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If X is the sum of first 50 positive even integers and Y is [#permalink]
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11 Sep 2012, 05:11
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If X is the sum of first 50 positive even integers and Y is the sum of first 50 positive odd integers, what is the value of xy? A. 0 B. 25 C. 50 D. 75 E. 100  Please try to explain your answers
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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nishtil wrote: If X is the sum of first 50 positive even integers and Y is the sum of first 50 positive odd integers, what is the value of xy?
A. 0 B. 25 C. 50 D. 75 E. 100
 Please try to explain your answers We can solve this question even if we don't know any formula for such sums: First even minus first odd = 21 = 1; The sum of first 2 even integers minus the sum of first 2 odd integers = (2+4)(1+3) = 2; The sum of first 3 even integers minus the sum of first 3 odd integers = (2+4+6)(1+3+5) = 3; ... We can see the patterns here, so the sum of first 50 positive even integers minus the sum of first 50 positive odd integers will be 50. Answer: C. OR: each even minus its preceding odd is one, so xy=50 (xy=(even1+even2+...+even50)(odd1+odd2+..+odd50)=(even1odd1)+(even2odd2)+...+(even50odd50)=1+1+...+1=50). Answer: C. Hope it's clear.
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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Nice way bunuel..!! and i did it in this way.. first even and last even ..2 and 100 respectively, 100+2=102/2=51..51 is average of first 50 even integers=num of terms*average =50*51=2550..so sum of first 50 integers is 2550 first odd and last odd of first 50 intergers is =99 and 1..99+1=100/2=50.. so 50*50=2500..sum of first 50 odd integers is 2500.. 25502500=50... ans c.
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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12 Sep 2012, 10:37
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nishtil wrote: If X is the sum of first 50 positive even integers and Y is the sum of first 50 positive odd integers, what is the value of xy?
A. 0 B. 25 C. 50 D. 75 E. 100
 Please try to explain your answers Let's use this process, x= 2+ 4+ 6+ ...... (up to 50th term) y= 1+ 3 +5+ .......(up to 50th term)  (xy)= 1+1+1+ .......... up to 50th term = 50
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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12 Sep 2012, 11:03
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Shortcut way without even a single calculation: Sum of first 'n' even integers is given by  n(n+1) Sum of first 'n' odd integers is given by  n^2 x = n(n+1) = 50 x 51 y= n^2 = 50 x 50 xy = 50 (5150) = 50 (1) = 50 Answer C Hope It helps
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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26 Mar 2013, 15:12
fameatop wrote: Shortcut way without even a single calculation:
Sum of first 'n' even integers is given by  n(n+1) Sum of first 'n' odd integers is given by  n^2
x = n(n+1) = 50 x 51 y= n^2 = 50 x 50 xy = 50 (5150) = 50 (1) = 50 Answer C
Hope It helps do these above formulas always hold true? And what is the formula for the first 100 positive integers?



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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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27 Mar 2013, 04:33
jmuduke08 wrote: fameatop wrote: Shortcut way without even a single calculation:
Sum of first 'n' even integers is given by  n(n+1) Sum of first 'n' odd integers is given by  n^2
x = n(n+1) = 50 x 51 y= n^2 = 50 x 50 xy = 50 (5150) = 50 (1) = 50 Answer C
Hope It helps do these above formulas always hold true? And what is the formula for the first 100 positive integers? Sum of n first positive integers: \(1+2+...+n=\frac{1+n}{2}*n\). So, the sum of 100 first positive integers is (1+100)/2*100. Sum of n first positive odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n1\). Given \(n=5\) first odd positive integers, then their sum equals to \(1+3+5+7+9=5^2=25\). Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n\)\(=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\). For more check here: mathnumbertheory88376.htmlHope it helps.
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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30 Mar 2013, 13:06
Bunuel wrote: jmuduke08 wrote: fameatop wrote: Shortcut way without even a single calculation:
Sum of first 'n' even integers is given by  n(n+1) Sum of first 'n' odd integers is given by  n^2
x = n(n+1) = 50 x 51 y= n^2 = 50 x 50 xy = 50 (5150) = 50 (1) = 50 Answer C
Hope It helps do these above formulas always hold true? And what is the formula for the first 100 positive integers? Sum of n first positive integers: \(1+2+...+n=\frac{1+n}{2}*n\). So, the sum of 100 first positive integers is (1+100)/2*100. Sum of n first positive odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n1\). Given \(n=5\) first odd positive integers, then their sum equals to \(1+3+5+7+9=5^2=25\). Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n\)\(=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\). For more check here: mathnumbertheory88376.htmlHope it helps. Bunuel, if I wanted to find the sum of the even integers between 26 and 62, inclusive, then the formula above for even integers n(n+1) would not work, because this formula is only for the first n even positive integers meaning we would need to start at 2. Is this the correct way to think about it?



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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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[quote= Bunuel, if I wanted to find the sum of the even integers between 26 and 62, inclusive, then the formula above for even integers n(n+1) would not work, because this formula is only for the first n even positive integers meaning we would need to start at 2. Is this the correct way to think about it?[/quote] Sum of x consecutive even integers = 2xn + x(x+1)/2 (n = (First term of series/2)  1) Sum of x consecutive odd integers = (x+n)^2  (n)^2 (n=(first term of series  1)/2)
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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16 Apr 2013, 11:56
X = 2+4+6+8+.... +100 Y = 1+3+5+7+.... +99
XY= (21) + (43) + (65) + .... +(10099) there are 50 terms XY= 1 + 1 + 1 + .... +1 50 terms => XY=50 Answer C



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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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25 Apr 2013, 01:30
sum of the 1st even integers =n(n+2)=25*27 sum of the 1st odd integers =k^2=25*25 25*2725*25=25*(2725)=50
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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I'll post two ways to solve this question : The formal way (in order for you guys to better understand the theory behind it) and the GMAT way (within the 2minute scope). Let's start. 1st method : The GMAT wayBunuel actually proposed the easiest and fastest method to solve this question. That is you should look for patterns through examples. First 2 even numbers : 2, 4 => summed up : (2+4) = 6 First 2 odd numbers : 1, 3 => summed up : (1+3) = 4 Their difference will be 2. First 3 even numbers : 2, 4, 6 => summed up : (2+4+6) = 12 First 3 odd numbers : 1, 3, 5 => summed up : (1+3+5) = 9 Their difference will be 3. And so forth. So eventually the difference between the sum of the first 50 even integers and the first 50 odd integers will be 50. Which is answer choice C. 2nd method : The formal wayTo use this method you should be familiar and comfortable with :  The general form of an even integer which is 2n ;  The general form of an odd integer which is 2n+1 ;  Counting the number of consecutive integers within a list which is given by the following formula : (Last number  First number) + 1 ;  The sum operator and manipulating it. Therefore the difference between the sum of the 50 first even integers and the 50 first odd integers is written as such : Attachment:
pic3.jpg [ 6.93 KiB  Viewed 14486 times ]
You'll notice two things :  I chose to index the sums from 0 to 49 since the first even integer is 0 and the first odd integer is 1 ;  I inverted the difference since I've written YX instead of XY. This is due to the general form of the odd integer which if left as the original question stem suggests would leave me with a (1) instead of 1. If we develop the difference above we get : Attachment:
pic4.jpg [ 10.55 KiB  Viewed 14482 times ]
Which, unsuprisingly, yields 50 which is answer choice C. Note that you can combine two sums if and only if they have the same index range (0 to 49 in both cases). Hope that helped.



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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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08 Dec 2013, 22:38
Y not consider zero as zero is also an even number..... correct me if i m wrong Kindly let me know



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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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09 Dec 2013, 01:16
Question: Is the number 0 even or odd? Answer: 0/2 = 0. The result is an integer, so the number 0 is divisible by 2. As a result, the number 0 is even i read this in one of the manhatttan forums...hence the doubt..... could you throw some light based on the above info



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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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09 Dec 2013, 01:23
iyersu wrote: Question: Is the number 0 even or odd? Answer: 0/2 = 0. The result is an integer, so the number 0 is divisible by 2. As a result, the number 0 is even i read this in one of the manhatttan forums...hence the doubt..... could you throw some light based on the above info As written above: yes, zero is an even number but the questions talks about positive even numbers and since zero is neither positive nor negative, we do not consider 0 for this question. THEORY: 1. EVEN/ODDAn even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. An odd number is an integer that is not evenly divisible by 2. According to the above both negative and positive integers can be even or odd. 2. ZEROZero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself). Hope it helps.
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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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08 Aug 2016, 13:49
sum of first 5 even integers=30 sum of first 5 odd integers=25 3025=5 5*(50/5)=50



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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
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12 Oct 2017, 18:14
nishtil wrote: If X is the sum of first 50 positive even integers and Y is the sum of first 50 positive odd integers, what is the value of xy?
A. 0 B. 25 C. 50 D. 75 E. 100 The sum of the first 50 positive even integers is: sum = average x quantity sum = (100 + 2)/2 x 50 = 51 x 50 The sum of the first 50 positive odd integers is: sum = (99 + 1)/2 x 50 = 50 x 50 Thus, x  y is 51 x 50  50 x 50 = 50(51  50) = 50. Alternate solution: The first 50 positive even integers are: 2, 4, 6, 8, …, 98, 100. The first 50 positive odd integers are: 1, 3, 5, 7, …, 97, 99. We see that each even integer is 1 more than its odd counterpart (2 is 1 more than 1, 4 is 1 more than 3, etc). Since there are 50 numbers in each set, the sum of the even integers will be 50 x 1 = 50 more than the sum of the odd integers. Answer: C
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