A botanist selects n^2 trees on an island and studies (2n +

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

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

Is there any other approach or style to solve this question.

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?

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)

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

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+1-1)(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 = k-1,

put in (1) and we get 3K+1.

In my opinion there is no direct algebraic way of solving.


DJ

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

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 36-13-13=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

(n-1)^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

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

Kudos is a nice way to say thank you

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?

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.

This question can be solved by division of equation rule. So when you are dividing n^2 by 2n+1 the remainder is [(3n/2)+1] . Now just see which of the options are matching the equation .

Division :
2n+1 ) [n^2] ( [n/2]
[n^2] + [n/2]
________________
- [n/2] -----> Remainder
Since n is postive, -[n/2] becomes negative.

So Postive remainder = -[n/2] + [2n+1] = [(3n/2) + 1]

Now ,

1. if [(3n/2)+1] = 13 , then n is integer hence not possible .
2. if [(3n/2)+1] = 28 , then n is integer hence not possible .
3.if [(3n/2)+1] = 17 , then n is non-integer hence this is the correct answer.
4.if [(3n/2)+1] = 31 , then n is integer hence not possible .
5.if [(3n/2)+1] = 79 , then n is integer hence not possible .

Took me a bit to digest what the question was after.

It doesn’t seem all that clear that the no. of trees shouldn’t be uniform each day.


Basically, we Total number or trees = (n)^2

Each day he studies (2n + 1) until the very last day, when either:

There isn’t (2n + 1) trees left to study

Or

There are another (2n + 1) trees left to study

The first option is what is going on in this question.

There will be some amount less than (2n + 1) that he studies on the last day ———> because (2n + 1) does not evenly divided into (n)^2 total trees. Thus, there will be some remainder of trees left over on the last day that is less than (2n + 1).

Understanding that, the question becomes conquerable.

When the question is written in such a way to say that (2n + 1) trees are studied everyday, it comes as across as every single day (including the last day) (2n + 1) trees will be studied.

Now off to YouTube, to see Princeton Review’s ad for the 9 millionth time in between videos....

Posted from my mobile device

Such a misleading question. He only selects the n^2 once (that at the beginning). Everyday he processes 2n + 1 trees!

Only if the wording of the question had improved

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