discussion topic: Algebra

yes i am here

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

yes punita??

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

As you know,
logaN=x then a^x=N

So, in given case, LOG2X=Squareroot(X)

hence, 2 ^ (Squareroot(X))= X
As X is less than 10, putting all values, the equation is satisfied for X=4

SO X=4.

For more questions like this, check the lesson Logarithms in Maths Zone.

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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@tulip and Sama
no where in the question it is mentioned that value of x,y,z,w should be unique wrt each other or that no two set of {x,y,z,w} should be jumbled up !
for example the solution set (1,2,1,2) is different from (2,2,1,1)

since x,y,z, abd w are odd integers ..let x=2a+1 ,y=2b+1 ,z=2c+1 and w=2d+1

substituting the value for x,y,z and w in the equation we get:
2(a+b+c+d)+4 =30
or a+b+c+d = 13

now this question reduces to assigning 13 things into 4 different groups (a,b,c,d) where each group has zero or more than zero element (why ? ;) )
the solution would be given be (n+r-1)C(r-1) = 16C3 = 560 !!

also if the constrain is such that each group has at least one element than the solution is given by (n-1) C (r-1)
for example consider the question :
How many solutions are possible for the equation :
x+y+z+w =30
when
1. x,y,z,w are non negative
2.x,y,z,w are positive

in the case 1 apply the formula (n+r-1) C (r-1) as zero is a valid value for (x,y,z,w)
in the case 2 apply the formula (n-1) C (r-1) as zero is not a valid value for (x,y,z,w)

How many integral solutions are there to the equation:
x+y+z+w =29 where x=>1 ,y=>1,z=>3 ,t=> 0 ?

Here is the solution for Avinash question

32^32^32 would leave a remainder of (4)^32^32 ( as 32/7 would leave 4 as remainder)

Now 32^32 would be a number in the form 3k+1
Why , ( try dividing 32^32 by 3 and it would leave reminder 1
as it would be (33-1)^32

So now, the question is reduced to 4^ (3k+1) to be divided by 7
or 4^(3k)x4 /7
now 3k would be divisible by 3 for sure so, we shall have
64^kx4/7

64^k divided by 7 would leave reminder 1 as 64=63+1
and 4/7 would leave remainder 4
so 1x4 would give remainder=4.

way off the target hemant!

hi
any one there???

can anyone help me out regarding the books for peparation for data interpretation.

plz tell me some gud books name for quatitative aptitude also..

vivek sir is right, it is really good. Even i bougt that

hi any body there..

A train moving at 60 km/hr passes through a man standing near a railway line in 15 seconds.The length of the train is?

x+y+z+w= 30
how many solutions are possible such that the variables in the Question are odd positive numbers?
Ans:
the required no of solutions are "264"
Let me explain..
the odd positive no's between "30" are 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29
If we are assuming the variables are not same..
here 23,25,27,29 are not valid..then by testing with each no we will get..
21+5+3+1=30,
19+7+3+1=30,
17+1+5+7=30,
17+9+3+1=30,
15+7+3+5=30,
15+11+1+3=30,
15+9+5+1=30,
13+9+7+1=30,
13+11+5+1=30,
13+3+5+9=30,
11+3+7+9=30,
Here we have 11 solutions..if we observe clearly the no's in each solution may change there places (for example 21+5+3+1=30 each no may be at any place like 5+1+3+21=30)so each solution mentioned above will have "24" solutions in it so we have total "11" solutions like this then 24*11=264..

what is the digit in units place?? if
x^2567 where 15>x>5

Given that x + y + z + w = 29 and x >=1, y >=1, z>=3 and w>=0

Let x = p + 1, where p >=0

y = q + 1, where q >=0

z = r + 3, where r >=0

Now we have p + 1 + q + 1 + r +3 + w = 29

i.e. p + q + r + w = 24, where p, q, r, w > = 0

Hence the total number of solutions will be 27C3 = 2925

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