Integral Solutions!
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For set A , find the number of positive integral solution of ( x , y , z ) that satisfy the condition x y z ∈ a .
Too easy? Try this problem .
The answer is 64.
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1 solution
Yes, there is. This deserves to be a solution but I unfortunately lost my attempts:
Clearly, a is the set of divisors of 3 0 .
However, 3 0 = 2 × 3 × 5
Now, take three boxes: x , y , z
For each of 2 , 3 , 5 , you have four choices: To put it in x , to put it in y , to put it in z , to put it throw it away.
Thus, there are a total of 4 3 ways to do the things.
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ah nice, to put in same box is equivalent to forming their product, nice , i shall use this method from now on, thankyou :)
Yes , Same But I Reason it in slight different way : Let \xyzt = 2 ∗ 3 ∗ 5 where x,y,z are 3 Real beggars and 't' is false beggar , So you have 3 different books (2,3,5) which can be distributed in 4 3 ways where 1st Book (say 2) has 4 options 2nd Book (say 3) has 4 options 3rd Book (say 5) has 4 options
So total ways are = 4 ∗ 4 ∗ 4 And Thanks For Adding Link of Your Question , and also for Editing Now It looks Better ! ! ⌣ ¨
Great approach. We create this "fake variable" to help us solve questions like
Find the total number of ordered triples of non-negative integers such that a + b + c ≤ 1 0 0 .
is there a better way than to observe that all numbers must be powers of 2 and 5 and 3 and their products, from which 1 comes in one way , 2,3,5 in 3 ways each, 6,10,15 in 9 ways each(products of two taken at a time) and 30 in 27 ways (product of all 3) so total 64