neutrino_super =>can you solve this ?????????????????????????????????????????????
The natural 'n' for which 3^9+3^12+3^15+3^n is a perfect cube of an integer is
The answer is 14.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Could you format stuff please next time? I'm not getting any of that xD
( 3 3 ) 3 + ( 3 3 ) 4 + ( 3 3 ) 5 + 3 n
= ( 3 3 ) 3 + ( 3 3 ) 3 ( 3 3 ) + ( 3 3 ) 3 ( 3 3 ) 2 + 3 n
Taking 3 n = x ,
= ( ( 3 3 ) 3 ) ( 1 + 2 7 + 7 2 9 + x )
= ( ( 3 3 ) 3 ) ( 7 5 7 + x )
We need to make ( 7 5 7 + x ) to the next nearest perfect cube, which is 1000, so
x = 2 4 3 = 3 5
Then,
3 n = ( ( 3 3 ) 3 ) ( x )
3 n = ( ( 3 3 ) 3 ) ( 3 5 )
3 n = 3 1 4
n = 1 4
(3^3)^3+(3^3)^4+(3^3)^5+3^n
=(3^3)^3+(3^3)^3(3^3)+(3^3)^3(3^3)^2+3^n
=(3^3)^3(1+27+729+x)
=(3^3)^3(757+x)
we need to make (757+x) to the next nearest perfect cube which is 1000, so x=243=3^5
then,
3^n=(3^3)^3(x)
3^n=(3^3)^3(3^5)
n=3^14