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Algebra

New kid on the Block

 Joined: 15 Jun 2007 Post: 7
4 Jul 2007 17:54:38 IST
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Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

find the no.of +ve integral solutions of the eqn
abc=210
give the method 2 proceed
also tell if the qn asks 2 find the no. of +ve unequal integral solutions

Scorching goIITian

Joined: 14 Apr 2007
Posts: 222
4 Jul 2007 18:17:22 IST
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write 210 as a product of prime no.s=7X5X3X2.
select 2 of these no.s(can be done in 4c2=6ways) and multiply them.let their product be p.then {a b c}={other 2 numbers,p}.each of a,b and c can be alloted to all 3 numbers.This again can be done in 6 ways.
hence total 6X6=36 solutions.

Blazing goIITian

Joined: 1 May 2007
Posts: 2830
5 Jul 2007 00:29:40 IST
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You made a slight mistake. You havent considered the cases where a single variable can take all the four numbers, ie, (7x5x3x2), 1, 1. etc.

So the solution is :

Case one : We select two numbers out of the given 4 in 4c2 ways, and we can assign the new product and the remaining 2 numbers to the 3 variables in 36 ways, as cooldude rightly pointed out.

Case two: We select three out of the given four numbers, in 4c3 = 4 ways, and let the product be p. Then, [a, p, 1] (a is the remaining number) can be arranged in 6 ways, ie total no of ways here is 6x4 = 24 ways.

Case three: We select all four of the numbers in 1 way, and assign it to one variable, ie, it becomes [p, 1, 1]. This arrange ment can be done in 3!/2! ways = 6/2 = 3 ways.

Total number of solutions is 36+24+3 = 63 integral solutions.

The general algorithm for such problems is:

1. Let the problem be something like abcde....z = n (n  I)
2. Prime factorize n.
3. Now, simply find the number of ways in which you can arrange the individual factors of n among the given variables.
4. Remember to include all cases. For example, if you have something like abc = 2x3x7, then you will have 3 cases. The first case will be when a variable, ie, a, b, or c takes a single factor, ie, it can take 2 or 3 or 7. Here there are 3! ways for that to happen.
The second case will be when a variable takes the product of 2 factors, ie, here, it will be like (2x3), 7, 1. So the number of arrange ments here will be 3! again.
The third case will be when a variable takes the product of 3 factors, ie (2x3x7), 1, 1. Here no of arrangements is 3!/2! (Just like arranging letters of a word where the words are repeated).
Keep doing this procedure till you exhaust all cases.

Hope it helped!

New kid on the Block

Joined: 15 Jun 2007
Posts: 7
5 Jul 2007 15:45:43 IST
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cooldude why have u selected 2 nos. only. make ur answer alittle more explnatory.plz

Blazing goIITian

Joined: 1 May 2007
Posts: 2830
5 Jul 2007 23:59:43 IST
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what is the actual answer to the question?

Forum Expert
Joined: 29 Dec 2006
Posts: 5524
8 Jul 2007 16:00:57 IST
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Karthik solution is perfect, and cool dude has chosen only two numbers because once two numbers are fixed third is automatically fixed

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