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# Two Difficult Problem

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1) The number of ordered paurs of positive integer (m,n), such that 1/n + 1/m = 1/15

2) Consider the equation 2u + v + w + x + y + z = 3. How many solutions (u , v , w , x , y , z) of non-integers does this equation have?

Oct 21, 2017

#1
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1) We can rewrite this as (n + m)/mn = 1/15, which also is mn = 15m + 15n.

We move everything to one side:

mn - 15m - 15n = 0

There is this method of adding a special number to both sides to make this doable.

In this case, we do + ( - 225) to both sides. mn - 15m - 15n - 225 = -225

We can rewrite this as -n*(m-15)-15(m-15)=-225, which also is (m-15)*(n-15) = 225

Using a factoring trick, 225 can be written as (1,225);(3,75);(5,45);(9,25);(15,15)

So there are nine ordered pairs because 15,15 is the same as 15,15. Others can be doubled.

Oct 21, 2017
#3
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Very crafty.......!!!!!

I like this one.....!!!!

CPhill  Oct 21, 2017
edited by CPhill  Oct 21, 2017
#2
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I am not sure with #2, can someone help me with that too?

Oct 21, 2017