Question

Show that there is an infinity of integers n such that the base-ten development of 2^n begins with 7.

Answer 1

thank you in advance for your help

Answer 2

what i mean by base-ten development ,example:

123 827 = 1×10^5 + 2×10^4 + 3×10^3 + 8×10^2 + 2×10^1 + 7×10^0

Answer 3

But that's impossible. 2^n is always even.

Answer 4

yes 2^n is even . So what's the problem ?

Answer 5

oh I see,123 827 is an example. and 7 was just a chance. what I mean by 'starts with 7' is that: \[2^{n}=7\times10^{q}+s \times10^{q -1}+............\]

Answer 6

Oh... I misinterpreted the question. -_-

Answer 7

okay, but if you have an idea or you know someone to do it, do not hesitate

Answer 8

what i know is that's generaly:we can show that there are infinitely many n such that p ^ n starts with d.

Answer 9

n = 46 + 10 k, k=0,1,2, 3 , ......

Answer 10

how??

Answer 11

Here are the first 6:

{{46, 70368744177664},
{56, 72057594037927936},
{66, 73786976294838206464},
{76, 75557863725914323419136},
{86, 77371252455336267181195264},
{96, 79228162514264337593543950336}}

It is your job to finish the proof.

Answer 12

thank you very much

Answer 13

even if the proof interests me the most

Answer 14

but at least I know what I need to find

Answer 15

You can prove that if 2^k starts with 7, then 2^{k+10} also starts with 7. If you can show that, then you are done if you know that
10^{46}=0368744177664

Answer 16

nice i'll try that

Answer 17

Also notice that\[
2^{10} \times 7 = 10 24 \times 7 = 7168
\]

Answer 18

You can prove that if 2^k starts with 7, then 2^{k+10} also starts with 7.
If you can show that, then you are done if you know that
10^{46}=70368744177664
I forgot the 7 in my previous post.