$2^n$

Number Theory Level 3

Can a number $2^n$ ever have four equal right most digits?

For example 842222 or 346262222.

*Bonus : Prove your answer

No Yes

1 solution

Zk Lin
Feb 14, 2016

The answer is no.

Since a power of $2$ can only end in $2,4,6$ or $8$ , write any such number as $2^{n}=10000a+1111b$ , where $b=2,4,6$ or $8$ with $a>0$

Suppose that $b=2$ , then

$2^{n}=10000a+2222$

$2^{n-1}=5000a+1111$ . $(n > 1$ since $5000a>1)$

The L.H.S. is even, but R.H.S. is not, a contradiction.

Suppose that $b=4$ , then

$2^{n}=10000a+4444$

$2^{n-2}=2500a+1111$ . $(n > 2$ since $2500a>1)$

The L.H.S. is even, but R.H.S. is not, a contradiction.

Suppose that $b=6$ , then

$2^{n}=10000a+6666$

$2^{n-1}=5000a+3333$ . $(n > 1$ since $5000a>1)$

The L.H.S. is even, but R.H.S. is not, a contradiction.

Suppose that $b=8$ , then

$2^{n}=10000a+8888$

$2^{n-3}=1250a+1111$ . $(n > 3$ since $1250a>1)$

The L.H.S. is even, but R.H.S. is not, a contradiction.

The conclusion follows.

Great solution. Using Chinese Remainder Theorem, we simply need to show that for $n \geq 5$ ,

$0 \equiv 2 ^n \equiv 1111 \times k \pmod{2^4 }$

has no solution where $1 \leq k \leq 9$ . This is obvious.

Nice solution!

Soumava Pal - 5 years, 3 months ago

Exactly Same Way.

Kushagra Sahni - 5 years, 4 months ago