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# Mathematic Riddle

Sun 15 Aug, 2010 12:45 am
The difference between two numbers equals one.
The product of the same two numbers equals one.
What is the difference of the cubes of the number?
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RealEyes

1
Sun 15 Aug, 2010 02:30 pm
Smith98 wrote:
The difference between two numbers equals one.
The product of the same two numbers equals one.
What is the difference of the cubes of the number?

Either you forgot to pluralize "the number" in the last sentence of the riddle, or both numbers are the same - which seems very paradoxical. I'll have to think on this one.
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RealEyes

1
Sun 15 Aug, 2010 02:33 pm
Smith98 wrote:
The difference between two numbers equals one.
The product of the same two numbers equals one.
What is the difference of the cubes of the number?

The difference between two numbers is always one number;
The product of two numbers is always equates to one number;
Therefore the difference of cubes of any two numbers is going to be "one."
contrex

1
Sun 15 Aug, 2010 03:38 pm
@RealEyes,
RealEyes wrote:

Smith98 wrote:
The difference between two numbers equals one.
The product of the same two numbers equals one.
What is the difference of the cubes of the number?

The difference between two numbers is always one number;
The product of two numbers is always equates to one number;
Therefore the difference of cubes of any two numbers is going to be "one."

I suspect English is not your native language. It's not "one number", it's the number one (1).

1
Sun 15 Aug, 2010 03:59 pm
@contrex,
x-y=1
x*y=1

x=1+y

(1+y)y=1

y=~0.618033989 (square root of 1.25 - .5)
x=~1.618033989 (square root of 1.25 + .5)

x^3-y^3=4
markr

2
Sun 15 Aug, 2010 10:47 pm
@Smith98,
For what it's worth, the numbers in question are the golden ratio and its inverse.
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AM1

1
Sun 12 Dec, 2010 12:29 pm
Hello,

I'm a new member of this forum. Although a long time has elapsed since
this question was posted, I would like to represent an other way to solve
this problem(Maybe you've thought about it anf have not wriiten it yet)

The way you've written, to solve an appropriate quadric equation is good.
This way that does not necessiate solving an equation:

I use the identity:
x^3-y^3=(x-y)(x^2+xy+y^2)
It is known that x-y=1, therefore we can substitute x-y in the identity by 1,
the term above becomes:
x^3-y^3=1*(x^2+xy+y^2)

It is known that xy=1
and
(x-y)^2=x^2-2xy+y^2=1^2=1
hence
x^2+y^2=1+2xy=1+2=3
Now the term above becomes:
x^3-y^3=(x-y)(x^2+xy+y^2) =(x-y)(x^2+y^2+xy )=1*(3+1)=4

I'll be happy to hear if you think that there is a better way.
Amos
E-mail: amos_s12@zahav.net.il

engineer

1
Sun 12 Dec, 2010 12:43 pm
@AM1,
Welcome to A2K! I really like that solution. Thanks for posting it.
0 Replies

uvosky

1
Thu 24 May, 2012 01:47 am
@AM1,
Long time has elapsed since you posted the solution AM1 , but since you wrote that you will be happy to hear if there is a better way to obtain the
solution , I post one here ;
you used the identity x^3 - y^3 = (x-y) (x^2 + xy + y^2 ) , instead I use the
identity x^3 - y^3 = ( x - y )^3 + 3 xy ( x - y ) , and since it is given that
xy = 1 and x - y = 1 it readily follows that x^3 - y^3 = 1 + 3 = 4
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raprap

1
Thu 24 May, 2012 10:47 am
@Smith98,
a-b=1
ab=1
a^3-b^3=(a-b)(a^2+ab+b^2)=(a^2+1+b^2)
Now (a-b)^2=a^2-2ab+b^2=a^2-2+b^2=1
So a^2+b^2=3
a^3-b^3=(a^2+b^2)+1=3+1=4

Rap
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