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If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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04 Apr 2007, 20:30
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If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a  b ? A. 1 B. 10 C. 19 D. 20 E. 21
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Last edited by Bunuel on 04 Nov 2014, 04:44, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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04 Apr 2007, 20:36
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The sum is 10.
The fastest way is to notice that since we're taking ab, all we are going to end up with is addings ten '1' since each even number is one larger than the odd.
Like 21 = 1, then 43 = 1.... 2019 = 1
That's just adding up a bunch of 1s.



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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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04 Apr 2007, 23:33
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faifai0714 wrote: Is there a faster way to calculate this other than adding the even numbers in A and adding the odd numbers in B?
A = 2+4+6 ... +20
B = 1+3+5 ...+19
We want AB
essentially 21 + 43 + 65 + ... + 2019
thus ... 1 + 1 + 1 .. + 1 (ten times)
Ans: 10 B



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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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05 Apr 2007, 01:04
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Answer is 10
Yes! there is really a faster way to solve it.
Sum of consecutive odd or even integers = (no. of odd or even ints) * (first int + last int) / 2
Here A = sum of even ints from 2 to 20, inclusive
number of even ints = 10,
first int + last int = 2+20 = 22
A = 10*22 / 2 = 110
B = sum of odd ints from 1 to 19, inclusive
number of odd ints = 10,
first int + last int = 1+19 = 20
A = 10*20 / 2 = 100
AB = 110  100 = 10
Hope it helps



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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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05 Apr 2007, 02:17
Good explanation ywilfred....



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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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12 Jan 2016, 06:02
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This is a solution from Beatthegmat: http://www.beatthegmat.com/ifaequals ... 38885.htmleven numbers: (202)/2 + 1 = 10 even integers. (20+2)/2 = 11 is the average of the even set. sum = avg*(#of elements) = 11*10 = 110 = a odd numbers: (191)/2 + 1 = 10 odd integers. (19+1)/2 = 10 is the average of the odd set. sum = avg*(#of elements) = 10*10 = 100 = b ab = 110  100 = 10. (B)
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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12 Jan 2016, 07:43
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faifai0714 wrote: If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a  b ?
A. 1 B. 10 C. 19 D. 20 E. 21 Hi, Another way.. Without getting into the sum etc, the answer can be found easily... 220 contains 10 integers and similarily 119 also has 10 integers.. smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers.. so ab=1*10=10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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07 Feb 2016, 03:18
chetan2u wrote: faifai0714 wrote: If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a  b ?
A. 1 B. 10 C. 19 D. 20 E. 21 Hi, Another way.. Without getting into the sum etc, the answer can be found easily... 220 contains 10 integers and similarily 119 also has 10 integers.. smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers.. so ab=1*10=10 how to count number of odd integers from 1 to n inclusive? Please advise
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If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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07 Feb 2016, 03:29
smartguy595 wrote: chetan2u wrote: faifai0714 wrote: If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a  b ?
A. 1 B. 10 C. 19 D. 20 E. 21 Hi, Another way.. Without getting into the sum etc, the answer can be found easily... 220 contains 10 integers and similarily 119 also has 10 integers.. smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers.. so ab=1*10=10 how to count number of odd integers from 1 to n inclusive? Please advise Hi, If n is odd, add 1to it and divide by 2... If n is even, straight way divide by 2... In this example, n is 19.. Add one,so 20.. 20/2=10.. smartguy595, at times it might be that you are given some integer in between to another integer ahead.. example, it may not be 1 to n, but n to x.. few points to note in that scenario.. 1) both x and n are eventotal integers will be (xn)+1... even will be 1 more than odd.. even= (xn)/2+1 and odd=(xn)/2.. 2) x is even and n is odd..total integers = xn +1.. even will be equal to odd.. even= odd={(xn)+1}/2 .. 3)x is odd and n is even..total integers = xn +1.. even will be equal to odd.. even= odd={(xn)+1}/2 .. 4) both x and n are oddtotal integers will be (xn)+1... odd will be 1 more than even.. odd= (xn)/2+1 and even=(xn)/2.. Your query on 1 to 19 will fall under 3) and 4) above.. hope this helps..
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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26 Mar 2016, 18:21
Here's a visual that should help.
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Screen Shot 20160326 at 6.18.49 PM.png [ 103.38 KiB  Viewed 11998 times ]
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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29 Aug 2016, 00:33
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faifai0714 wrote: If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a  b ?
A. 1 B. 10 C. 19 D. 20 E. 21 Solution : a = Sum of all even integers between 2 to 20 Theory : Sum of n even integers = n(n+1) Therefore a = n^2+n b = sum of all odd integers between 1 to 19 Theory : Sum of n odd integers = n^2 Therefore ab = n^2+nn^2 = n Here n = no of even integers between 2 to 20 = no of odd integers between 1 to 19 = 10 Therefore, ab = n = 10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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29 Aug 2016, 01:26
faifai0714 wrote: If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a  b ?
A. 1 B. 10 C. 19 D. 20 E. 21 Option B sum of even integers = 10*22/2= 110 sum of odd integers = 10*20/2 =100 a  b= 110100 =10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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02 Sep 2017, 05:14
We can list numbers as 2 4 6 .....20 and odd numbers 1 3 5 ....19 There are a total of 10 even 10 odd numbers so if we subtract each corresponding odd number from even numbers we will get 1 so total will be 10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]
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19 Mar 2018, 11:21
Hi All, Sequence questions are almost always based on a pattern. If you can spot the pattern, then you can avoid a lot of heavy calculations and deduce the correct answer in a much faster way. Here, if we compare the first term of A and the first term of B, we have… 21 = 1 Next, let's compare the second term of A and second term of B… 43 = 1 This pattern will continue all the way up to the 10th term of A and tenth term of B… 2019 = 1 We have 10 of these identical results: 10(1) = 10 Final Answers: GMAT assassins aren't born, they're made, Rich
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