discussion topic: Algebra

hi is anyone there

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Hi usha..r u preparing for CAT 2010??

hi i need some information

yes punita??

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hows your prepartion going on......

hello guys please drop questions here , it will be real good to discuss

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hi....... guys

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x+y+z+w= 30
how many solutions are possible such that the variables in the Question are odd positive numbers?

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Let me try.

( Assuming x,y,z,w can't be same)

As the sum of 4 odd positive integers is required to be 30, no number can be equal to 29,27, 25 or 23.

Now,
21+1+3+5=30 is one solution
17+1+5+7
17+1+9+3
17+5+1+7
15+1+5+9
15+1+3+11
15+9+1+5
15+3+5+7
13+1+5+11
13+1+7+9
13+3+5+9

So 11 Solutions

(32^32^32)/7 den wat will be remainder........solve it guys

The multiplication rule would hold. So it would be 4^4^4=64. This when divided by 7 would give a remainder of 1.

Check the lesson 'Number System-Remainders & Factors ' in Maths Zone for the entire concept and more questions like these

x+y+z+w= 30
how many solutions are possible such that the variables in the Question are odd positive numbers?

ANS: ONLY 9 ARE POSIBLE IF V TAKE IT AS NO JUMBLING OF NO. SHOULD B DONE.
hey Sama ..u repeated 2 combinations.

what is the distance between two cities A and B in integral number of kilometers
if x satisfies the equation log x base2=sqrroot x and x<=10 km

It will be 4