A "New" Operation

Algebra Level 3

To every pair of positive real numbers x x and y y , we assign a positive real number x y x\star y satisfying the two properties that x x = 1 x\star x=1 and x ( y z ) = ( x y ) × z x\star (y\star z)=(x\star y) \times z (where × \times represents standard multiplication). Find k k if 61 2013 = 1 k 61\star 2013=\frac{1}{k} .


The answer is 33.

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2 solutions

Skye Rzym
Jun 23, 2017

Substitute x = y = z = 61 x=y=z=61 , we get 61 ( 61 61 ) = ( 61 61 ) × 61 61 \star (61 \star 61) = (61 \star 61) \times 61 61 1 = 1 × 61 = 61 61 \star 1 = 1 \times 61 = 61 And then, substitute x = 61 , y = z = 2013 x=61, y=z=2013 , we get 61 ( 2013 2013 ) = ( 61 2013 ) × 2013 61 \star (2013 \star 2013) = (61 \star 2013) \times 2013 61 1 = 1 k × 2013 = 2013 k 61 \star 1 = \frac{1}{k} \times 2013 = \frac{2013}{k} Thus, we get 2013 k = 61 \frac{2013}{k} = 61 k = 33 k = \boxed{33}

Wonderful!

To generalize your steps, you showed that:
1. With x = y = z x= y = z , x 1 = x x \star 1 = x .
2. With x = x , y = z x=x, y = z , x 1 = ( x y ) × y x \star 1 = ( x \star y ) \times y .
3. Hence x y = x y x \star y = \frac{ x } { y } .


Calvin Lin Staff - 3 years, 11 months ago

Out of everyone that I have showed this problem to, you are the only person that I know of to solve it without assuming \star was the division operation. Take an upvote.

Adam Hufstetler - 3 years, 11 months ago
Marta Reece
Jun 23, 2017

The new operation is division. x / x = 1 x/x=1 and x / ( y / z ) = x y z = x y × z x/(y/z)=\dfrac x{\frac yz}=\dfrac xy\times z

therefore 1 k = 61 2013 = 1 33 \dfrac 1k=\dfrac{61}{2013}=\dfrac 1{33}

k = 33 k=\boxed{33}

How can we prove that the only possibility is division?

Calvin Lin Staff - 3 years, 11 months ago

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I don't think it is even true that the only possibility is division. Not sure how to show it though ; I guess we could exploit the uncountability of reals in some way , similar to what we do in Cauchy equation.

Shourya Pandey - 3 years, 11 months ago

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See Rizky's solution above.

Calvin Lin Staff - 3 years, 11 months ago

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@Calvin Lin Oh yeah. I got that immediately after I wrote my comment :P Thanks.

Shourya Pandey - 3 years, 11 months ago

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