number cubed and sqaured diff

[umm]

I need help please

Gine thinks of a number. The difference between its square and its cube is 3840 what is the number

 

[pka]

I need help please

Gine thinks of a number. The difference between its square and its cube is 3840 what is the number

Look at this webpage.

 

[JeffM]

I need help please

Gine thinks of a number. The difference between its square and its cube is 3840 what is the number

A non-graphical way to solve this is using the integral root theorem.

\(\displaystyle n^3 - n^2 = 3840 \implies n^3 - n^2 - 3840 = 0.\)

3840 = 2^8 * 3 * 5.

Obviously, \(\displaystyle 32^3 - 32^2 = 32768 - 1024 = 31744 >> 3840 \implies n < 32.\)

And \(\displaystyle 10^3 - 10^2 = 1000 - 100 = 900 << 3840 \implies 10 < n < 32.\)

\(\displaystyle 15^3 - 15^2 = 3375 - 225 = 3150 < 3840 \implies 15 < n < 32.\) Getting close.

\(\displaystyle 16^3 - 16^2 = 4096 - 256 = 3840 \implies n = 16.\)

This method works only if the solution is an integer.

 

[Deleted member 4993]

You can solve this using "brute force" - with a spread-sheet like excel

 

[umm]

A non-graphical way to solve this is using the integral root theorem.

\(\displaystyle n^3 - n^2 = 3840 \implies n^3 - n^2 - 3840 = 0.\)

3840 = 2^8 * 3 * 5.

Obviously, \(\displaystyle 32^3 - 32^2 = 32768 - 1024 = 31744 >> 3840 \implies n < 32.\)

And \(\displaystyle 10^3 - 10^2 = 1000 - 100 = 900 << 3840 \implies 10 < n < 32.\)

\(\displaystyle 15^3 - 15^2 = 3375 - 225 = 3150 < 3840 \implies 15 < n < 32.\) Getting close.

\(\displaystyle 16^3 - 16^2 = 4096 - 256 = 3840 \implies n = 16.\)

This method works only if the solution is an integer.

and That is the answer ! THANKS infinitely JeffM..# but this was a question for level 6 exam for 11 year olds I mean !!!!!!!!!!!!!!!!! what are they expecting ! You are an absolute GEM

 

[pka]

a question for level 6 exam for 11 year olds I mean what are they expecting ! You are an absolute GEM

It would have been nice if you had told us the background to begin with.

Note that \(\displaystyle n^3-n^2=n^2(n-1)\).

Factor and get \(\displaystyle 2^8\cdot 3\cdot 5=(16)^2(15)\).

 

[JeffM]

It would have been nice if you had told us the background to begin with.

Note that \(\displaystyle n^3-n^2=n^2(n-1)\).

Factor and get \(\displaystyle 2^8\cdot 3\cdot 5=(16)^2(15)\).

Elegant.