Question

The number of positve integral solution of the equation
x1x2x3x4x5=1050 is
a)1800
b)1600
c)1400
d)none

Answer 1

@ganeshie8

Answer 2

atlast a problem that i can attempt confidently xD

Answer 3

familiar with stars and bars ?

Answer 4

enlighten me :)

Answer 5

stars and bars??

Answer 6

oly i know here is that the prime factors is 5^2,2,3,7 and this can be used in the problem

Answer 7

okay lets see if we can avoid stars and bars
yes prime factorization is a good start

Answer 8

Next notice \(x_1, x_2, x_3, x_4\) are the divisors of \(1050\)

Answer 9

yes

Answer 10

\[\large x_1\cdot x_2\cdot x_3\cdot x_4 = 2^13^15^27^1\]

so it is easy to see that each divisor, \(x_i\), on the left hand side also has to be of form \(\large 2^{a_i}3^{b_i}5^{c_i}7^{d_i}\)

Answer 11

can u please explain a bit more the previous step

Answer 12

Oh, this is familiar with one of Kainui's posts... I recall this notation.

Answer 13

i searched the net randomly and got hte answer http://snag.gy/ddki1.jpg
but in this i have a simple doubt

Answer 14

http://snag.gy/qohAU.jpg

Answer 15

int the answer how can you insert 5^2 in 5 ways (5^2 is not a prime number )

Answer 16

Oh you have five factors on left hand side is it, let me go through the solution

Answer 17

5^2 can be assigned to any "one" of the five factors on left hand side in 5 ways

Answer 18

another way to say it is :
any one of the five factors on left hand side can be 25

Answer 19

how 25 is not a prime number and even if u have 25 then there will be nly 4 factors in 5 varibles(x1 ,x2....)???

Answer 20

we don't need x1, x2, x3, x4, x5 to be prime

Answer 21

one solution is : 25*2*3*7*1

Answer 22

ohhh i confused with positive with prime

Answer 23

happens.. haha! I see now what you're asking.. that answer you found looks neat to me :)

Answer 24

hmm i got, so we have 15 ways in which we can assign 5 ,5 and 25 in the 5 variable , am i right???

Answer 25

thats right!

Answer 26

thanx u certainly deserve a medal

Answer 27

you can choose "one" of the 5 factors to be 25 in `5C1` ways

you can choose "two" of the 5 factors to be 5 in `5C2` ways

Answer 28

if you're interested in a direct formula, go through this
http://openstudy.com/users/ganeshie8#/updates/5465b806e4b04662b98a0518

Answer 29

\[\large x_{1}\cdot x_2\cdot x_3\cdot x_4 \cdot x_5= 2^{\color{red}{1}}3^{\color{red}{1}}5^{\color{red}{2}}7^{\color{red}{1}}\]

solving this problem is equivalent to finding the number of non negative solutions to below equations and multiplying them :
\[ a_1+a_2+a_3+a_4 + a_5 = \color{red}{1} \\
b_1+b_2+b_3+b_4 + b_5 = \color{red}{1} \\
c_1+c_2+c_3+c_4 + c_5 = \color{red}{2} \\
d_1+d_2+d_3+d_4 + d_5 = \color{red}{1} \\
\]

Answer 30

hmm i understand this now sorry for late reply @ganeshie8

Answer 31

All good :) that method works in general for any number of factors on left hand side!