If $2^n=(… underbrace{aa…aaa}_{xtext{ times}})_{10}$ what is maximum of $x$?
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If $2^n=(... underbrace{aa...aaa}_{xtext{ times}})_{10}$ for positive integer $n$ then we have
$$2^n=(... underbrace{aa...aaa}_{xtext{ times}})_{10}=(bunderbrace{aa...aaa}_{xtext{ times}})_{10}=10^x b+a(10^{x-1}+...+1)$$
Where $a$ is a digit between $0$ and $9$ and $b$ is a natural number.
Simplifying more gives
$$9 times2^n=9b times 10^x + a (10^x-1)$$
Now if $n>3$ and $x>3$ then taking mod 2 gives $v_2(a)>3$ which is impossible. So one $x$ or $n$ is not greater than $3$. But the book says answer is $x=125$. Am I interpret this wrong?
elementary-number-theory
asked Nov 15 at 22:31
Ghartal
2,86211335
add a comment |
up vote
3
down vote
favorite
If $2^n=(... underbrace{aa...aaa}_{xtext{ times}})_{10}$ for positive integer $n$ then we have
$$2^n=(... underbrace{aa...aaa}_{xtext{ times}})_{10}=(bunderbrace{aa...aaa}_{xtext{ times}})_{10}=10^x b+a(10^{x-1}+...+1)$$
Where $a$ is a digit between $0$ and $9$ and $b$ is a natural number.
Simplifying more gives
$$9 times2^n=9b times 10^x + a (10^x-1)$$
Now if $n>3$ and $x>3$ then taking mod 2 gives $v_2(a)>3$ which is impossible. So one $x$ or $n$ is not greater than $3$. But the book says answer is $x=125$. Am I interpret this wrong?
elementary-number-theory
asked Nov 15 at 22:31
Ghartal
2,86211335
2
Yeah, $x=125$ definitely seems wrong.
– Thomas Andrews
Nov 15 at 22:47
1
I see that your solution is right. For a different approach, one can prove that $2^{n+500} equiv 2^n pmod{10^4}$ for all $n geq 4$ (by using $varphi(5^4)=500$, for instance), and up to $n leq 503$, $x$ only takes the values ${1, 2, 3}$, with $2^{39} = 549755813888$ the smallest case with $x=3$.
– Sangchul Lee
Nov 15 at 22:57
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If $2^n=(... underbrace{aa...aaa}_{xtext{ times}})_{10}$ for positive integer $n$ then we have
$$2^n=(... underbrace{aa...aaa}_{xtext{ times}})_{10}=(bunderbrace{aa...aaa}_{xtext{ times}})_{10}=10^x b+a(10^{x-1}+...+1)$$
Where $a$ is a digit between $0$ and $9$ and $b$ is a natural number.
Simplifying more gives
$$9 times2^n=9b times 10^x + a (10^x-1)$$
Now if $n>3$ and $x>3$ then taking mod 2 gives $v_2(a)>3$ which is impossible. So one $x$ or $n$ is not greater than $3$. But the book says answer is $x=125$. Am I interpret this wrong?
elementary-number-theory
asked Nov 15 at 22:31
Ghartal
2,86211335
If $2^n=(... underbrace{aa...aaa}_{xtext{ times}})_{10}$ for positive integer $n$ then we have
$$2^n=(... underbrace{aa...aaa}_{xtext{ times}})_{10}=(bunderbrace{aa...aaa}_{xtext{ times}})_{10}=10^x b+a(10^{x-1}+...+1)$$
Where $a$ is a digit between $0$ and $9$ and $b$ is a natural number.
Simplifying more gives
$$9 times2^n=9b times 10^x + a (10^x-1)$$
Now if $n>3$ and $x>3$ then taking mod 2 gives $v_2(a)>3$ which is impossible. So one $x$ or $n$ is not greater than $3$. But the book says answer is $x=125$. Am I interpret this wrong?
elementary-number-theory
elementary-number-theory
asked Nov 15 at 22:31
Ghartal
2,86211335
asked Nov 15 at 22:31
Ghartal
2,86211335
asked Nov 15 at 22:31
Ghartal
2,86211335
asked Nov 15 at 22:31
Ghartal
2,86211335
asked Nov 15 at 22:31
Ghartal
2,86211335
2,86211335
2
Yeah, $x=125$ definitely seems wrong.
– Thomas Andrews
Nov 15 at 22:47
1
I see that your solution is right. For a different approach, one can prove that $2^{n+500} equiv 2^n pmod{10^4}$ for all $n geq 4$ (by using $varphi(5^4)=500$, for instance), and up to $n leq 503$, $x$ only takes the values ${1, 2, 3}$, with $2^{39} = 549755813888$ the smallest case with $x=3$.
– Sangchul Lee
Nov 15 at 22:57
add a comment |
2
Yeah, $x=125$ definitely seems wrong.
– Thomas Andrews
Nov 15 at 22:47
1
I see that your solution is right. For a different approach, one can prove that $2^{n+500} equiv 2^n pmod{10^4}$ for all $n geq 4$ (by using $varphi(5^4)=500$, for instance), and up to $n leq 503$, $x$ only takes the values ${1, 2, 3}$, with $2^{39} = 549755813888$ the smallest case with $x=3$.
– Sangchul Lee
Nov 15 at 22:57
2
2
Yeah, $x=125$ definitely seems wrong.
– Thomas Andrews
Nov 15 at 22:47
Yeah, $x=125$ definitely seems wrong.
– Thomas Andrews
Nov 15 at 22:47
1
1
I see that your solution is right. For a different approach, one can prove that $2^{n+500} equiv 2^n pmod{10^4}$ for all $n geq 4$ (by using $varphi(5^4)=500$, for instance), and up to $n leq 503$, $x$ only takes the values ${1, 2, 3}$, with $2^{39} = 549755813888$ the smallest case with $x=3$.
– Sangchul Lee
Nov 15 at 22:57
I see that your solution is right. For a different approach, one can prove that $2^{n+500} equiv 2^n pmod{10^4}$ for all $n geq 4$ (by using $varphi(5^4)=500$, for instance), and up to $n leq 503$, $x$ only takes the values ${1, 2, 3}$, with $2^{39} = 549755813888$ the smallest case with $x=3$.
– Sangchul Lee
Nov 15 at 22:57
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If $nge 4$, then $2^n$ must be a multiple of $16$, therefore so are its last four digits. There are no factors of $5$ so the last four digits cannot be $0000$. Now, which other possibilities for four identical last digits will give a multiple of $16$? Try them out and see.
The comments identify $2^{39}=...888$. If you work out the part above you are left with what must be the answer.
answered Nov 16 at 1:21
Oscar Lanzi
11.8k11936
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $nge 4$, then $2^n$ must be a multiple of $16$, therefore so are its last four digits. There are no factors of $5$ so the last four digits cannot be $0000$. Now, which other possibilities for four identical last digits will give a multiple of $16$? Try them out and see.
The comments identify $2^{39}=...888$. If you work out the part above you are left with what must be the answer.
answered Nov 16 at 1:21
Oscar Lanzi
11.8k11936
add a comment |
up vote
1
down vote
accepted
If $nge 4$, then $2^n$ must be a multiple of $16$, therefore so are its last four digits. There are no factors of $5$ so the last four digits cannot be $0000$. Now, which other possibilities for four identical last digits will give a multiple of $16$? Try them out and see.
The comments identify $2^{39}=...888$. If you work out the part above you are left with what must be the answer.
answered Nov 16 at 1:21
Oscar Lanzi
11.8k11936
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $nge 4$, then $2^n$ must be a multiple of $16$, therefore so are its last four digits. There are no factors of $5$ so the last four digits cannot be $0000$. Now, which other possibilities for four identical last digits will give a multiple of $16$? Try them out and see.
The comments identify $2^{39}=...888$. If you work out the part above you are left with what must be the answer.
answered Nov 16 at 1:21
Oscar Lanzi
11.8k11936
If $nge 4$, then $2^n$ must be a multiple of $16$, therefore so are its last four digits. There are no factors of $5$ so the last four digits cannot be $0000$. Now, which other possibilities for four identical last digits will give a multiple of $16$? Try them out and see.
The comments identify $2^{39}=...888$. If you work out the part above you are left with what must be the answer.
answered Nov 16 at 1:21
Oscar Lanzi
11.8k11936
edited Nov 16 at 1:30
answered Nov 16 at 1:21
Oscar Lanzi
11.8k11936
answered Nov 16 at 1:21
Oscar Lanzi
11.8k11936
answered Nov 16 at 1:21
Oscar Lanzi
11.8k11936
11.8k11936
add a comment |
add a comment |
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2
Yeah, $x=125$ definitely seems wrong.
– Thomas Andrews
Nov 15 at 22:47
1
I see that your solution is right. For a different approach, one can prove that $2^{n+500} equiv 2^n pmod{10^4}$ for all $n geq 4$ (by using $varphi(5^4)=500$, for instance), and up to $n leq 503$, $x$ only takes the values ${1, 2, 3}$, with $2^{39} = 549755813888$ the smallest case with $x=3$.
– Sangchul Lee
Nov 15 at 22:57