OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

thank you in advance for your help

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (blockcolder):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (blockcolder):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

how??

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

thank you very much

5 years ago
OpenStudy (anonymous):

even if the proof interests me the most

5 years ago
OpenStudy (anonymous):

but at least I know what I need to find

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

nice i'll try that

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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.

5 years ago
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