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

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.

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?

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!!

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.

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.