IIT JEE , AIEEE Forum - Mathematics , Algebra

rik_mad

WAY TO GO SG_90 ...I NEVER THOUGHT OF THAT
HERE'S A SALUTE

animesh.manglik

xyz=24

now we take combinations

first of all the factors-----1*2*2*2*3

now combinations....th enumbers can be taken by (x,y or z)

_ _ _ can be 1*2*12 .... (total combinations 6 i.e 3C1 )

_ _ _ can be 1*24*1... (total combinations 3 3c1/2!

_ _ _ can be 8*3*1 (total combinations 3 3c1

_ _ _ can be 4*6*1 (total combinations 3 3c1

_ _ _ can be 4*3*2 (total combinations 3 3c1

_ _ _ can be 6*2*2 (total combinations 3 3c1/2!

all summing up gives up 30 cases( note this is the answer only for positive integers_

Now for integral solutions it can be state when x&y are negative or Y&Z are negative or x&z are negative and a case when no one is negative ....hence 4 cases

Thus the solution for integral solution is ....30 *4 =120 which is the answer

pink_ele

experts please comment

konichiwa2x

Hey, these types of problems are common. The general method for solving it is this:

if you were to have:

where the p's are prime, then

each of the

's could be "distributed" among the k factors in

(ie,^mj+k-1 C _k-1 ) ways.

x1.x2.x3 = 2^3.3

Hence, ^(3+3-1)C_{3 ^x} ^(3+1-1)C_{1^{. = 30.}}

Notice that this only gives you the number of POSITIVE integral solutions, whereas what is asked is number of integral solutions. If any 2 of x1,x2,x3 are negative, their product still comes out to be 24.

The number of ways any 2 of x1, x2,x3 can be negative is equal to ³C_2.

Thus the total number of solutions = 30 x 4 = 120.

If you are still unclear, check out this post: http://www.goiit.com/posts/list/2402.htm#9606 (scroll to the bottom)

magiclko

animesh method's is easiset (since thr's no formula to apply)..... and its absolutely correct, salute for u animesh

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