Powers of an integer

Number Theory Level 3

Three positive integers x , x 2 x, x^{2} and x 3 x^{3} all begin with the same digit. Does this imply that the common first digit is 1?

Yes No

Ppk K by ppk k , 20, India

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.

2 solutions

Marta Reece
Mar 2, 2017

That digit could be a 9 9 , for example 999 , 99 9 2 = 998001 , 99 9 3 = 997002999. 999, 999^2=998 001, 999^3=997 002 999.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

yes, basically the answer is no.

ppk k - 4 years, 3 months ago
Ppk K
Feb 27, 2017

Answer: NOT.

We can always find such k k that x = 1 0 k − 1 x= 10^k-1 and x , x 2 , . . . , x n x,x^2,...,x^n all begin with 9 9 .

Proof: Obvious, that enough to prove, that $x^n$ begin with $9$ so $(10^k-1)^n>9*10^{kn-1}$ $(10^k-1)^n=10^{kn}-n10^{k(n-1)}+...>10^{kn}-n10^{k(n-1)}-C {[\frac{n}{2}]}^{n}(1+10^k+...+1+10^{k(n-2)})>10^{kn}-n10^{k(n-1)}-C {[\frac{n}{2}]}^{n} 10^{k(n-1)}$

Set $k> \log {10}{(C {[\frac{n}{2}]}^{n}+n)}+1$ Then $10^{k-1}>C {[\frac{n}{2}]}^{n}+n$ and $10^{kn}-n10^{k(n-1)}-C {[\frac{n}{2}]}^{n} 10^{k(n-1)}>10^{kn}-10^{k-1}10^{k(n-1)}=9*10^{kn-1}$

1 Helpful 0 Interesting 0 Brilliant 0 Confused

FYI To use Latex, we use " \ ( \ ) \backslash ( \quad \backslash ) " instead of "$ $". I've edited the first sentence for your reference.

Calvin Lin Staff - 4 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...