mini_india said – Sun, 22 Mar 2009 19:35:57 -0000 ( Link )
I found a book that has a bunch of pages missing. If the sum of the page numbers of the missing pages is 5296, what are the pages?
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Math q : interesting one (Discussion)
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oLahav said – Mon, 23 Mar 2009 13:55:45 -0000 ( Link )
Hmm… this seems like an interesting question. I’m not sure it’s possible to solve this without more information… let’s see- we have the sum of the missing pages as 5296. And since these are pages in a book they’ve got to be consecutive, i.e. x, x + 1, x + 2, etc. But we don’t know how many pages we have.
Well, we know it says a bunch of pages, so it can’t be one. Can it be 2 pages? in that case we get x + x + 1 = 5296, or 2x = 5295 which will result in half a page, so that can’t be right. How about 3 pages? we get 3x + 3 = 5296, which doesn’t work out either since 5296 isn’t divisible by 3. What about 4? 4x + 6 = 5296, this doesn’t work out either.
Can you spot the pattern? We’ll always deal with nx + (1 + ... + (n – 1) ), which as we all know can also be written as nx + n(n – 1)/2 = 5296. Therefore n must be a factor of 5296 * 2 (n is the number of pages by the way). The factoring gives you 32 * 5296, and a quick check shows n = 32 is the only one that works with x = 150.
Therefore the pages missing are 150, 151, ... 180, 181. Sweet!
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i_m_wat_i_m said – Tue, 24 Mar 2009 09:18:06 -0000 ( Link )
Well my approch was similar to yours, but i arrived at the result by using Arithematic progression and factrorization. Almost similar to your method. Thanks for a great explaination Olahav
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oLahav said – Tue, 24 Mar 2009 13:51:45 -0000 ( Link )
You’re welcome! The thing is, on the GMAT, I probably wouldn’t spend a lot of time on this kind of question, because it looks like it’ll take long to solve right away (it did take me a while). Especially if it’s one of the later questions, it can indicate I’m doing pretty well so far since if you get hard questions it usually means you answered the last ones right. But since not finishing on time will have a major effect on the score, I’ll probably take a guess on this kind of question rather than solve it. It would depend on overall timing of course, but it’s just a thought. Never spend too much time on a question.
By the way, I made a small mistake- the factoring should show up as 32 * 331.
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Sureshbala said – Thu, 26 Mar 2009 10:53:10 -0000 ( Link )
Folks, let me make this more clear to all of you…..
Let us suppose the number of missing pages to be n, starting with x+1. (oLahav started with x instead)
Given that
Since n, the number of pages is a positive integer, it can be either 32 or 331. But if n =331, 2x+(n+1) cannot be 32. So the possibility that n = 331 is ruled out.
Now n can be either 32 or a factor of 32.
If you suppose that n =16, then the value of x will not be an integer. Of course from here you can conclude that x must be 32.
Hence number of pages =32 and x = 149.
So the missing pages are 150, 151,.............181
Hope this helps
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Sureshbala said – Thu, 26 Mar 2009 11:04:29 -0000 ( Link )
Now let’s look at a big quicker way of solving this….
Since 5296=16×331, we are looking for 16 pairs of numbers that add up to 331. So totally we are looking for 32 numbers.
Also, of these numbers, the middle most 2 numbers which are consecutive must add up to 331. Hence the middle two numbers are 165 and 166.
So now we will have 15 numbers before 165 and another 15 numbers after 166.
Hence the numbers must be
150, 151, 152,......165,166,..............179,180,181.
Observe this series, we have totally 32 numbers. Pair up the first number with the last number and you will have their sum as 331. Similarly pair up second from the left and second from the right i.e 151+180 which will add up to 331.
So totally we have 16 pairs and sum of each pair is 331. Hence their sum = 16×331=5296
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rneve said – Tue, 24 Mar 2009 07:20:52 -0000 ( Flag Edit Link )
brillient way of solving the complicated maths problems !! Keep it up !!
hrnan said – Thu, 26 Mar 2009 06:35:06 -0000 ( Flag Edit Link )
Can you please tell me how did you arrive at the generalisation of “nx + (1 + ... + (n – 1) ) ”?