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No. of integral solutions????
by Abhi S - Tuesday, 2 October 2007, 10:16 AM
  find the number of integral sloutions to IxI + IyI + IzI = 15?
answer is 902.
dunno hw to get it?
help!!
Re: No. of integral solutions????
by Abhi S - Tuesday, 2 October 2007, 01:10 PM
  its mod(x) + mod(y) + mod(z).....
Re: No. of integral solutions????
by the underdog - Tuesday, 2 October 2007, 04:39 PM
  2x + 3y = 100

I read this method in one of the posts:

Now take y = 0, we get x= 50

take y = 32, x = 2

So x ranges from 2 to 50 in steps of 3 (This has been explained in one of TG's lessons)

So no. of values = (50-2)/3 + 1 = 17



How would one extend this to more than two variables?
Re: No. of integral solutions????
by sandeep gupta - Tuesday, 2 October 2007, 05:15 PM
 

First I solve without taking modulus as and considering only non-negative values for x,y,z

x+y+z=15

taking z=0 , x+y=15 , no. of solution can be 16 (am I correct!)

taking z=1 , x+y=14  ,no. of solution can be 15

similarily take value ffor z=2,3...15 which give solutions from 14,13....1

so total solutions for x+y+z=15(taking only positive values) is

16+15+...+2+1=136

 

 

Re: No. of integral solutions????
by sandeep gupta - Tuesday, 2 October 2007, 05:19 PM
 

Now,as modulus is given,values will always become positive.

Now take z as negative from -1 to -15,which will be same as taking z from 1 to 15 so total solution will be 15+14+...+2+1=120

so total solutions (taking x and y positive)=136+120=256

Re: No. of integral solutions????
by sandeep gupta - Tuesday, 2 October 2007, 05:37 PM
 

Now we will take y negative,in above mention 256 cases,all the cases will repeat except when y was taken 0.

so,total number of solutions taking y negative = 15+14+....+1 and 14+13+...+1 = 120+105 = 225 (am I correct!)

now take cases with x negative,

when y and z positive = 15+...+1=120

when y positive and z negative=14+..+1=105

when y negative and z positive=14+..+2=104

when y and z negative=13+...+2=90

now for x negative,total=120+105+104+90=419

Now final total will be=256+225+419=900

I may have done some mistake some-where!

 

 

 

 

Re: No. of integral solutions????
by Abhi S - Tuesday, 2 October 2007, 07:42 PM
  hi sandeep. i cudnt understand your approach....an its too cumbersome....
my approach s as follows....aldo i cudnt get d answer..
nw since x, y, z can take values ranging frm -15 to 15.
therefore we need to find the co-fficient of a15 in the equation,
(a-15 + a-14 +---------------------+  a14 + a15)(a-15 + a-14 +---------------------+  a14 + a15)(a-15 + a-14 +---------------------+  a14 + a15)....
nw i think solving this we get the right answer....
i dont knw hw to solve this...
TG cud u plz tell us wud this approach work...and how to solve this....
help!!
Re: No. of integral solutions????
by Abhi S - Tuesday, 2 October 2007, 11:07 PM
  TG sir where r u....?
Re: No. of integral solutions????
by Total Gadha - Wednesday, 3 October 2007, 12:39 AM
  First, let each of |x|, |y| and |z| be equal to 1 at least. Then we need to find the  number of positive integral solutions of a + b + c = 12 where a = |x|, b = |y|, c = |z|. The number of solutions are 14C2 = 91. Now for each value of |x|, |y|, and |z| we will have two values of x, y and z [if |x| = m --> x = ±m]. Therefore, the total number of solutions = 91 × 2 × 2 × 2 = 728.

Now let one of the variables be equal to 0. For example, let x = 0 and |y| and |z| be at least equal to 1. Therefore, we need the solution of a + b = 13, where a = |y| and b = |z|. The number of solutions are 14C1 = 14. Each of these solutions will give two values and there are 3 ways in which we can keep one of the variables equal to 0. Therefore, total number of ways are 14 × 2 × 2 × 3 = 168.

Now let two of the variables be equal to 0. In this case, the total number of solutions is equal to 6.

Therefore, the total number of solutions = 728 + 168 + 6 = 902.

Note: The total number of whole number solutions for a1 + a2 + a3 + ... + ar = n is equal to n + r - 1Cr - 1
Re: No. of integral solutions????
by fundoo bond - Wednesday, 3 October 2007, 02:12 PM
  awesome solution TG smile
Re: No. of integral solutions????
by Kunal Gupta - Wednesday, 3 October 2007, 02:19 PM
  a + b + c = 12 .. Is it a typo..shuld it be 15 or explain..
Re: No. of integral solutions????
by fundoo bond - Wednesday, 3 October 2007, 03:41 PM
  kunal,TG has taken a,b,c's value as 1....hence a+b+c = 12
Re: No. of integral solutions????
by Arun prasath - Wednesday, 3 October 2007, 04:12 PM
 

if none of the x,y,z are zeros then

case1 :

we can divide 15 into 3 persons  in 14c2 ways = 14*13/2 = 91

(x,y,z)  could take (+/- ,+/- ,+/-) so under each element in 91 we have 8 diff x,y,z values

so 91*8 = 728

case 2:

    one of the x,y,z could be zeros

here we have   13+1  c 1  =  14

(0,+/-,+/-)   = 4* 3  { 3 ,since x could be zero or y could be zero or z could be}

 

12*14 = 168

 

case 3:

15,0,0  here we have 3*2 = 6

so answer is:

===========

728+168+6 = 902

 

Re: No. of integral solutions????
by pranav tendolkar - Wednesday, 3 October 2007, 09:35 PM
  ARUN Prakash is rit
Re: No. of integral solutions????
by Swapnil Chauhan - Wednesday, 5 August 2015, 02:03 PM
  SHORTCUT:
Number of integral solutions for |x|+|y|+|z|=p is (4P^2)+2, So the answer is 902.
ENJOY
Re: No. of integral solutions????
by Viral Kanabar - Sunday, 5 June 2016, 08:24 PM
  what will be the formula when we have four varaibles
eg: |a|+|b|+|c|+|d|= p