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Are there any non-trivial examples of this decimal-binary property
The 2019 Stack Overflow Developer Survey Results Are InDo the digits of $pi$ contain every possible finite-length digit sequence?Can we factor an integer N if we know some of the digits of an associated number M?Is there a formula to calculate base digit swaps given two specific bases?Can a number be palindrome in 4 consecutive number bases?For which bases is the prime digit products question still not settled?What percentage of proper fractions using n digits or less are in their simplest form?Find $c=atimes b$ such that $c$ has digits from either $a$ or $b$Palindromes in multiple basesSecond digit of square numbers in binary yields $sqrt2$Are there more decimal or binary numbers in the world?
$begingroup$
My birthday is 10th of October or 1010 in MMDD format.
I just realized that 1010 contains two copies of the number 10 and if spelled out in binary,
$1010_2=10$
I was wondering how many other numbers have this property
In general we can use the following formula
$N=sum_j=0^n a_j*2^j$ + $sum_j=n+1^2n+1a_j-n-1*2^j$
$N=sum_k=0^n a_k*10^k$
(In both equations all the $a_i$ are either $0$ or $1$, n is one less than the number of digits in the repeated digit sequence, so for $1010_2$, $n=1$, $a_0=0$, $a_1=1$)
Subtracting the second from the first and factoring coefficients,
$0=sum_d=0^n a_d*(2^d+2^n+d+1-10^d)$
$0=sum_d=0^n a_d*2^d*(2^n+1+1-5^d)$
I am not sure how to solve from here on out
Also would like to know if there are any other solutions besides 0 and 1 in other bases.
elementary-number-theory binary decimal-expansion
edited Mar 23 at 5:27
Arturo Magidin
266k34590921
asked Mar 23 at 5:16
amanaman
34310
$endgroup$
add a comment |
$begingroup$
My birthday is 10th of October or 1010 in MMDD format.
I just realized that 1010 contains two copies of the number 10 and if spelled out in binary,
$1010_2=10$
I was wondering how many other numbers have this property
In general we can use the following formula
$N=sum_j=0^n a_j*2^j$ + $sum_j=n+1^2n+1a_j-n-1*2^j$
$N=sum_k=0^n a_k*10^k$
(In both equations all the $a_i$ are either $0$ or $1$, n is one less than the number of digits in the repeated digit sequence, so for $1010_2$, $n=1$, $a_0=0$, $a_1=1$)
Subtracting the second from the first and factoring coefficients,
$0=sum_d=0^n a_d*(2^d+2^n+d+1-10^d)$
$0=sum_d=0^n a_d*2^d*(2^n+1+1-5^d)$
I am not sure how to solve from here on out
Also would like to know if there are any other solutions besides 0 and 1 in other bases.
elementary-number-theory binary decimal-expansion
edited Mar 23 at 5:27
Arturo Magidin
266k34590921
asked Mar 23 at 5:16
amanaman
34310
$endgroup$
$begingroup$
A small comment on notation: you set this up to equate an $(n+1)$-digit decimal numeral with a $(2n+2)$-digit binary numeral. It might be simpler to compare an $n$-digit number with a $2n$-digit number.
$endgroup$
– David K
Mar 23 at 20:09
1
$begingroup$
Requiring a leading $0$ in binary & base $10$ expansions results in relatively few values to check (& I believe the answer is there are no more, but I haven't confirmed) as shown by the answers below. However, you did not explicitly state this is required. Thus, for example, if you allow the use of leading zeros, a valid value to check on could have a base $2$ expansion of $000101000101_2$. This more general case is, I believe, quite interesting. There'll be few values which fit your criteria, but I suspect there may still be infinitely many. If you're interested in this, please specify it.
$endgroup$
– John Omielan
Mar 24 at 1:26
$begingroup$
Yes, that is interesting, though I had not considered it initially
$endgroup$
– aman
Mar 24 at 3:00
add a comment |
$begingroup$
My birthday is 10th of October or 1010 in MMDD format.
I just realized that 1010 contains two copies of the number 10 and if spelled out in binary,
$1010_2=10$
I was wondering how many other numbers have this property
In general we can use the following formula
$N=sum_j=0^n a_j*2^j$ + $sum_j=n+1^2n+1a_j-n-1*2^j$
$N=sum_k=0^n a_k*10^k$
(In both equations all the $a_i$ are either $0$ or $1$, n is one less than the number of digits in the repeated digit sequence, so for $1010_2$, $n=1$, $a_0=0$, $a_1=1$)
Subtracting the second from the first and factoring coefficients,
$0=sum_d=0^n a_d*(2^d+2^n+d+1-10^d)$
$0=sum_d=0^n a_d*2^d*(2^n+1+1-5^d)$
I am not sure how to solve from here on out
Also would like to know if there are any other solutions besides 0 and 1 in other bases.
elementary-number-theory binary decimal-expansion
edited Mar 23 at 5:27
Arturo Magidin
266k34590921
asked Mar 23 at 5:16
amanaman
34310
$endgroup$
My birthday is 10th of October or 1010 in MMDD format.
I just realized that 1010 contains two copies of the number 10 and if spelled out in binary,
$1010_2=10$
I was wondering how many other numbers have this property
In general we can use the following formula
$N=sum_j=0^n a_j*2^j$ + $sum_j=n+1^2n+1a_j-n-1*2^j$
$N=sum_k=0^n a_k*10^k$
(In both equations all the $a_i$ are either $0$ or $1$, n is one less than the number of digits in the repeated digit sequence, so for $1010_2$, $n=1$, $a_0=0$, $a_1=1$)
Subtracting the second from the first and factoring coefficients,
$0=sum_d=0^n a_d*(2^d+2^n+d+1-10^d)$
$0=sum_d=0^n a_d*2^d*(2^n+1+1-5^d)$
I am not sure how to solve from here on out
Also would like to know if there are any other solutions besides 0 and 1 in other bases.
elementary-number-theory binary decimal-expansion
elementary-number-theory binary decimal-expansion
edited Mar 23 at 5:27
Arturo Magidin
266k34590921
asked Mar 23 at 5:16
amanaman
34310
edited Mar 23 at 5:27
Arturo Magidin
266k34590921
asked Mar 23 at 5:16
amanaman
34310
edited Mar 23 at 5:27
Arturo Magidin
266k34590921
edited Mar 23 at 5:27
Arturo Magidin
266k34590921
edited Mar 23 at 5:27
Arturo Magidin
266k34590921
266k34590921
asked Mar 23 at 5:16
amanaman
34310
asked Mar 23 at 5:16
amanaman
34310
asked Mar 23 at 5:16
amanaman
34310
34310
$begingroup$
A small comment on notation: you set this up to equate an $(n+1)$-digit decimal numeral with a $(2n+2)$-digit binary numeral. It might be simpler to compare an $n$-digit number with a $2n$-digit number.
$endgroup$
– David K
Mar 23 at 20:09
1
$begingroup$
Requiring a leading $0$ in binary & base $10$ expansions results in relatively few values to check (& I believe the answer is there are no more, but I haven't confirmed) as shown by the answers below. However, you did not explicitly state this is required. Thus, for example, if you allow the use of leading zeros, a valid value to check on could have a base $2$ expansion of $000101000101_2$. This more general case is, I believe, quite interesting. There'll be few values which fit your criteria, but I suspect there may still be infinitely many. If you're interested in this, please specify it.
$endgroup$
– John Omielan
Mar 24 at 1:26
$begingroup$
Yes, that is interesting, though I had not considered it initially
$endgroup$
– aman
Mar 24 at 3:00
add a comment |
$begingroup$
A small comment on notation: you set this up to equate an $(n+1)$-digit decimal numeral with a $(2n+2)$-digit binary numeral. It might be simpler to compare an $n$-digit number with a $2n$-digit number.
$endgroup$
– David K
Mar 23 at 20:09
1
$begingroup$
Requiring a leading $0$ in binary & base $10$ expansions results in relatively few values to check (& I believe the answer is there are no more, but I haven't confirmed) as shown by the answers below. However, you did not explicitly state this is required. Thus, for example, if you allow the use of leading zeros, a valid value to check on could have a base $2$ expansion of $000101000101_2$. This more general case is, I believe, quite interesting. There'll be few values which fit your criteria, but I suspect there may still be infinitely many. If you're interested in this, please specify it.
$endgroup$
– John Omielan
Mar 24 at 1:26
$begingroup$
Yes, that is interesting, though I had not considered it initially
$endgroup$
– aman
Mar 24 at 3:00
$begingroup$
A small comment on notation: you set this up to equate an $(n+1)$-digit decimal numeral with a $(2n+2)$-digit binary numeral. It might be simpler to compare an $n$-digit number with a $2n$-digit number.
$endgroup$
– David K
Mar 23 at 20:09
$begingroup$
A small comment on notation: you set this up to equate an $(n+1)$-digit decimal numeral with a $(2n+2)$-digit binary numeral. It might be simpler to compare an $n$-digit number with a $2n$-digit number.
$endgroup$
– David K
Mar 23 at 20:09
1
1
$begingroup$
Requiring a leading $0$ in binary & base $10$ expansions results in relatively few values to check (& I believe the answer is there are no more, but I haven't confirmed) as shown by the answers below. However, you did not explicitly state this is required. Thus, for example, if you allow the use of leading zeros, a valid value to check on could have a base $2$ expansion of $000101000101_2$. This more general case is, I believe, quite interesting. There'll be few values which fit your criteria, but I suspect there may still be infinitely many. If you're interested in this, please specify it.
$endgroup$
– John Omielan
Mar 24 at 1:26
$begingroup$
Requiring a leading $0$ in binary & base $10$ expansions results in relatively few values to check (& I believe the answer is there are no more, but I haven't confirmed) as shown by the answers below. However, you did not explicitly state this is required. Thus, for example, if you allow the use of leading zeros, a valid value to check on could have a base $2$ expansion of $000101000101_2$. This more general case is, I believe, quite interesting. There'll be few values which fit your criteria, but I suspect there may still be infinitely many. If you're interested in this, please specify it.
$endgroup$
– John Omielan
Mar 24 at 1:26
$begingroup$
Yes, that is interesting, though I had not considered it initially
$endgroup$
– aman
Mar 24 at 3:00
$begingroup$
Yes, that is interesting, though I had not considered it initially
$endgroup$
– aman
Mar 24 at 3:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The number of digits in the decimal expression of a number is $lfloor log_10 N rfloor + 1$, and the number of bits in its binary expansion is $lfloor log_2 N rfloor + 1$, so any number $N$ with the given reduplication property must satisfy
$$lfloor log_2 N rfloor + 1 = 2 (lfloor log_10 N rfloor + 1) .$$
Using twice that $log_b N geq lfloor log_b N rfloor > log_b N - 1$ gives
$$log_2 N < 2 log_10 N + 2 = frac2log_2 10 log_2 N + 2.$$
Rearranging gives
$$log_2 N leq frac2 log_2 10log_2 10 - 2 < 6,$$
so $N < 2^6 ,$ but this leaves only $0, 1, 10, 11$ to check.
answered Mar 23 at 19:13
TravisTravis
64.1k769151
$endgroup$
1
$begingroup$
Very nice. Sometimes it's better to work with easier necessary conditions than the exact statement.
$endgroup$
– amsmath
Mar 23 at 20:13
1
$begingroup$
Another way to look at it: setting $n$ as in the original question, the decimal number is at least $10^n,$ and the binary number is less than $4cdot4^n,$ so you can have a solution only if $10^n<4cdot4^n,$ or $2.5^n<4.$ Since $2.5^2 = 6.25,$ we must have $n<2,$ so the decimal number can have only $1+1 =2$ digits at most. Basically the same logic but with exponents instead of logarithms.
$endgroup$
– David K
Mar 23 at 20:16
$begingroup$
Nicely put, David.
$endgroup$
– Travis
Mar 23 at 20:30
$begingroup$
Thanks for the answer. However, N should be used rather than n, that confused me initially
$endgroup$
– aman
Mar 24 at 3:04
$begingroup$
You're welcome. I've changed the notation to match that in the question.
$endgroup$
– Travis
Mar 24 at 3:28
add a comment |
$begingroup$
This is only a partial answer. For $n>1$ there is always a smallest number $d_0le n$ such that $2^n+1+1-5^d_0 < 0$. So, your equation becomes
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = sum_d=d_0^na_dcdot 2^d(5^d-(2^n+1+1)).
$$
Assume that one of $a_d_0+1,ldots,a_n$ is non-zero. Then the RHS is at least as large as $2^d_0+1(5^d_0+1-(2^n+1+1))$, which by definition of $d_0$ is larger than $2^d_0+1(5(2^n+1+1)-(2^n+1+1)) = 2^d_0+3(2^n+1+1)$. The LHS can be estimated by
beginalign*
LHS
&le sum_d=0^d_0-12^d(2^n+1+1-5^d) = (2^n+1+1)(2^d_0-1) - frac10^d_0-19\
&le (2^n+1+1)(2^d_0-1) - frac(2^n+1+1)92^d_0 + frac 19\
&= (2^n+1+1)left(frac 89,2^d_0-1right) + frac 19 < (2^n+1+1)left(2^d_0-1right) + frac 19\
&< (2^n+1+1)2^d_0+3 < RHS.
endalign*
Hence, we conclude that $a_d_0+1=cdots=a_n = 0$. This reduces your equation to
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = 2^d_0(5^d_0-(2^n+1+1)).
$$
answered Mar 23 at 18:33
amsmathamsmath
3,287421
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number of digits in the decimal expression of a number is $lfloor log_10 N rfloor + 1$, and the number of bits in its binary expansion is $lfloor log_2 N rfloor + 1$, so any number $N$ with the given reduplication property must satisfy
$$lfloor log_2 N rfloor + 1 = 2 (lfloor log_10 N rfloor + 1) .$$
Using twice that $log_b N geq lfloor log_b N rfloor > log_b N - 1$ gives
$$log_2 N < 2 log_10 N + 2 = frac2log_2 10 log_2 N + 2.$$
Rearranging gives
$$log_2 N leq frac2 log_2 10log_2 10 - 2 < 6,$$
so $N < 2^6 ,$ but this leaves only $0, 1, 10, 11$ to check.
answered Mar 23 at 19:13
TravisTravis
64.1k769151
$endgroup$
1
$begingroup$
Very nice. Sometimes it's better to work with easier necessary conditions than the exact statement.
$endgroup$
– amsmath
Mar 23 at 20:13
1
$begingroup$
Another way to look at it: setting $n$ as in the original question, the decimal number is at least $10^n,$ and the binary number is less than $4cdot4^n,$ so you can have a solution only if $10^n<4cdot4^n,$ or $2.5^n<4.$ Since $2.5^2 = 6.25,$ we must have $n<2,$ so the decimal number can have only $1+1 =2$ digits at most. Basically the same logic but with exponents instead of logarithms.
$endgroup$
– David K
Mar 23 at 20:16
$begingroup$
Nicely put, David.
$endgroup$
– Travis
Mar 23 at 20:30
$begingroup$
Thanks for the answer. However, N should be used rather than n, that confused me initially
$endgroup$
– aman
Mar 24 at 3:04
$begingroup$
You're welcome. I've changed the notation to match that in the question.
$endgroup$
– Travis
Mar 24 at 3:28
add a comment |
$begingroup$
The number of digits in the decimal expression of a number is $lfloor log_10 N rfloor + 1$, and the number of bits in its binary expansion is $lfloor log_2 N rfloor + 1$, so any number $N$ with the given reduplication property must satisfy
$$lfloor log_2 N rfloor + 1 = 2 (lfloor log_10 N rfloor + 1) .$$
Using twice that $log_b N geq lfloor log_b N rfloor > log_b N - 1$ gives
$$log_2 N < 2 log_10 N + 2 = frac2log_2 10 log_2 N + 2.$$
Rearranging gives
$$log_2 N leq frac2 log_2 10log_2 10 - 2 < 6,$$
so $N < 2^6 ,$ but this leaves only $0, 1, 10, 11$ to check.
answered Mar 23 at 19:13
TravisTravis
64.1k769151
$endgroup$
1
$begingroup$
Very nice. Sometimes it's better to work with easier necessary conditions than the exact statement.
$endgroup$
– amsmath
Mar 23 at 20:13
1
$begingroup$
Another way to look at it: setting $n$ as in the original question, the decimal number is at least $10^n,$ and the binary number is less than $4cdot4^n,$ so you can have a solution only if $10^n<4cdot4^n,$ or $2.5^n<4.$ Since $2.5^2 = 6.25,$ we must have $n<2,$ so the decimal number can have only $1+1 =2$ digits at most. Basically the same logic but with exponents instead of logarithms.
$endgroup$
– David K
Mar 23 at 20:16
$begingroup$
Nicely put, David.
$endgroup$
– Travis
Mar 23 at 20:30
$begingroup$
Thanks for the answer. However, N should be used rather than n, that confused me initially
$endgroup$
– aman
Mar 24 at 3:04
$begingroup$
You're welcome. I've changed the notation to match that in the question.
$endgroup$
– Travis
Mar 24 at 3:28
add a comment |
$begingroup$
The number of digits in the decimal expression of a number is $lfloor log_10 N rfloor + 1$, and the number of bits in its binary expansion is $lfloor log_2 N rfloor + 1$, so any number $N$ with the given reduplication property must satisfy
$$lfloor log_2 N rfloor + 1 = 2 (lfloor log_10 N rfloor + 1) .$$
Using twice that $log_b N geq lfloor log_b N rfloor > log_b N - 1$ gives
$$log_2 N < 2 log_10 N + 2 = frac2log_2 10 log_2 N + 2.$$
Rearranging gives
$$log_2 N leq frac2 log_2 10log_2 10 - 2 < 6,$$
so $N < 2^6 ,$ but this leaves only $0, 1, 10, 11$ to check.
answered Mar 23 at 19:13
TravisTravis
64.1k769151
$endgroup$
The number of digits in the decimal expression of a number is $lfloor log_10 N rfloor + 1$, and the number of bits in its binary expansion is $lfloor log_2 N rfloor + 1$, so any number $N$ with the given reduplication property must satisfy
$$lfloor log_2 N rfloor + 1 = 2 (lfloor log_10 N rfloor + 1) .$$
Using twice that $log_b N geq lfloor log_b N rfloor > log_b N - 1$ gives
$$log_2 N < 2 log_10 N + 2 = frac2log_2 10 log_2 N + 2.$$
Rearranging gives
$$log_2 N leq frac2 log_2 10log_2 10 - 2 < 6,$$
so $N < 2^6 ,$ but this leaves only $0, 1, 10, 11$ to check.
answered Mar 23 at 19:13
TravisTravis
64.1k769151
edited Mar 24 at 3:26
answered Mar 23 at 19:13
TravisTravis
64.1k769151
answered Mar 23 at 19:13
TravisTravis
64.1k769151
answered Mar 23 at 19:13
TravisTravis
64.1k769151
64.1k769151
1
$begingroup$
Very nice. Sometimes it's better to work with easier necessary conditions than the exact statement.
$endgroup$
– amsmath
Mar 23 at 20:13
1
$begingroup$
Another way to look at it: setting $n$ as in the original question, the decimal number is at least $10^n,$ and the binary number is less than $4cdot4^n,$ so you can have a solution only if $10^n<4cdot4^n,$ or $2.5^n<4.$ Since $2.5^2 = 6.25,$ we must have $n<2,$ so the decimal number can have only $1+1 =2$ digits at most. Basically the same logic but with exponents instead of logarithms.
$endgroup$
– David K
Mar 23 at 20:16
$begingroup$
Nicely put, David.
$endgroup$
– Travis
Mar 23 at 20:30
$begingroup$
Thanks for the answer. However, N should be used rather than n, that confused me initially
$endgroup$
– aman
Mar 24 at 3:04
$begingroup$
You're welcome. I've changed the notation to match that in the question.
$endgroup$
– Travis
Mar 24 at 3:28
add a comment |
1
$begingroup$
Very nice. Sometimes it's better to work with easier necessary conditions than the exact statement.
$endgroup$
– amsmath
Mar 23 at 20:13
1
$begingroup$
Another way to look at it: setting $n$ as in the original question, the decimal number is at least $10^n,$ and the binary number is less than $4cdot4^n,$ so you can have a solution only if $10^n<4cdot4^n,$ or $2.5^n<4.$ Since $2.5^2 = 6.25,$ we must have $n<2,$ so the decimal number can have only $1+1 =2$ digits at most. Basically the same logic but with exponents instead of logarithms.
$endgroup$
– David K
Mar 23 at 20:16
$begingroup$
Nicely put, David.
$endgroup$
– Travis
Mar 23 at 20:30
$begingroup$
Thanks for the answer. However, N should be used rather than n, that confused me initially
$endgroup$
– aman
Mar 24 at 3:04
$begingroup$
You're welcome. I've changed the notation to match that in the question.
$endgroup$
– Travis
Mar 24 at 3:28
1
1
$begingroup$
Very nice. Sometimes it's better to work with easier necessary conditions than the exact statement.
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– amsmath
Mar 23 at 20:13
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Very nice. Sometimes it's better to work with easier necessary conditions than the exact statement.
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– amsmath
Mar 23 at 20:13
1
1
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Another way to look at it: setting $n$ as in the original question, the decimal number is at least $10^n,$ and the binary number is less than $4cdot4^n,$ so you can have a solution only if $10^n<4cdot4^n,$ or $2.5^n<4.$ Since $2.5^2 = 6.25,$ we must have $n<2,$ so the decimal number can have only $1+1 =2$ digits at most. Basically the same logic but with exponents instead of logarithms.
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– David K
Mar 23 at 20:16
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Another way to look at it: setting $n$ as in the original question, the decimal number is at least $10^n,$ and the binary number is less than $4cdot4^n,$ so you can have a solution only if $10^n<4cdot4^n,$ or $2.5^n<4.$ Since $2.5^2 = 6.25,$ we must have $n<2,$ so the decimal number can have only $1+1 =2$ digits at most. Basically the same logic but with exponents instead of logarithms.
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– David K
Mar 23 at 20:16
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Nicely put, David.
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– Travis
Mar 23 at 20:30
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Nicely put, David.
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– Travis
Mar 23 at 20:30
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Thanks for the answer. However, N should be used rather than n, that confused me initially
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– aman
Mar 24 at 3:04
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Thanks for the answer. However, N should be used rather than n, that confused me initially
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– aman
Mar 24 at 3:04
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You're welcome. I've changed the notation to match that in the question.
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– Travis
Mar 24 at 3:28
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You're welcome. I've changed the notation to match that in the question.
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– Travis
Mar 24 at 3:28
add a comment |
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This is only a partial answer. For $n>1$ there is always a smallest number $d_0le n$ such that $2^n+1+1-5^d_0 < 0$. So, your equation becomes
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = sum_d=d_0^na_dcdot 2^d(5^d-(2^n+1+1)).
$$
Assume that one of $a_d_0+1,ldots,a_n$ is non-zero. Then the RHS is at least as large as $2^d_0+1(5^d_0+1-(2^n+1+1))$, which by definition of $d_0$ is larger than $2^d_0+1(5(2^n+1+1)-(2^n+1+1)) = 2^d_0+3(2^n+1+1)$. The LHS can be estimated by
beginalign*
LHS
&le sum_d=0^d_0-12^d(2^n+1+1-5^d) = (2^n+1+1)(2^d_0-1) - frac10^d_0-19\
&le (2^n+1+1)(2^d_0-1) - frac(2^n+1+1)92^d_0 + frac 19\
&= (2^n+1+1)left(frac 89,2^d_0-1right) + frac 19 < (2^n+1+1)left(2^d_0-1right) + frac 19\
&< (2^n+1+1)2^d_0+3 < RHS.
endalign*
Hence, we conclude that $a_d_0+1=cdots=a_n = 0$. This reduces your equation to
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = 2^d_0(5^d_0-(2^n+1+1)).
$$
answered Mar 23 at 18:33
amsmathamsmath
3,287421
$endgroup$
add a comment |
$begingroup$
This is only a partial answer. For $n>1$ there is always a smallest number $d_0le n$ such that $2^n+1+1-5^d_0 < 0$. So, your equation becomes
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = sum_d=d_0^na_dcdot 2^d(5^d-(2^n+1+1)).
$$
Assume that one of $a_d_0+1,ldots,a_n$ is non-zero. Then the RHS is at least as large as $2^d_0+1(5^d_0+1-(2^n+1+1))$, which by definition of $d_0$ is larger than $2^d_0+1(5(2^n+1+1)-(2^n+1+1)) = 2^d_0+3(2^n+1+1)$. The LHS can be estimated by
beginalign*
LHS
&le sum_d=0^d_0-12^d(2^n+1+1-5^d) = (2^n+1+1)(2^d_0-1) - frac10^d_0-19\
&le (2^n+1+1)(2^d_0-1) - frac(2^n+1+1)92^d_0 + frac 19\
&= (2^n+1+1)left(frac 89,2^d_0-1right) + frac 19 < (2^n+1+1)left(2^d_0-1right) + frac 19\
&< (2^n+1+1)2^d_0+3 < RHS.
endalign*
Hence, we conclude that $a_d_0+1=cdots=a_n = 0$. This reduces your equation to
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = 2^d_0(5^d_0-(2^n+1+1)).
$$
answered Mar 23 at 18:33
amsmathamsmath
3,287421
$endgroup$
add a comment |
$begingroup$
This is only a partial answer. For $n>1$ there is always a smallest number $d_0le n$ such that $2^n+1+1-5^d_0 < 0$. So, your equation becomes
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = sum_d=d_0^na_dcdot 2^d(5^d-(2^n+1+1)).
$$
Assume that one of $a_d_0+1,ldots,a_n$ is non-zero. Then the RHS is at least as large as $2^d_0+1(5^d_0+1-(2^n+1+1))$, which by definition of $d_0$ is larger than $2^d_0+1(5(2^n+1+1)-(2^n+1+1)) = 2^d_0+3(2^n+1+1)$. The LHS can be estimated by
beginalign*
LHS
&le sum_d=0^d_0-12^d(2^n+1+1-5^d) = (2^n+1+1)(2^d_0-1) - frac10^d_0-19\
&le (2^n+1+1)(2^d_0-1) - frac(2^n+1+1)92^d_0 + frac 19\
&= (2^n+1+1)left(frac 89,2^d_0-1right) + frac 19 < (2^n+1+1)left(2^d_0-1right) + frac 19\
&< (2^n+1+1)2^d_0+3 < RHS.
endalign*
Hence, we conclude that $a_d_0+1=cdots=a_n = 0$. This reduces your equation to
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = 2^d_0(5^d_0-(2^n+1+1)).
$$
answered Mar 23 at 18:33
amsmathamsmath
3,287421
$endgroup$
This is only a partial answer. For $n>1$ there is always a smallest number $d_0le n$ such that $2^n+1+1-5^d_0 < 0$. So, your equation becomes
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = sum_d=d_0^na_dcdot 2^d(5^d-(2^n+1+1)).
$$
Assume that one of $a_d_0+1,ldots,a_n$ is non-zero. Then the RHS is at least as large as $2^d_0+1(5^d_0+1-(2^n+1+1))$, which by definition of $d_0$ is larger than $2^d_0+1(5(2^n+1+1)-(2^n+1+1)) = 2^d_0+3(2^n+1+1)$. The LHS can be estimated by
beginalign*
LHS
&le sum_d=0^d_0-12^d(2^n+1+1-5^d) = (2^n+1+1)(2^d_0-1) - frac10^d_0-19\
&le (2^n+1+1)(2^d_0-1) - frac(2^n+1+1)92^d_0 + frac 19\
&= (2^n+1+1)left(frac 89,2^d_0-1right) + frac 19 < (2^n+1+1)left(2^d_0-1right) + frac 19\
&< (2^n+1+1)2^d_0+3 < RHS.
endalign*
Hence, we conclude that $a_d_0+1=cdots=a_n = 0$. This reduces your equation to
$$
sum_d=0^d_0-1a_dcdot 2^d(2^n+1+1-5^d) = 2^d_0(5^d_0-(2^n+1+1)).
$$
answered Mar 23 at 18:33
amsmathamsmath
3,287421
edited Mar 23 at 19:03
answered Mar 23 at 18:33
amsmathamsmath
3,287421
answered Mar 23 at 18:33
amsmathamsmath
3,287421
answered Mar 23 at 18:33
amsmathamsmath
3,287421
3,287421
add a comment |
add a comment |
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$begingroup$
A small comment on notation: you set this up to equate an $(n+1)$-digit decimal numeral with a $(2n+2)$-digit binary numeral. It might be simpler to compare an $n$-digit number with a $2n$-digit number.
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– David K
Mar 23 at 20:09
1
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Requiring a leading $0$ in binary & base $10$ expansions results in relatively few values to check (& I believe the answer is there are no more, but I haven't confirmed) as shown by the answers below. However, you did not explicitly state this is required. Thus, for example, if you allow the use of leading zeros, a valid value to check on could have a base $2$ expansion of $000101000101_2$. This more general case is, I believe, quite interesting. There'll be few values which fit your criteria, but I suspect there may still be infinitely many. If you're interested in this, please specify it.
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– John Omielan
Mar 24 at 1:26
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Yes, that is interesting, though I had not considered it initially
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– aman
Mar 24 at 3:00