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A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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06 Jun 2013, 20:20
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A botanist selects n^2 trees on an island and studies (2n + 1) trees everyday where n is an even integer. He does not study the same tree twice. Which of the following cannot be the number of trees that he studies on the last day of his exercise? A. 13 B. 28 C. 17 D. 31 E. 79
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Re: A botanist select n^2 trees ... [#permalink]
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06 Jun 2013, 21:22
First of all, two options(31,79) are very poorly framed & thus creating confusion. Anyways, the logic that the author wants to check is No of books that can be read on any day = 2n+1 , where n is even integer. 2n+1 = even + odd = odd So 2n+1 can never be even. Option C fits this bill. Hence the answer. Note botanist can not read 31, 79 books as well. But i am assuming there is sth wrong with these two options. Fame
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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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07 Jun 2013, 00:54
TirthankarP wrote: A botanist selects n^2 trees on an island and studies (2n + 1) trees everyday where n is an even integer. He does not study the same tree twice. Which of the following cannot be the number of trees that he studies on the last day of his exercise? A. 13 B. 28 C. 17 D. 31 E. 79 n=2 > n^2=4 trees total> 2n+1=5 trees studied everyday > (last day)=4 (4 is the remainder when 4 is divided by 5); n=4 > n^2=16 trees total > 2n+1=9 trees studied everyday > (last day)=7 (7 is the remainder when 16 is divided by 9); n=6 > n^2=36 trees total > 2n+1=13 trees studied everyday > (last day)=10 (10 is the remainder when 36 is divided by 13); n=8 > n^2=64 trees total > 2n+1=17 trees studied everyday > (last day)=13 (13 is the remainder when 64 is divided by 17); n=10 > n^2=100 trees total > 2n+1=21 trees studied everyday > (last day)=16 (16 is the remainder when 100 is divided by 21); ... (last day) = 4, 7, 10, 13, 16, ... a multiples of 3 plus 1. Only option C (17) does not fit. Answer: C. Hope it's clear.
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Re: A botanist select n^2 trees ... [#permalink]
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08 Sep 2013, 11:32
fameatop wrote: First of all, two options(31,79) are very poorly framed & thus creating confusion. Anyways, the logic that the author wants to check is No of books that can be read on any day = 2n+1 , where n is even integer.
2n+1 = even + odd = odd So 2n+1 can never be even. Option C fits this bill. Hence the answer.
Note botanist can not read 31, 79 books as well. But i am assuming there is sth wrong with these two options.
Fame Hi fameatop there is nothing wrong with the options.. think of it this way: the total no of trees to be studied is n^2, divisor is (2n+1).. now the question asks which of the following CANNOT be the remainder? (13, 28, 17, 31, 79)... if we just divide n^2 by (2n+1) quotient will be (n/2) and remainder will be (n/2).. this remainder of (n/2) can also be written as (2n + 1  n/2) or (3n/2 + 1)... which means the remainder (or the no of trees on last day) will always be of the form (3n/2 + 1) where n is even (so 2 in denominator will be reduced/cancel out) which means this number will be of the form 3K + 1 where k is an integer... so whichever option does not satisfy this will be our answer... that is only one option, option C (now i know some of you would be thinking 'how the hell is the quotient n/2 and remainder n/2'... well that is a mathematical concept and i am not yet prepared to explain how it comes.... for you explanation by Bunuel is the best (anyday)..



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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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16 Sep 2013, 02:59
Bunuel wrote: TirthankarP wrote: A botanist selects n^2 trees on an island and studies (2n + 1) trees everyday where n is an even integer. He does not study the same tree twice. Which of the following cannot be the number of trees that he studies on the last day of his exercise? A. 13 B. 28 C. 17 D. 31 E. 79 n=2 > n^2=4 trees total> 2n+1=5 trees studied everyday > (last day)=4 (4 is the remainder when 4 is divided by 5); n=4 > n^2=16 trees total > 2n+1=9 trees studied everyday > (last day)=7 (7 is the remainder when 16 is divided by 9); n=6 > n^2=36 trees total > 2n+1=13 trees studied everyday > (last day)=10 (10 is the remainder when 36 is divided by 13); n=8 > n^2=64 trees total > 2n+1=17 trees studied everyday > (last day)=13 (13 is the remainder when 64 is divided by 17); n=10 > n^2=100 trees total > 2n+1=21 trees studied everyday > (last day)=16 (16 is the remainder when 100 is divided by 21); ... (last day) = 4, 7, 10, 13, 16, ... a multiples of 3 plus 1. Only option C (17) does not fit. Answer: C. Hope it's clear. Is there any other approach or style to solve this question.
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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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09 Jan 2014, 07:59
TirthankarP wrote: A botanist selects n^2 trees on an island and studies (2n + 1) trees everyday where n is an even integer. He does not study the same tree twice. Which of the following cannot be the number of trees that he studies on the last day of his exercise? A. 13 B. 28 C. 17 D. 31 E. 79 There must be a more elegant way to solve this question than just plugging numbers and eliminating answer choices We have that n^2 (2k^2) must of course be even while (2n+1) must be odd and a multiple of 4k + 1 We are basically asked for the remainder When I divide 2k^2 / 4k + 1 I am then a bit stuck with the algebra cause I can't get rid of the 1 in the denominator to find possible remainders Would anybody venture on to this one? Cheers! J



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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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09 Jan 2014, 14:00
Let the no of trees be 4k^2
No of trees studied every day = 4k +1
Let the no of days it takes scientist to study all trees = x+1
No of trees studied on last day = 4k^2  (x+11)(4k+1)..... (1)
Now we have to define x in terms of k, which can only be done by putting values 1,2,3 etc. For k=1, x=0, For k=2, x=1, For k=3, x=2 So x = k1,
put in (1) and we get 3K+1.
In my opinion there is no direct algebraic way of solving.
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Re: A botanist select n^2 trees ... [#permalink]
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12 Jan 2014, 15:35
amanvermagmat wrote: fameatop wrote: First of all, two options(31,79) are very poorly framed & thus creating confusion. Anyways, the logic that the author wants to check is No of books that can be read on any day = 2n+1 , where n is even integer.
2n+1 = even + odd = odd So 2n+1 can never be even. Option C fits this bill. Hence the answer.
Note botanist can not read 31, 79 books as well. But i am assuming there is sth wrong with these two options.
Fame Hi fameatop there is nothing wrong with the options.. think of it this way: the total no of trees to be studied is n^2, divisor is (2n+1).. now the question asks which of the following CANNOT be the remainder? (13, 28, 17, 31, 79)... if we just divide n^2 by (2n+1) quotient will be (n/2) and remainder will be (n/2).. this remainder of (n/2) can also be written as (2n + 1  n/2) or (3n/2 + 1)... which means the remainder (or the no of trees on last day) will always be of the form (3n/2 + 1) where n is even (so 2 in denominator will be reduced/cancel out) which means this number will be of the form 3K + 1 where k is an integer... so whichever option does not satisfy this will be our answer... that is only one option, option C (now i know some of you would be thinking 'how the hell is the quotient n/2 and remainder n/2'... well that is a mathematical concept and i am not yet prepared to explain how it comes.... for you explanation by Bunuel is the best (anyday).. Excuse me sir could you explain this part? " if we just divide n^2 by (2n+1) quotient will be (n/2) and remainder will be (n/2).. this remainder of (n/2) can also be written as (2n + 1  n/2) or (3n/2 + 1)... which means the remainder (or the no of trees on last day) will always be of the form (3n/2 + 1) where n is even (so 2 in denominator will be reduced/cancel out) which means this number will be of the form 3K + 1 where k is an integer" How do you get thos when you divide? Thanks! Cheers! J



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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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13 Jan 2014, 18:18
How can he ever study an even number of trees when n is an even integer? Won't (2n + 1) always be odd leaving B as the answer?



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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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14 Jan 2014, 01:49
HCalum11 wrote: How can he ever study an even number of trees when n is an even integer? Won't (2n + 1) always be odd leaving B as the answer? The question asks "which of the following cannot be the number of trees that he studies on the last day of his exercise?" So, even though 2n+1 is odd, last day there can be even number of trees left to study. For example, if n=6, then there are total of n^2=36 trees and each day he studies 2n+1=13 trees. Thus on the first day he studies 13 trees, on the second day also 13 trees but on the last, 3rd day, there are only 361313=10 trees left. Therefore on the last day he studies 10 trees. Hope it's clear.
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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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14 Jan 2014, 04:17
n=2: There are n^2=4 trees in total Botanist studies (2n + 1)=5 trees everyday Last day=First day=4 trees remaining
n=4: There are n^2=16 trees in total Botanist studies 2n+1=9 trees everyday Last day=7
n=6: There are n^2=36 trees in total Botanist studies 2n+1=13 trees everyday Last day=36(13trees x 2days)=10 (10 is the remainder when 36 is divided by 13); Note that (13trees x 3days) is bigger than 36
n=8: There are n^2=64 trees in total Botanist studies 2n+1=17 trees everyday Last day=64(17trees x 3days)=13 (13 is the remainder when 64 is divided by 17); Note that (17trees x 4days) is bigger than 64
n=10: There are n^2=100 trees in total Botanist studies 2n+1=21 trees everyday Last day=100(21trees x 4days)=16 (16 is the remainder when 100 is divided by 21); Note that (21trees x 5days) is bigger than 100
...
Last days = {4, 7, 10, 13, 16, ... } = {4, (4+3), (4+3*2), (4+3*3), (4+3*4), ... } = = 4 + {0, 3, (3*2), (3*3), (3*4), ... } = 4 + 3*{0, 1, 2, 3, 4, ... } = 4+3n
A. 13 = 4+3*3  OK  NOT THE ANSWER B. 28 = 4+3*8  OK  NOT THE ANSWER C. 17 CANNOT BE WRITTEN AS 4+3n  THIS IS THE ANSWER D. 31 E. 79



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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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12 Mar 2014, 14:27
Bunuel wrote: HCalum11 wrote: How can he ever study an even number of trees when n is an even integer? Won't (2n + 1) always be odd leaving B as the answer? The question asks "which of the following cannot be the number of trees that he studies on the last day of his exercise?" So, even though 2n+1 is odd, last day there can be even number of trees left to study. For example, if n=6, then there are total of n^2=36 trees and each day he studies 2n+1=13 trees. Thus on the first day he studies 13 trees, on the second day also 13 trees but on the last, 3rd day, there are only 361313=10 trees left. Therefore on the last day he studies 10 trees. Hope it's clear. Could one do something like the following? n^2 = (2n+1) + r n^2  2n +1 = r+2 (n1)^2 = r+2 Now we are being asked about the remainder, so remainder would be a perfect square minus 2 But it doesn't seem to fit with the number choices



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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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14 Apr 2014, 08:55
I gave it another shot. Here's another way one can solve
So we can begin with n=4 so 16/9 will have remainder 7 Then let's try n=6 so 36/13 remainder 10 Finally n=8, then 64/17 will yield a remainder of 13.
So we can see a pattern here. Answer will always be in the form 7+3k. Thus subtracting 7 from each answer choice should give a multiple of 3.
C is the only answer choice that doesn't fit the bill
Hope this helps Cheers! J
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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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14 Feb 2016, 23:55
Bunuel wrote: TirthankarP wrote: A botanist selects n^2 trees on an island and studies (2n + 1) trees everyday where n is an even integer. He does not study the same tree twice. Which of the following cannot be the number of trees that he studies on the last day of his exercise? A. 13 B. 28 C. 17 D. 31 E. 79 n=2 > n^2=4 trees total> 2n+1=5 trees studied everyday > (last day)=4 (4 is the remainder when 4 is divided by 5); n=4 > n^2=16 trees total > 2n+1=9 trees studied everyday > (last day)=7 (7 is the remainder when 16 is divided by 9); n=6 > n^2=36 trees total > 2n+1=13 trees studied everyday > (last day)=10 (10 is the remainder when 36 is divided by 13); n=8 > n^2=64 trees total > 2n+1=17 trees studied everyday > (last day)=13 (13 is the remainder when 64 is divided by 17); n=10 > n^2=100 trees total > 2n+1=21 trees studied everyday > (last day)=16 (16 is the remainder when 100 is divided by 21); ... (last day) = 4, 7, 10, 13, 16, ... a multiples of 3 plus 1. Only option C (17) does not fit. Answer: C. Hope it's clear. Should it not be mentioned in the question that he completes studying all books by last day? Or he studies the same number of books everyday but last? I found it a little confusing! Can I expect the same sort of Q from GMAT?



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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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01 Nov 2017, 11:20
amanvermagmat wrote: fameatop wrote: First of all, two options(31,79) are very poorly framed & thus creating confusion. Anyways, the logic that the author wants to check is No of books that can be read on any day = 2n+1 , where n is even integer.
2n+1 = even + odd = odd So 2n+1 can never be even. Option C fits this bill. Hence the answer.
Note botanist can not read 31, 79 books as well. But i am assuming there is sth wrong with these two options.
Fame Hi fameatop there is nothing wrong with the options.. think of it this way: the total no of trees to be studied is n^2, divisor is (2n+1).. now the question asks which of the following CANNOT be the remainder? (13, 28, 17, 31, 79)... if we just divide n^2 by (2n+1) quotient will be (n/2) and remainder will be (n/2).. this remainder of (n/2) can also be written as (2n + 1  n/2) or (3n/2 + 1)... which means the remainder (or the no of trees on last day) will always be of the form (3n/2 + 1) where n is even (so 2 in denominator will be reduced/cancel out) which means this number will be of the form 3K + 1 where k is an integer... so whichever option does not satisfy this will be our answer... that is only one option, option C (now i know some of you would be thinking 'how the hell is the quotient n/2 and remainder n/2'... well that is a mathematical concept and i am not yet prepared to explain how it comes.... for you explanation by Bunuel is the best (anyday).. Could someone explain how the remainder of (n/2) can also be written as (2n + 1  n/2) or (3n/2 +1) ? It is the only thing I got stuck in this explanation. Thank you.



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Re: A botanist selects n^2 trees on an island and studies (2n + [#permalink]
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10 Feb 2018, 01:32
gbatistoti wrote: amanvermagmat wrote: fameatop wrote: First of all, two options(31,79) are very poorly framed & thus creating confusion. Anyways, the logic that the author wants to check is No of books that can be read on any day = 2n+1 , where n is even integer.
2n+1 = even + odd = odd So 2n+1 can never be even. Option C fits this bill. Hence the answer.
Note botanist can not read 31, 79 books as well. But i am assuming there is sth wrong with these two options.
Fame Hi fameatop there is nothing wrong with the options.. think of it this way: the total no of trees to be studied is n^2, divisor is (2n+1).. now the question asks which of the following CANNOT be the remainder? (13, 28, 17, 31, 79)... if we just divide n^2 by (2n+1) quotient will be (n/2) and remainder will be (n/2).. this remainder of (n/2) can also be written as (2n + 1  n/2) or (3n/2 + 1)... which means the remainder (or the no of trees on last day) will always be of the form (3n/2 + 1) where n is even (so 2 in denominator will be reduced/cancel out) which means this number will be of the form 3K + 1 where k is an integer... so whichever option does not satisfy this will be our answer... that is only one option, option C (now i know some of you would be thinking 'how the hell is the quotient n/2 and remainder n/2'... well that is a mathematical concept and i am not yet prepared to explain how it comes.... for you explanation by Bunuel is the best (anyday).. Could someone explain how the remainder of (n/2) can also be written as (2n + 1  n/2) or (3n/2 +1) ? It is the only thing I got stuck in this explanation. Thank you. You asked this question long back  somehow i missed. I will try to explain now. This is a concept of positive/negative remainders, which can be explained simply with an example. Lets say you divide 14 by 8, giving you a quotient of 1, and remainder of 6. That's because 14 = 8*1 + 6 Now interesting thing is this same remainder of '6' can also be said to be a remainder of '2'. How you may ask? Thats because you can also write 14 as: 14 = 8*2  2 (I have increased the quotient by 1, from '1' earlier to '2' now) 14 can be written as a 8K + 6 (a multiple of 8 plus 6) and also be written as 8m  2 (a multiple of 8 minus 2). Similarly if we divide 20 by 7, we can say the remainder is '1' or we can say that the remainder is '6'. Because 20 = 7*3  1 or 20 = 7*2 + 6 (I have decreased the quotient by 1, from '3' earlier to '2' now) By similar logic, when we divide n^2 by 2n+1 we can say remainder is 'n/2' because n^2 can be written as = (2n+1)*(n/2)  n/2 But we can also say that the remainder is '3n/2 + 1' because n^2 can also be written as = (2n+1)*(n/2  1) + (3n/2 + 1) (I have just decreased the quotient by 1, from n/2 to n/2  1)




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