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Algebra

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15 Apr 2010 17:23:24 IST

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no of integral solution of x1*x2*x3*x4*x5=1050 is?

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no of integral solution of x1*x2*x3*x4*x5=1050 is?

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HIMANSHU JAIN

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16 Apr 2010 17:35:51 IST

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see it's very easy to analyze:

first find it's factors: 1050 = 2*3*5*5*7

Now, as we can change the values between different xs.

Soo, we get ans: 5!

but we have two 5s. That's why, ans is 5!/2! = 60.

Anant Kumar

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17 Apr 2010 16:06:23 IST

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That's not correct.
First, the prime factorization 1050:
$1050 = 2\cdot 3 \cdot 5^2\cdot 7$
Each factor of 1050 can contain only these primes. So the general form of each of the $x_i$ is
$x_i=2^{a_i}\cdot 3^{b_i}\cdot 5^{c_i}\cdot 7^{d_i}$ , where $a_i$ , $b_i$ , $c_i$ , $d_i$ are non-negative integers and $i=1,\ 2,\ 3,\ 4,\ 5$ .
Accordingly,
$x_1x_2 x_3x_4 x_5 = 2^{\sum a_i}\cdot 3^{\sumb_i}\cdot 5^{\sum c_i} \cdot 7^{\sum d_i}$
We must assign the $a_i$ 's, $b_i$ 's, $c_i$ 's and $d_i$ 's so that the following set of equalities hold:
$a_1+a_2+a_3+a_4+a_5=1$ ,
$b_1+b_2+b_3+b_4+b_5=1$ ,
$c_1+c_2+c_3+c_4+c_5=2$
$d_1+d_2+d_3+d_4+d_5=1$
The number of possible ordered lists for these equations are:
$a_1+a_2+a_3+a_4+a_5=1$ : $^5\mathrm{C}_1=5$
$b_1+b_2+b_3+b_4+b_5=1$ : $^5\mathrm{C}_1=5$
$c_1+c_2+c_3+c_4+c_5=2$ : $^6\mathrm{C}_2=15$
$d_1+d_2+d_3+d_4+d_5=1$ : $^5\mathrm{C}_1=5$
Since each list $(a_i,\,b_i,\,c_i,\,d_i)$ correspond to a factor, the number of solutions of the given equation
$x_1x_2x_3x_4x_5=1050$ is $5\times 5\times 15\times 5= 1875$

dhruv.tara

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Joined: 16 Dec 2006

Posts: 68

18 Apr 2010 12:59:47 IST

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I didn't exactly get what you did here... can you please elaborate.. sorry for such an ignorance... :)

Anant Kumar

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26 Apr 2010 10:04:47 IST

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May be you were not able to get it because you could not read it well. Try again:

First, the prime factorization 1050:
$1050 = 2\cdot 3 \cdot 5^2\cdot 7$
Each factor of 1050 can contain only these primes. So the general form of each of the $x_i$ is
$x_i=2^{a_i}\cdot 3^{b_i}\cdot 5^{c_i}\cdot 7^{d_i}$ , where $a_i, b_i, c_i, d_i$ are non-negative integers and $i=1,\ 2,\ 3,\ 4,\ 5$ .
Accordingly,
$x_1x_2 x_3x_4 x_5 = 2^{\sum a_i}\cdot3^{\sumb_i}\cdot 5^{\sum c_i} \cdot 7^{\sum d_i}$
We must assign the $a_i$ 's, $b_i$ 's, $c_i$ 's and $d_i$ 's so that the following set of equalities hold:
$a_1+a_2+a_3+a_4+a_5=1$ ,
$b_1+b_2+b_3+b_4+b_5=1$ ,
$c_1+c_2+c_3+c_4+c_5=2$
$d_1+d_2+d_3+d_4+d_5=1$
The number of possible ordered lists for these equations are:
$a_1+a_2+a_3+a_4+a_5=1: ^5\mathrm{C}_1=5$
$b_1+b_2+b_3+b_4+b_5=1: ^5\mathrm{C}_1=5$
$c_1+c_2+c_3+c_4+c_5=2: ^6\mathrm{C}_2=15$
$d_1+d_2+d_3+d_4+d_5=1: ^5\mathrm{C}_1=5$
Since each list $(a_i,\,b_i,\,c_i,\,d_i)$ correspond to a factor, the number of solutions of the given equation
$x_1x_2x_3x_4x_5=1050$ is $5\times 5\times 15\times 5= 1875$ .

HIMANSHU JAIN

Scorching goIITian

Joined: 8 Feb 2007

Posts: 270

27 Apr 2010 22:16:41 IST

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but in your case, x1..x5 will not remain integer.

They all need to be integer, so how there powers can be reduced less than 1.

We are asked integral solution, so all x1..x5 need to be integers.

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