even integers between 1 to n is n/2 . ( eg 1 to 48 : no. of even integers 24 ; 1 to 100 : no. of even integers is 50) . Here since (n1) is the last even integer (between 1 to n) , hence the no. of even integers is (n1)/2 Just a thought..kindly guide me if i am wrong Thanks!



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Re: The sum of the even numbers between 1 and n is 79*80, where [#permalink]
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11 Jan 2016, 08:49
mohan514 wrote: The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=? sum of the numbers = 2+4+6+...+(n1)=79*80 this is an arithmetic series with first term 2 and last term n1. therefore, their sum = [(n1)/2][(2(n1)/2]=6320 n^2  1 = 6320*4 = 25280 n^2 = 25281 n = + or  159 n must be 159 as it must be positive.



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The sum of the even numbers between 1 and n is 79*80, where [#permalink]
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25 Jan 2017, 10:14
chan4312 wrote: The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?
(A) 79 (B) 80 (C) 81 (D) 157 (E) 159 The sum of the first k positive even numbers is k(k + 1). For example, the sum of the first 5 positive even numbers is 5(6) = 30. We can verify this by actually adding the first 5 positive even numbers: 2 + 4 + 6 + 8 + 10 = 30. Therefore, to solve this problem, we need to find the number of even numbers between 1 and n, in which n is an odd number. In this range, the smallest even number is 2 and the largest even number is (n  1) since n is odd. Thus, the number of even numbers between 1 and n is: [(n  1)  2]/2 + 1 = (n  1)/2  1 + 1 = (n  1)/2 Thus, the the sum of the even numbers between 1 and n is (n  1)/2 * [(n  1)/2 + 1]. We are given that this sum is equal to 79*80, so we can set (n  1)/2 * [(n  1)/2 + 1] = 79*80 and solve for n. However, since (n  1)/2 + 1 is 1 more than (n  1)/2 and 80 is one more than 79, (n  1)/2 must be 79 and (n  1)/2 + 1 must be 80. Now we can determine n: (n  1)/2 = 79 n  1 = 158 n = 159 Answer: E
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The sum of the even numbers between 1 and n is 79*80, where [#permalink]
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02 Feb 2017, 05:52



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The sum of the even numbers between 1 and n is 79*80, where [#permalink]
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29 Dec 2017, 04:50
Bunuel wrote: nonameee wrote: Can I ask someone to take a look at this one and provide a short explanation? Thank you. The sum of the even numbers between 1 and n is 79*80, where k is an odd number, then n=? (A) 79 (B) 80 (C) 81 (D) 157 (E) 159 \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\) (check this: http://gmatclub.com/forum/totallybasic94862.html and this: http://gmatclub.com/forum/mathnumbertheory88376.html). Now, as as \(n\) is odd then the last even number between 1 and \(n\) will be \(n1\) and the first even number will be 2. So, there are \(\frac{(n1)2}{2}+1=\frac{n1}{2}\) even numbers (multiples of 2) between 1 and \(n\). Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms > \(\frac{2+(n1)}{2}*\frac{n1}{2}=79*80\) > \((n1)(n+1)=158*160\) > \(n=159\) Answer E. Hello Bunuel how did you get this n1/2 from this (n1)2/2+1 ? did you simplify this (n1)2/2+1 if yes, please show how you did. Thanks!



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The sum of the even numbers between 1 and n is 79*80, where [#permalink]
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29 Dec 2017, 05:20
dave13 wrote: Bunuel wrote: nonameee wrote: Can I ask someone to take a look at this one and provide a short explanation? Thank you. The sum of the even numbers between 1 and n is 79*80, where k is an odd number, then n=? (A) 79 (B) 80 (C) 81 (D) 157 (E) 159 \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\) (check this: http://gmatclub.com/forum/totallybasic94862.html and this: http://gmatclub.com/forum/mathnumbertheory88376.html). Now, as as \(n\) is odd then the last even number between 1 and \(n\) will be \(n1\) and the first even number will be 2. So, there are \(\frac{(n1)2}{2}+1=\frac{n1}{2}\) even numbers (multiples of 2) between 1 and \(n\). Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms > \(\frac{2+(n1)}{2}*\frac{n1}{2}=79*80\) > \((n1)(n+1)=158*160\) > \(n=159\) Answer E. Hello Bunuel how did you get this n1/2 from this (n1)2/2+1 ? did you simplify this (n1)2/2+1 if yes, please show how you did. Thanks! First of all, brackets do matter. It's (n  1)/2, which is NOT the same as n  1/2. The same way ((n1)2)/2+1 is NOT the same as (n1)2/2+1. \(\frac{(n1)2}{2}+1=\frac{n3}{2}+1=\frac{n3+2}{2}=\frac{n3}{2}+\frac{2}{2}=\frac{n1}{2}\).
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