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Help me solve this PS question...

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Source: — Problem Solving |

by aarati » Thu Aug 05, 2010 3:28 am
I too came across this tricky question,but i can't solve it..........
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by kvcpk » Thu Aug 05, 2010 3:38 am
jamesmith42u wrote:I found an interesting and a tough PS question in the coming soon page of gmatmaths.com website.

here is the question

sqrt[(132*131*130*129)+1]

tried a lot using algebra..but couldn't solve this
sqrt[(132*131*130*129)+1] = x
x^2-1 = 132*131*130*129
(x+1)(x-1) = 132*131*130*129
let x+1= 130*131
x-1 = 130*131-2
x-1 = 130(130+1) -2
x-1 = 130^2 +130 -2

132*129 = (130+2)(130-1)
=130^2 -130+260-2
=130^2+130-2
=x-1

Hence x+1 = 130*131
hence x = 130*131-1 = 17030-1 = 17029.

I really doubt if this can be a GMAT problem if it needs to be solved this way.
Options would have made life easy.
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by sheelanadh » Thu Aug 05, 2010 3:50 am
kvcpk wrote:
jamesmith42u wrote:I found an interesting and a tough PS question in the coming soon page of gmatmaths.com website.

here is the question

sqrt[(132*131*130*129)+1]

tried a lot using algebra..but couldn't solve this
sqrt[(132*131*130*129)+1] = x
x^2-1 = 132*131*130*129
(x+1)(x-1) = 132*131*130*129
let x+1= 130*131
x-1 = 130*131-2
x-1 = 130(130+1) -2
x-1 = 130^2 +130 -2

132*129 = (130+2)(130-1)
=130^2 -130+260-2
=130^2+130-2
=x-1

Hence x+1 = 130*131
hence x = 130*131-1 = 17030-1 = 17029.

I really doubt if this can be a GMAT problem if it needs to be solved this way.
Options would have made life easy.
Yes I agree ..... If this is a GMAT Question there must definitely be an alternate solution...
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by aarati » Thu Aug 05, 2010 3:51 am
the solution is good but is there any other way to solve this.......
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by jaskaran » Thu Aug 05, 2010 5:17 am
I just skimmed through kvcp's explanation, but did not quite get it, but I think if we have the answers, we can make a better guess.

My method would be to multiply the unit digits of the nos. which would result in something ending with a 0, because 130 is one of the numbers, and adding '1' will give us a value ending with a '1', the square root of which should end with a '1'

Would that be a fair method of guesstimation?

PS- KVCP's method did not result in an answer with '1', so am I missing something?
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by kvcpk » Thu Aug 05, 2010 6:29 am
jaskaran wrote:I just skimmed through kvcp's explanation, but did not quite get it, but I think if we have the answers, we can make a better guess.

My method would be to multiply the unit digits of the nos. which would result in something ending with a 0, because 130 is one of the numbers, and adding '1' will give us a value ending with a '1', the square root of which should end with a '1'

Would that be a fair method of guesstimation?

PS- KVCP's method did not result in an answer with '1', so am I missing something?
You missed the square root. A number ending in 1 will have a square root ending with 9 or 1.

Hope this helps!!
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by jaskaran » Thu Aug 05, 2010 6:53 am
kvcpk wrote:
jaskaran wrote:I just skimmed through kvcp's explanation, but did not quite get it, but I think if we have the answers, we can make a better guess.

My method would be to multiply the unit digits of the nos. which would result in something ending with a 0, because 130 is one of the numbers, and adding '1' will give us a value ending with a '1', the square root of which should end with a '1'

Would that be a fair method of guesstimation?

PS- KVCP's method did not result in an answer with '1', so am I missing something?
You missed the square root. A number ending in 1 will have a square root ending with 9 or 1.

Hope this helps!!
right 9 or 1, thanks!.... well at least we can get it down it to 2!...
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by sheelanadh » Thu Aug 05, 2010 10:20 pm
Hi, there's another interesting question today on the coming soon page of gmatmaths.com website

Here is the question.......

2^301 = x^y + y^x where x & y are distinct natural numbers. k = max possible value of x+y. Find 2^301 - k

What is the answer for this? I am not getting any one of the answer choices given there...
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by Ian Stewart » Thu Aug 05, 2010 11:31 pm
sheelanadh wrote:Hi, there's another interesting question today on the coming soon page of gmatmaths.com website

Here is the question.......

2^301 = x^y + y^x where x & y are distinct natural numbers. k = max possible value of x+y. Find 2^301 - k

What is the answer for this? I am not getting any one of the answer choices given there...
This is absolutely *not* a GMAT question - it doesn't even remotely resemble what you might see on the test - so you should ignore it. But if you're curious, it's really just a trick question. If you let

x = 1
y = 2^(301) - 1

then we simply have

x^y = 1
y^x = 2^301 - 1

and x^y + y^x = 2^301

as we want. This is easily going to give us the maximum value for x+y (and I'd be very surprised if there were any other integer solutions, though that would be a crazy thing to try to prove). So 2^301 - x - y = 0.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by sheelanadh » Thu Aug 05, 2010 11:38 pm
Ian Stewart wrote:
sheelanadh wrote:Hi, there's another interesting question today on the coming soon page of gmatmaths.com website

Here is the question.......

2^301 = x^y + y^x where x & y are distinct natural numbers. k = max possible value of x+y. Find 2^301 - k

What is the answer for this? I am not getting any one of the answer choices given there...
This is absolutely *not* a GMAT question - it doesn't even remotely resemble what you might see on the test - so you should ignore it. But if you're curious, it's really just a trick question. If you let

x = 1
y = 2^(301) - 1

then we simply have

x^y = 1
y^x = 2^301 - 1

and x^y + y^x = 2^301

as we want. This is easily going to give us the maximum value for x+y (and I'd be very surprised if there were any other integer solutions, though that would be a crazy thing to try to prove). So 2^301 - x - y = 0.
haa....now this seems very simple...not sure whether it would appear on GMAT or not .....
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by aarati » Fri Aug 06, 2010 12:11 am
Ian Stewart wrote:
sheelanadh wrote:Hi, there's another interesting question today on the coming soon page of gmatmaths.com website

Here is the question.......

2^301 = x^y + y^x where x & y are distinct natural numbers. k = max possible value of x+y. Find 2^301 - k

What is the answer for this? I am not getting any one of the answer choices given there...
This is absolutely *not* a GMAT question - it doesn't even remotely resemble what you might see on the test - so you should ignore it. But if you're curious, it's really just a trick question. If you let

x = 1
y = 2^(301) - 1

then we simply have

x^y = 1
y^x = 2^301 - 1

and x^y + y^x = 2^301

as we want. This is easily going to give us the maximum value for x+y (and I'd be very surprised if there were any other integer solutions, though that would be a crazy thing to try to prove). So 2^301 - x - y = 0.
This is really a very simple solution...Dear Ian, I am curious to know how can we decide on whether a particular question is of GMAT style or not...I am novice in GMAT Prep and from the trick involved in this question I felt that there is a scope for questions like these to appear on GMAT, especially after seeing your solution...
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by ven4gmat » Fri Aug 06, 2010 7:40 pm
Hi, there's another question today on the coming soon page of gmatmaths.com

In triangle XYZ, XY=12, YZ=18, YM is the angular bisector of angle XYZ, and N is the mid point of YM. If area of the triangle XYZ is 90, find area of triangle XYN? (You can visit gmatmaths.com for the figure)

A. 18
B. 27
C.12
D.24
E.None
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