[Kashyap Rastogi]

Originally Posted by naga25french

(i)All are non zero integers
So total number of soln: 14C2*8=728

(ii)One zero two non zero integers
So total number of soln: 14C1*4*3=56*3=168

(iii)One non zero integer
So total number of solution: 6

Total number of solutions of the equation are
728+168+6=902

Can you please explain why 14C2, 14C1?? I have never gone thru such questions.. Pardon my ignorance..

[brajesh.pandey]

Originally Posted by Kashyap Rastogi

Can you please elaborate a bit.. please !!!

Originally Posted by brajesh.pandey
14C2*8 all non zero solution..
For zeroes..
1 Zero (15C2-14C2)*4
2 zero (16C2-14C2)*2
3 Zeroes 1

Total=728+56+34=818

If all the solutions are non zeroe so two fro each + and - 2*2*2*2=8
14C2*8 all non zero integral ssolution.

1 Zero --> Total solution with one zero - all non-zeroes
(15C2-14C2)*4 for eac +ve and -ive numbers.
Similarly 2 Zeroes
and last 3 zeroes will not be there.

==================================================
Please scrutnize the process...

[amitshanky]

Let X = (√3 + √5)^222. If X is written in decimal form, what will be the first digit after the decimal point?

OPTIONS 1)5 2)6 3)7 4)8 5)9

[brajesh.pandey]

Originally Posted by Kashyap Rastogi

Can you please explain why 14C2, 14C1?? I have never gone thru such questions.. Pardon my ignorance..

If x+y+z=a
and we neeeed to find the solutions which are non negative non zeres
Think of a balls and if u place 2 partitions to divide alll the balls.Count the number of balls thats the solution.

a-1C3-1 will give the soln.
Similary if zeroes are allowed put ssom other dummy balls whose value is zero..
so
a+2C3-1(note not all the balls could be zero so only 2

So general formulae m-1Cn-1 non zero solution

So general formulae m+n-1Cn-1 solutions along with zeroes

where n is number of unknown

[uhhh...]

Originally Posted by crack_2009

dude if neither of x and y is divisible by z then there xy can't be divisble by z..
hope this ans ur query

x = 15, y =14, z =6

as you can see x is not divisible by z, y is not divisible by z.. but xy is divisible by z

[bhupi3149]

Consider the set G of all integers between 100 and 9999 (including the extremes). Call two integers a and b in G to be in the same category if the digits appearing in a and b are the same. For example, if a = 101, b = 100, c = 1000 and d = 120, then a, b and c are in the same category but c and d are not. Find the number of different categories that can be formed out of G.

[amitshanky]

Originally Posted by shreyans_nitt

A quick guess from my side (going on the shop floor )

20*21/2 = 210

so atmost 20 consecutive numbers added give 210

x + x+1 + x+2 + ...... + x+n-1 = 210

n/2 (2x + (n-1)) = 210

2x = (420/n) - n + 1

n = 2,3,4,5,6,7,10,12,14,15,20 are possible for which 420/n is integer

n=2,6,10,14 do not give x as integer.....

so 7 ways...

please explain me the bold part iam not getting it..

[Vineetverma]

Originally Posted by jaidev_iyer

Saw a problem somewhere. I dont remember the exact Q, but will try to recollect.

Q) Say a = 100, b= 101,c=102,d= 301.Here it is defined that
"a" and "b" fall into one category because they have the same digits, "c" falls in another category and so falls "d".How many different categories of numbers exist from 100 to 9999.

Originally Posted by uhhh...

a category can be formed by 1 - 4 digits

10C1 + 10C2 + 10C3 +10C4 -1

Originally Posted by bhupi3149

Consider the set G of all integers between 100 and 9999 (including the extremes). Call two integers a and b in G to be in the same category if the digits appearing in a and b are the same. For example, if a = 101, b = 100, c = 1000 and d = 120, then a, b and c are in the same category but c and d are not. Find the number of different categories that can be formed out of G.

refer to the post above, uhhh.. has solved it

[akshaypirate]

Originally Posted by amitshanky

Let X = (√3 + √5)^222. If X is written in decimal form, what will be the first digit after the decimal point?

OPTIONS 1)5 2)6 3)7 4)8 5)9

refer to this post:
http://www.pagalguy.com/forum/quanti...ystem-685.html (Number System)

An ingenious solution given there.

[rahul20001]

if 2 equations x 2 +px+q=0 & x 2 +lx+m=0 have one common root,then how many of following 3 statements can be true?
1) Common root is (pm-ql)/(q-m)
2)Common root is ( q-m )/(l-p)
3) Square of common root is (pm-ql)/(l-p)
1. 0 2. 1 3. 2 4. 3
please post the sol along...
source- powerprep