Five is Prime
x y z × y z x × z x y = 5 x y z
Find the number of ordered pairs of distinct positive integers x , y , z satisfying the above equation.
The answer is 3.
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1 solution
Your thinking is correct but the required triplets are ( 1 , 5 , 2 ) , ( 5 , 2 , 1 ) and ( 2 , 1 , 5 )
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F o r ( 1 , 5 , 2 ) w e h a v e 1 5 2 − 1 ∗ 5 2 1 − 1 ∗ 2 1 5 − 1 = 1 2 4 ∗ 5 1 ∗ 2 0 ∗ = 5 So yours is the correct answer since this a cyclic equation. Congratulations.
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can you recount to some specific property you must me considering with your reference to cyclic equation!!!
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@Gaurav Jain – The first article in red is the one that gives very good idea. From google "cyclic algebraic expressions " gives you many articles. Two of them are more interesting.......... [ 1 ] [ P D F ] " F a c t o r i z a t i o n o f c y c l i c a n d s y m m e t r i c p o l y n o m i a l s " [ 2 ] " f a c t o r i s a t i o n o f c y c l i c e x p r e s s i o n s u s i n g f a c t o r t h e o r e m "
Yeah right. Thanks :)
To Omkar Kulkarni :-
F o r ( 1 , 1 , 5 ) w e h a v e 1 1 5 − 1 ∗ 1 5 1 − 1 ∗ 5 1 1 − 1 = 1 0 ∗ 1 4 ∗ 5 0 ∗ = 1 N O T 5 ? ? ? ?
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But the question was for distinct numbers. So the above won't work.
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I only gave source for cyclic expression. In my solution cyclic refers to two other solutions, since I had given only one.
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@Niranjan Khanderia – Yes it is Sir. Certainly
The only triplets satisfying this are ( 1 , 5 , 2 ) , ( 5 , 2 , 1 ) , ( 2 , 1 , 5 ) , because the equation becomes
x y z − 1 . y z x − 1 . z x y − 1 = 5
which can only have the above three solutions, as the only factors of 5 are 1 and 5 . 5 is a prime number.