solution is =3 with the condition p!=0 and q!=0 and r!=0
Ashima
M.Sc.(Tech)Information Systems
4th year
BITS Pilani
Rajasthan
On Thu, Sep 15, 2011 at 10:38 PM, Piyush Grover
wrote:
> @abhinav...
>
> it's not about being over smart or to show someone or to prove someone
> anything. It's just th
@abhinav...
it's not about being over smart or to show someone or to prove someone
anything. It's just that
you should not take any assumptions by yourself or if you do you should
specify clearly.
If u r asked this question in an interview and you give the answer 3 without
telling your assumption,
Right, and in every proof above, at some point there is a possible
division by zero. Therefore the proof is not valid in cases where R or
P or Q are zero, and there are infinitely many such cases.
The problem states P+Q+R=0 as the only constraint. There are
infinitely many cases which fit that cons
dude dats outside the domain of the qs...dont be oversmart.
On Thu, Sep 15, 2011 at 9:49 AM, Don wrote:
> No, not at all. Here is a trivial counterexample:
>
> P = Q = R = 0
>
> Don
>
> On Sep 15, 11:46 am, abhinav gupta wrote:
> > Shut up...its 3,,
> >
> >
> >
> > On Thu, Sep 15, 2011 at 9:43
u cnt divide a number by 0..that thing is self undrstod
On Thu, Sep 15, 2011 at 9:49 AM, Piyush Grover wrote:
> Don is right
>
> if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!!
>
>
>
> On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta > wrote:
>
>> Shut up...its 3,,
>>
No, not at all. Here is a trivial counterexample:
P = Q = R = 0
Don
On Sep 15, 11:46 am, abhinav gupta wrote:
> Shut up...its 3,,
>
>
>
> On Thu, Sep 15, 2011 at 9:43 AM, Don wrote:
> > It might be 3, but it doesn't have to be 3.
> > Don
>
> > On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN wrote:
>
Don is right
if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!!
On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta
wrote:
> Shut up...its 3,,
>
>
> On Thu, Sep 15, 2011 at 9:43 AM, Don wrote:
>
>> It might be 3, but it doesn't have to be 3.
>> Don
>>
>> On Sep 14, 11:56 pm,
Shut up...its 3,,
On Thu, Sep 15, 2011 at 9:43 AM, Don wrote:
> It might be 3, but it doesn't have to be 3.
> Don
>
> On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN wrote:
> > if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ??
> >
> > how to solve this??
>
> --
> You received this message because you are
It might be 3, but it doesn't have to be 3.
Don
On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN wrote:
> if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ??
>
> how to solve this??
--
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3x+4y = 60 can be expressed as 15 -y = 3x+3y -45
i.e, 15-y = 3(x+y-15)
which implies tht for every value of x,y in the above eq 15-y is divisible
by 3
On Sun, Aug 28, 2011 at 10:03 PM, Dave wrote:
> @Harhsit: Normally, 0 is not considered positive.
>
> Dave
>
> On Aug 28, 10:45 am, harshit set
@Harhsit: Normally, 0 is not considered positive.
Dave
On Aug 28, 10:45 am, harshit sethi wrote:
> sorry 6 solutions y=15,12,9,6,3,0
> and x=0,4,8,12,16,20 respectively
>
> On 8/28/11, harshit sethi wrote:
>
>
>
> > maximum value of y satisfying this is y=15 and for that x=0;
>
> > now decreas
sorry 6 solutions y=15,12,9,6,3,0
and x=0,4,8,12,16,20 respectively
On 8/28/11, harshit sethi wrote:
> maximum value of y satisfying this is y=15 and for that x=0;
>
> now decrease y by 3 and increase x by 4 ,you will have x and y
> satisfying the equation.
>
> keep on doing this till you reach
maximum value of y satisfying this is y=15 and for that x=0;
now decrease y by 3 and increase x by 4 ,you will have x and y
satisfying the equation.
keep on doing this till you reach minimum value of y i.e 0
this you can do 5 times decreasing y=15 by 3 every time
so there will be 5 solutions .
3x+4y = 60
it's a straight line equation whose x intercept is 20 and y intercept is 15.
Draw it in first quadrant
(as x, y are positive integers)
now x = (60 - 4y)/3 = 4(15-y)/3
now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3 or
not. It's simple y = 3, 6, 9, 12
-Piyush
On
@Sivaviknesh: The smallest values of x and y are 1. The largest value
of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is
an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a
multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3.
Thus, y = 3, 6, 9, 12. Then y
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