sqrt[(132*131*130*129)+1] = xjamesmith42u 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
Yes I agree ..... If this is a GMAT Question there must definitely be an alternate solution...kvcpk wrote:sqrt[(132*131*130*129)+1] = xjamesmith42u 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
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.
You missed the square root. A number ending in 1 will have a square root ending with 9 or 1.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?
right 9 or 1, thanks!.... well at least we can get it down it to 2!...kvcpk wrote:You missed the square root. A number ending in 1 will have a square root ending with 9 or 1.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?
Hope this helps!!
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 letsheelanadh 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...
haa....now this seems very simple...not sure whether it would appear on GMAT or not .....Ian Stewart wrote: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 letsheelanadh 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...
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...Ian Stewart wrote: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 letsheelanadh 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...
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.
