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Multiples of $999$ have digit sum $geq 27$
Clash Royale CLAN TAG#URR8PPP
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How could we prove the following claim?
The sum of the digits of $kcdot 999$ is $ge 27$
I checked $k = 1$ up to $9$. And I found that if it's true of $d$ it's also true of $10cdot d$.
I also checked many values with a computer, it seems to always be the case. Further we can see that the digit sum must always be a multiple of 9.
I checked how to prove 'casting out nines', but I could not apply the same method here because it's just the digit sum not the digital root. and $27$ is bigger than our base $10$.
recreational-mathematics decimal-expansion
edited Aug 1 at 15:29
Asaf Karagila
291k31401731
asked Aug 1 at 14:29
rain1
835
add a comment |Â
up vote
16
down vote
favorite
How could we prove the following claim?
The sum of the digits of $kcdot 999$ is $ge 27$
I checked $k = 1$ up to $9$. And I found that if it's true of $d$ it's also true of $10cdot d$.
I also checked many values with a computer, it seems to always be the case. Further we can see that the digit sum must always be a multiple of 9.
I checked how to prove 'casting out nines', but I could not apply the same method here because it's just the digit sum not the digital root. and $27$ is bigger than our base $10$.
recreational-mathematics decimal-expansion
edited Aug 1 at 15:29
Asaf Karagila
291k31401731
asked Aug 1 at 14:29
rain1
835
this is how I prove that the digit sum of factorials diverge to infinity, it gives a lower bound that grows laughably slowly
– mercio
Aug 2 at 15:55
add a comment |Â
up vote
16
down vote
favorite
up vote
16
down vote
favorite
How could we prove the following claim?
The sum of the digits of $kcdot 999$ is $ge 27$
I checked $k = 1$ up to $9$. And I found that if it's true of $d$ it's also true of $10cdot d$.
I also checked many values with a computer, it seems to always be the case. Further we can see that the digit sum must always be a multiple of 9.
I checked how to prove 'casting out nines', but I could not apply the same method here because it's just the digit sum not the digital root. and $27$ is bigger than our base $10$.
recreational-mathematics decimal-expansion
edited Aug 1 at 15:29
Asaf Karagila
291k31401731
asked Aug 1 at 14:29
rain1
835
How could we prove the following claim?
The sum of the digits of $kcdot 999$ is $ge 27$
I checked $k = 1$ up to $9$. And I found that if it's true of $d$ it's also true of $10cdot d$.
I also checked many values with a computer, it seems to always be the case. Further we can see that the digit sum must always be a multiple of 9.
I checked how to prove 'casting out nines', but I could not apply the same method here because it's just the digit sum not the digital root. and $27$ is bigger than our base $10$.
recreational-mathematics decimal-expansion
edited Aug 1 at 15:29
Asaf Karagila
291k31401731
asked Aug 1 at 14:29
rain1
835
edited Aug 1 at 15:29
Asaf Karagila
291k31401731
edited Aug 1 at 15:29
Asaf Karagila
291k31401731
edited Aug 1 at 15:29
Asaf Karagila
291k31401731
291k31401731
asked Aug 1 at 14:29
rain1
835
asked Aug 1 at 14:29
rain1
835
asked Aug 1 at 14:29
rain1
835
835
this is how I prove that the digit sum of factorials diverge to infinity, it gives a lower bound that grows laughably slowly
– mercio
Aug 2 at 15:55
add a comment |Â
this is how I prove that the digit sum of factorials diverge to infinity, it gives a lower bound that grows laughably slowly
– mercio
Aug 2 at 15:55
this is how I prove that the digit sum of factorials diverge to infinity, it gives a lower bound that grows laughably slowly
– mercio
Aug 2 at 15:55
this is how I prove that the digit sum of factorials diverge to infinity, it gives a lower bound that grows laughably slowly
– mercio
Aug 2 at 15:55
add a comment |Â
5 Answers
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Lemma. Let $n$ be an integer $ge 1000$. Then there exists a positive integer $m$ such that $m<n$, $n-m$ is a multiple of $999$ and for the decimal digit sums, we have $q(m)le q(n)$.
Proof.
$n$ has a $k$-digit decimal expansion $n=overlinea_ka_k-1ldots a_1$ (with $kge 4$ and $a_kge1$), then $m:=n-999cdot 10^k-4$ is non-negative and has a decimal expansion $m=overlineb_kb_k-1ldots b_1$, where $b_j=a_j$ for all $j$ except
$$begincasesb_k=a_k-1,b_k-3=a_k-3+1&textif a_k-3<9\
b_k=a_k-1,b_k-3=0, b_k-2=a_k-2+1&textif a_k-2<a_k-3=9\
b_k=a_k-1,b_k-2=b_k-3=0, b_k-1=a_k-1+1&textif a_k-1<a_k-2=a_k-3=9\
b_k-1=b_k-2=b_k-3=0&textif a_k-1=a_k-2=a_k-3=9\
endcases $$
Then for the digit sum of $m$ we find accordingly
$$q(m)=begincasesq(n)\q(n)-9\q(n)-18\q(n)-27endcasesle q(n) $$
Hence if $m>0$, the claim follows. On the other hand, if $m=0$, it follows that $n=999cdot 10^k-4$, $q(n)=27$, and we can take $m=999$. $square$
Corollary. If $n$ is a positive multiple of $999$, then $q(n)ge 27$.
Proof. By the lemma, the set of positive multiples of $999$ with digit sum $<27$ has no smallest element. $square$
answered Aug 1 at 15:55
Hagen von Eitzen
265k20258475
1
$m$ is by construction always positive no? ($999cdot 10^k-4$ only has $k-1$ digits, not $k$).
– WimC
Aug 1 at 16:27
add a comment |Â
up vote
2
down vote
Let us consider some examples, all steps in the argumentation are then also applied on the examples:
3300652000033011
12345678987654321
(1) We start with a number written in base $10$, which is divisible by $999$. We break it in blocks of numbers of three digits, starting from the units digit, where we find the "first block". The last block may be incomplete", in this case we may add or not zeros in front of it. Because $1000$ is congruent to one modulo $999$, the sum of these blocks, considered as numbers between $0$ and $999$, is also divisible by $999$.
In our case, we separate the groups
3.300.652.000.033.011
12.345.678.987.654.321
obtain the blocks
003 and respectively 012
300 345
652 678
000 987
033 654
011 321
and the sum of the corresponding numbers is $999$, and respectively $2997$. It stays divisible by $999$. We want to show that the sum of the digits of the numbers in the blocks is at least $27$.
(2) We repeat this operation till we get a number of three digits. This number is of course $999$ in the first case. In the second one we group again 002 and 997, add, get $999$, and stop here.
(3) To finish the proof we note the fact that looking at the sum of the digits in the "blocks" before and after applying the step (1), the sum drops (by a multiple of $9$), it was before bigger than after. This has something to do with the algorithm we learn first in school. We put two numbers over each other. We add the unit digits. If the result is $le 9$, then the contribution of the digits to the sum of digits of the two numbers we start with is the same as the corresponding contribution in the result. Else we have a drop by $9$. This goes forward for the next digits...
Inductively we are done.
Note: There is "nothing special" about $999$, compared to $9$, $99$, ... , $underbrace99dots99_ntext digits$, the same works by building blocks of length $n$ (in the general case, the last one explicitly listed).
answered Aug 1 at 19:54
dan_fulea
4,0321211
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Just a partial answer
This is true for all 3 digit $k$.
Let $k=overlineabc$.
The $999k=overlineabc000-abc$.
When $cne0$:
For the difference:
Unit digit is $10-c$.
Tens digit is $9-b$.
Hundreds digit is $9-a$.
Thousands digit is $c-1$.
Ten thousands digit(?) is $b$.
Hundred thousands digit(?) is $a$.
Thus the sum of digit is exactly $27$.
A similar approach can prove for the case $c=0, b>0$ and $c=0,b=0$.
answered Aug 1 at 15:11
Szeto
3,8431421
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If you are looking for a divisibility test for $27$, take the sum of $3$-digit groups starting from the units digit and adding any needed initial zeroes to the leading group. The sum matches the original number modulo $999$, thus also congruent modulo $27$ since $27times 37=999$. For instance
$$1,485,069 implies 001+485+069=555=20×27+15$$
so this number fails divisibility by $27$. But since $37$ is also a factor of $999$ and $555=15times 37$, the above number passes divisibility by $37$.
edited Aug 1 at 15:47
Asaf Karagila
291k31401731
answered Aug 1 at 15:28
Oscar Lanzi
9,92111632
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$9|999$ so the sum of the digits of any multiple of $999$ is a multiple of $9$. So either the sum of the digits is $9$ or $18$ or $ ge 27$.
The sum of the digits of $999 = 27 ge 27$.
Let $k*999$ be the very lowest positive multiple in which the sums of the digits is $le 18$.
Bear with me:
Let $B = 999k = sum_i=0^n 10^ib_i$ are suppose there are two digits $b_j$ and $b_j + 3$ so that $b_j < 9$ and $b_j+3 > 0$.
Then $C = B - 10^i*999 = B -10^i*(1000 - 1) = sum_i= 0^n 10^i c_i$ where $c_j = b_j + 1$ and and $c_j+3 = b_j+3 - 1$ and $c_i = b_i; i ne j, j+3$.
So the sum of the digits of $C$ is the same of the digits of $B$ but that contradicts that $B$ is the lowest multiple of $999$ with digits adding to $18$ or less.
Now $b_n ne 0$ so that means $b_n-3 = 9$ and $18 > b_n + b_n-3 ge 10$ so none of the other digits may equal $9$. Which means if there is an non zero digit $b_j$ it must be that $j < 3$.
This also means the sum of the digits must be exactly $18$.
We don't have many possible choices for $B$. To begin with if $B$ is a multiple of $10$ then $frac B10$ is a smaller multiple of $999$ with the same sums of digits. So $b_0 ne 0$ with means either $b_3 = 0$ or $n =3$.
To spell out the options with have. $B = :$
$9009$ which isn't a multiple of $999$. or
$abc9$ where $a +b+c=9; a> 0$ (easily verified that none of the first nine multiples of $999$ are fo this form. They are all of the form $a99(9-a)$. Also $abc9 - 999 = (a-1)b(c+1)0$ and the sum is less not more.
$a0b9c$ where $a+b+c = 9; c>0; a > 0$. $a0b9c - 999= (a-1)9b9(c+1)$ so the sum of digits is 27. So $a0b9c = wv*999$ for some $wv$. We can verify know such numbers match those forms. (Probably.... It'll involve tedious case checking.)
Final option is $a009bc$ and we can probably verify no $wv*999$ or $wvz*999$ are of that form.
Theres probably a much slicker way to do this.
answered Aug 1 at 18:12
fleablood
60.1k22575
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Lemma. Let $n$ be an integer $ge 1000$. Then there exists a positive integer $m$ such that $m<n$, $n-m$ is a multiple of $999$ and for the decimal digit sums, we have $q(m)le q(n)$.
Proof.
$n$ has a $k$-digit decimal expansion $n=overlinea_ka_k-1ldots a_1$ (with $kge 4$ and $a_kge1$), then $m:=n-999cdot 10^k-4$ is non-negative and has a decimal expansion $m=overlineb_kb_k-1ldots b_1$, where $b_j=a_j$ for all $j$ except
$$begincasesb_k=a_k-1,b_k-3=a_k-3+1&textif a_k-3<9\
b_k=a_k-1,b_k-3=0, b_k-2=a_k-2+1&textif a_k-2<a_k-3=9\
b_k=a_k-1,b_k-2=b_k-3=0, b_k-1=a_k-1+1&textif a_k-1<a_k-2=a_k-3=9\
b_k-1=b_k-2=b_k-3=0&textif a_k-1=a_k-2=a_k-3=9\
endcases $$
Then for the digit sum of $m$ we find accordingly
$$q(m)=begincasesq(n)\q(n)-9\q(n)-18\q(n)-27endcasesle q(n) $$
Hence if $m>0$, the claim follows. On the other hand, if $m=0$, it follows that $n=999cdot 10^k-4$, $q(n)=27$, and we can take $m=999$. $square$
Corollary. If $n$ is a positive multiple of $999$, then $q(n)ge 27$.
Proof. By the lemma, the set of positive multiples of $999$ with digit sum $<27$ has no smallest element. $square$
answered Aug 1 at 15:55
Hagen von Eitzen
265k20258475
1
$m$ is by construction always positive no? ($999cdot 10^k-4$ only has $k-1$ digits, not $k$).
– WimC
Aug 1 at 16:27
add a comment |Â
up vote
3
down vote
accepted
Lemma. Let $n$ be an integer $ge 1000$. Then there exists a positive integer $m$ such that $m<n$, $n-m$ is a multiple of $999$ and for the decimal digit sums, we have $q(m)le q(n)$.
Proof.
$n$ has a $k$-digit decimal expansion $n=overlinea_ka_k-1ldots a_1$ (with $kge 4$ and $a_kge1$), then $m:=n-999cdot 10^k-4$ is non-negative and has a decimal expansion $m=overlineb_kb_k-1ldots b_1$, where $b_j=a_j$ for all $j$ except
$$begincasesb_k=a_k-1,b_k-3=a_k-3+1&textif a_k-3<9\
b_k=a_k-1,b_k-3=0, b_k-2=a_k-2+1&textif a_k-2<a_k-3=9\
b_k=a_k-1,b_k-2=b_k-3=0, b_k-1=a_k-1+1&textif a_k-1<a_k-2=a_k-3=9\
b_k-1=b_k-2=b_k-3=0&textif a_k-1=a_k-2=a_k-3=9\
endcases $$
Then for the digit sum of $m$ we find accordingly
$$q(m)=begincasesq(n)\q(n)-9\q(n)-18\q(n)-27endcasesle q(n) $$
Hence if $m>0$, the claim follows. On the other hand, if $m=0$, it follows that $n=999cdot 10^k-4$, $q(n)=27$, and we can take $m=999$. $square$
Corollary. If $n$ is a positive multiple of $999$, then $q(n)ge 27$.
Proof. By the lemma, the set of positive multiples of $999$ with digit sum $<27$ has no smallest element. $square$
answered Aug 1 at 15:55
Hagen von Eitzen
265k20258475
1
$m$ is by construction always positive no? ($999cdot 10^k-4$ only has $k-1$ digits, not $k$).
– WimC
Aug 1 at 16:27
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Lemma. Let $n$ be an integer $ge 1000$. Then there exists a positive integer $m$ such that $m<n$, $n-m$ is a multiple of $999$ and for the decimal digit sums, we have $q(m)le q(n)$.
Proof.
$n$ has a $k$-digit decimal expansion $n=overlinea_ka_k-1ldots a_1$ (with $kge 4$ and $a_kge1$), then $m:=n-999cdot 10^k-4$ is non-negative and has a decimal expansion $m=overlineb_kb_k-1ldots b_1$, where $b_j=a_j$ for all $j$ except
$$begincasesb_k=a_k-1,b_k-3=a_k-3+1&textif a_k-3<9\
b_k=a_k-1,b_k-3=0, b_k-2=a_k-2+1&textif a_k-2<a_k-3=9\
b_k=a_k-1,b_k-2=b_k-3=0, b_k-1=a_k-1+1&textif a_k-1<a_k-2=a_k-3=9\
b_k-1=b_k-2=b_k-3=0&textif a_k-1=a_k-2=a_k-3=9\
endcases $$
Then for the digit sum of $m$ we find accordingly
$$q(m)=begincasesq(n)\q(n)-9\q(n)-18\q(n)-27endcasesle q(n) $$
Hence if $m>0$, the claim follows. On the other hand, if $m=0$, it follows that $n=999cdot 10^k-4$, $q(n)=27$, and we can take $m=999$. $square$
Corollary. If $n$ is a positive multiple of $999$, then $q(n)ge 27$.
Proof. By the lemma, the set of positive multiples of $999$ with digit sum $<27$ has no smallest element. $square$
answered Aug 1 at 15:55
Hagen von Eitzen
265k20258475
Lemma. Let $n$ be an integer $ge 1000$. Then there exists a positive integer $m$ such that $m<n$, $n-m$ is a multiple of $999$ and for the decimal digit sums, we have $q(m)le q(n)$.
Proof.
$n$ has a $k$-digit decimal expansion $n=overlinea_ka_k-1ldots a_1$ (with $kge 4$ and $a_kge1$), then $m:=n-999cdot 10^k-4$ is non-negative and has a decimal expansion $m=overlineb_kb_k-1ldots b_1$, where $b_j=a_j$ for all $j$ except
$$begincasesb_k=a_k-1,b_k-3=a_k-3+1&textif a_k-3<9\
b_k=a_k-1,b_k-3=0, b_k-2=a_k-2+1&textif a_k-2<a_k-3=9\
b_k=a_k-1,b_k-2=b_k-3=0, b_k-1=a_k-1+1&textif a_k-1<a_k-2=a_k-3=9\
b_k-1=b_k-2=b_k-3=0&textif a_k-1=a_k-2=a_k-3=9\
endcases $$
Then for the digit sum of $m$ we find accordingly
$$q(m)=begincasesq(n)\q(n)-9\q(n)-18\q(n)-27endcasesle q(n) $$
Hence if $m>0$, the claim follows. On the other hand, if $m=0$, it follows that $n=999cdot 10^k-4$, $q(n)=27$, and we can take $m=999$. $square$
Corollary. If $n$ is a positive multiple of $999$, then $q(n)ge 27$.
Proof. By the lemma, the set of positive multiples of $999$ with digit sum $<27$ has no smallest element. $square$
answered Aug 1 at 15:55
Hagen von Eitzen
265k20258475
answered Aug 1 at 15:55
Hagen von Eitzen
265k20258475
answered Aug 1 at 15:55
Hagen von Eitzen
265k20258475
answered Aug 1 at 15:55
Hagen von Eitzen
265k20258475
265k20258475
1
$m$ is by construction always positive no? ($999cdot 10^k-4$ only has $k-1$ digits, not $k$).
– WimC
Aug 1 at 16:27
add a comment |Â
1
$m$ is by construction always positive no? ($999cdot 10^k-4$ only has $k-1$ digits, not $k$).
– WimC
Aug 1 at 16:27
1
1
$m$ is by construction always positive no? ($999cdot 10^k-4$ only has $k-1$ digits, not $k$).
– WimC
Aug 1 at 16:27
$m$ is by construction always positive no? ($999cdot 10^k-4$ only has $k-1$ digits, not $k$).
– WimC
Aug 1 at 16:27
add a comment |Â
up vote
2
down vote
Let us consider some examples, all steps in the argumentation are then also applied on the examples:
3300652000033011
12345678987654321
(1) We start with a number written in base $10$, which is divisible by $999$. We break it in blocks of numbers of three digits, starting from the units digit, where we find the "first block". The last block may be incomplete", in this case we may add or not zeros in front of it. Because $1000$ is congruent to one modulo $999$, the sum of these blocks, considered as numbers between $0$ and $999$, is also divisible by $999$.
In our case, we separate the groups
3.300.652.000.033.011
12.345.678.987.654.321
obtain the blocks
003 and respectively 012
300 345
652 678
000 987
033 654
011 321
and the sum of the corresponding numbers is $999$, and respectively $2997$. It stays divisible by $999$. We want to show that the sum of the digits of the numbers in the blocks is at least $27$.
(2) We repeat this operation till we get a number of three digits. This number is of course $999$ in the first case. In the second one we group again 002 and 997, add, get $999$, and stop here.
(3) To finish the proof we note the fact that looking at the sum of the digits in the "blocks" before and after applying the step (1), the sum drops (by a multiple of $9$), it was before bigger than after. This has something to do with the algorithm we learn first in school. We put two numbers over each other. We add the unit digits. If the result is $le 9$, then the contribution of the digits to the sum of digits of the two numbers we start with is the same as the corresponding contribution in the result. Else we have a drop by $9$. This goes forward for the next digits...
Inductively we are done.
Note: There is "nothing special" about $999$, compared to $9$, $99$, ... , $underbrace99dots99_ntext digits$, the same works by building blocks of length $n$ (in the general case, the last one explicitly listed).
answered Aug 1 at 19:54
dan_fulea
4,0321211
add a comment |Â
up vote
2
down vote
Let us consider some examples, all steps in the argumentation are then also applied on the examples:
3300652000033011
12345678987654321
(1) We start with a number written in base $10$, which is divisible by $999$. We break it in blocks of numbers of three digits, starting from the units digit, where we find the "first block". The last block may be incomplete", in this case we may add or not zeros in front of it. Because $1000$ is congruent to one modulo $999$, the sum of these blocks, considered as numbers between $0$ and $999$, is also divisible by $999$.
In our case, we separate the groups
3.300.652.000.033.011
12.345.678.987.654.321
obtain the blocks
003 and respectively 012
300 345
652 678
000 987
033 654
011 321
and the sum of the corresponding numbers is $999$, and respectively $2997$. It stays divisible by $999$. We want to show that the sum of the digits of the numbers in the blocks is at least $27$.
(2) We repeat this operation till we get a number of three digits. This number is of course $999$ in the first case. In the second one we group again 002 and 997, add, get $999$, and stop here.
(3) To finish the proof we note the fact that looking at the sum of the digits in the "blocks" before and after applying the step (1), the sum drops (by a multiple of $9$), it was before bigger than after. This has something to do with the algorithm we learn first in school. We put two numbers over each other. We add the unit digits. If the result is $le 9$, then the contribution of the digits to the sum of digits of the two numbers we start with is the same as the corresponding contribution in the result. Else we have a drop by $9$. This goes forward for the next digits...
Inductively we are done.
Note: There is "nothing special" about $999$, compared to $9$, $99$, ... , $underbrace99dots99_ntext digits$, the same works by building blocks of length $n$ (in the general case, the last one explicitly listed).
answered Aug 1 at 19:54
dan_fulea
4,0321211
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let us consider some examples, all steps in the argumentation are then also applied on the examples:
3300652000033011
12345678987654321
(1) We start with a number written in base $10$, which is divisible by $999$. We break it in blocks of numbers of three digits, starting from the units digit, where we find the "first block". The last block may be incomplete", in this case we may add or not zeros in front of it. Because $1000$ is congruent to one modulo $999$, the sum of these blocks, considered as numbers between $0$ and $999$, is also divisible by $999$.
In our case, we separate the groups
3.300.652.000.033.011
12.345.678.987.654.321
obtain the blocks
003 and respectively 012
300 345
652 678
000 987
033 654
011 321
and the sum of the corresponding numbers is $999$, and respectively $2997$. It stays divisible by $999$. We want to show that the sum of the digits of the numbers in the blocks is at least $27$.
(2) We repeat this operation till we get a number of three digits. This number is of course $999$ in the first case. In the second one we group again 002 and 997, add, get $999$, and stop here.
(3) To finish the proof we note the fact that looking at the sum of the digits in the "blocks" before and after applying the step (1), the sum drops (by a multiple of $9$), it was before bigger than after. This has something to do with the algorithm we learn first in school. We put two numbers over each other. We add the unit digits. If the result is $le 9$, then the contribution of the digits to the sum of digits of the two numbers we start with is the same as the corresponding contribution in the result. Else we have a drop by $9$. This goes forward for the next digits...
Inductively we are done.
Note: There is "nothing special" about $999$, compared to $9$, $99$, ... , $underbrace99dots99_ntext digits$, the same works by building blocks of length $n$ (in the general case, the last one explicitly listed).
answered Aug 1 at 19:54
dan_fulea
4,0321211
Let us consider some examples, all steps in the argumentation are then also applied on the examples:
3300652000033011
12345678987654321
(1) We start with a number written in base $10$, which is divisible by $999$. We break it in blocks of numbers of three digits, starting from the units digit, where we find the "first block". The last block may be incomplete", in this case we may add or not zeros in front of it. Because $1000$ is congruent to one modulo $999$, the sum of these blocks, considered as numbers between $0$ and $999$, is also divisible by $999$.
In our case, we separate the groups
3.300.652.000.033.011
12.345.678.987.654.321
obtain the blocks
003 and respectively 012
300 345
652 678
000 987
033 654
011 321
and the sum of the corresponding numbers is $999$, and respectively $2997$. It stays divisible by $999$. We want to show that the sum of the digits of the numbers in the blocks is at least $27$.
(2) We repeat this operation till we get a number of three digits. This number is of course $999$ in the first case. In the second one we group again 002 and 997, add, get $999$, and stop here.
(3) To finish the proof we note the fact that looking at the sum of the digits in the "blocks" before and after applying the step (1), the sum drops (by a multiple of $9$), it was before bigger than after. This has something to do with the algorithm we learn first in school. We put two numbers over each other. We add the unit digits. If the result is $le 9$, then the contribution of the digits to the sum of digits of the two numbers we start with is the same as the corresponding contribution in the result. Else we have a drop by $9$. This goes forward for the next digits...
Inductively we are done.
Note: There is "nothing special" about $999$, compared to $9$, $99$, ... , $underbrace99dots99_ntext digits$, the same works by building blocks of length $n$ (in the general case, the last one explicitly listed).
answered Aug 1 at 19:54
dan_fulea
4,0321211
answered Aug 1 at 19:54
dan_fulea
4,0321211
answered Aug 1 at 19:54
dan_fulea
4,0321211
answered Aug 1 at 19:54
dan_fulea
4,0321211
4,0321211
add a comment |Â
add a comment |Â
up vote
1
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Just a partial answer
This is true for all 3 digit $k$.
Let $k=overlineabc$.
The $999k=overlineabc000-abc$.
When $cne0$:
For the difference:
Unit digit is $10-c$.
Tens digit is $9-b$.
Hundreds digit is $9-a$.
Thousands digit is $c-1$.
Ten thousands digit(?) is $b$.
Hundred thousands digit(?) is $a$.
Thus the sum of digit is exactly $27$.
A similar approach can prove for the case $c=0, b>0$ and $c=0,b=0$.
answered Aug 1 at 15:11
Szeto
3,8431421
add a comment |Â
up vote
1
down vote
Just a partial answer
This is true for all 3 digit $k$.
Let $k=overlineabc$.
The $999k=overlineabc000-abc$.
When $cne0$:
For the difference:
Unit digit is $10-c$.
Tens digit is $9-b$.
Hundreds digit is $9-a$.
Thousands digit is $c-1$.
Ten thousands digit(?) is $b$.
Hundred thousands digit(?) is $a$.
Thus the sum of digit is exactly $27$.
A similar approach can prove for the case $c=0, b>0$ and $c=0,b=0$.
answered Aug 1 at 15:11
Szeto
3,8431421
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just a partial answer
This is true for all 3 digit $k$.
Let $k=overlineabc$.
The $999k=overlineabc000-abc$.
When $cne0$:
For the difference:
Unit digit is $10-c$.
Tens digit is $9-b$.
Hundreds digit is $9-a$.
Thousands digit is $c-1$.
Ten thousands digit(?) is $b$.
Hundred thousands digit(?) is $a$.
Thus the sum of digit is exactly $27$.
A similar approach can prove for the case $c=0, b>0$ and $c=0,b=0$.
answered Aug 1 at 15:11
Szeto
3,8431421
Just a partial answer
This is true for all 3 digit $k$.
Let $k=overlineabc$.
The $999k=overlineabc000-abc$.
When $cne0$:
For the difference:
Unit digit is $10-c$.
Tens digit is $9-b$.
Hundreds digit is $9-a$.
Thousands digit is $c-1$.
Ten thousands digit(?) is $b$.
Hundred thousands digit(?) is $a$.
Thus the sum of digit is exactly $27$.
A similar approach can prove for the case $c=0, b>0$ and $c=0,b=0$.
answered Aug 1 at 15:11
Szeto
3,8431421
answered Aug 1 at 15:11
Szeto
3,8431421
answered Aug 1 at 15:11
Szeto
3,8431421
answered Aug 1 at 15:11
Szeto
3,8431421
3,8431421
add a comment |Â
add a comment |Â
up vote
1
down vote
If you are looking for a divisibility test for $27$, take the sum of $3$-digit groups starting from the units digit and adding any needed initial zeroes to the leading group. The sum matches the original number modulo $999$, thus also congruent modulo $27$ since $27times 37=999$. For instance
$$1,485,069 implies 001+485+069=555=20×27+15$$
so this number fails divisibility by $27$. But since $37$ is also a factor of $999$ and $555=15times 37$, the above number passes divisibility by $37$.
edited Aug 1 at 15:47
Asaf Karagila
291k31401731
answered Aug 1 at 15:28
Oscar Lanzi
9,92111632
add a comment |Â
up vote
1
down vote
If you are looking for a divisibility test for $27$, take the sum of $3$-digit groups starting from the units digit and adding any needed initial zeroes to the leading group. The sum matches the original number modulo $999$, thus also congruent modulo $27$ since $27times 37=999$. For instance
$$1,485,069 implies 001+485+069=555=20×27+15$$
so this number fails divisibility by $27$. But since $37$ is also a factor of $999$ and $555=15times 37$, the above number passes divisibility by $37$.
edited Aug 1 at 15:47
Asaf Karagila
291k31401731
answered Aug 1 at 15:28
Oscar Lanzi
9,92111632
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you are looking for a divisibility test for $27$, take the sum of $3$-digit groups starting from the units digit and adding any needed initial zeroes to the leading group. The sum matches the original number modulo $999$, thus also congruent modulo $27$ since $27times 37=999$. For instance
$$1,485,069 implies 001+485+069=555=20×27+15$$
so this number fails divisibility by $27$. But since $37$ is also a factor of $999$ and $555=15times 37$, the above number passes divisibility by $37$.
edited Aug 1 at 15:47
Asaf Karagila
291k31401731
answered Aug 1 at 15:28
Oscar Lanzi
9,92111632
If you are looking for a divisibility test for $27$, take the sum of $3$-digit groups starting from the units digit and adding any needed initial zeroes to the leading group. The sum matches the original number modulo $999$, thus also congruent modulo $27$ since $27times 37=999$. For instance
$$1,485,069 implies 001+485+069=555=20×27+15$$
so this number fails divisibility by $27$. But since $37$ is also a factor of $999$ and $555=15times 37$, the above number passes divisibility by $37$.
edited Aug 1 at 15:47
Asaf Karagila
291k31401731
answered Aug 1 at 15:28
Oscar Lanzi
9,92111632
edited Aug 1 at 15:47
Asaf Karagila
291k31401731
edited Aug 1 at 15:47
Asaf Karagila
291k31401731
edited Aug 1 at 15:47
Asaf Karagila
291k31401731
291k31401731
answered Aug 1 at 15:28
Oscar Lanzi
9,92111632
answered Aug 1 at 15:28
Oscar Lanzi
9,92111632
answered Aug 1 at 15:28
Oscar Lanzi
9,92111632
9,92111632
add a comment |Â
add a comment |Â
up vote
1
down vote
$9|999$ so the sum of the digits of any multiple of $999$ is a multiple of $9$. So either the sum of the digits is $9$ or $18$ or $ ge 27$.
The sum of the digits of $999 = 27 ge 27$.
Let $k*999$ be the very lowest positive multiple in which the sums of the digits is $le 18$.
Bear with me:
Let $B = 999k = sum_i=0^n 10^ib_i$ are suppose there are two digits $b_j$ and $b_j + 3$ so that $b_j < 9$ and $b_j+3 > 0$.
Then $C = B - 10^i*999 = B -10^i*(1000 - 1) = sum_i= 0^n 10^i c_i$ where $c_j = b_j + 1$ and and $c_j+3 = b_j+3 - 1$ and $c_i = b_i; i ne j, j+3$.
So the sum of the digits of $C$ is the same of the digits of $B$ but that contradicts that $B$ is the lowest multiple of $999$ with digits adding to $18$ or less.
Now $b_n ne 0$ so that means $b_n-3 = 9$ and $18 > b_n + b_n-3 ge 10$ so none of the other digits may equal $9$. Which means if there is an non zero digit $b_j$ it must be that $j < 3$.
This also means the sum of the digits must be exactly $18$.
We don't have many possible choices for $B$. To begin with if $B$ is a multiple of $10$ then $frac B10$ is a smaller multiple of $999$ with the same sums of digits. So $b_0 ne 0$ with means either $b_3 = 0$ or $n =3$.
To spell out the options with have. $B = :$
$9009$ which isn't a multiple of $999$. or
$abc9$ where $a +b+c=9; a> 0$ (easily verified that none of the first nine multiples of $999$ are fo this form. They are all of the form $a99(9-a)$. Also $abc9 - 999 = (a-1)b(c+1)0$ and the sum is less not more.
$a0b9c$ where $a+b+c = 9; c>0; a > 0$. $a0b9c - 999= (a-1)9b9(c+1)$ so the sum of digits is 27. So $a0b9c = wv*999$ for some $wv$. We can verify know such numbers match those forms. (Probably.... It'll involve tedious case checking.)
Final option is $a009bc$ and we can probably verify no $wv*999$ or $wvz*999$ are of that form.
Theres probably a much slicker way to do this.
answered Aug 1 at 18:12
fleablood
60.1k22575
add a comment |Â
up vote
1
down vote
$9|999$ so the sum of the digits of any multiple of $999$ is a multiple of $9$. So either the sum of the digits is $9$ or $18$ or $ ge 27$.
The sum of the digits of $999 = 27 ge 27$.
Let $k*999$ be the very lowest positive multiple in which the sums of the digits is $le 18$.
Bear with me:
Let $B = 999k = sum_i=0^n 10^ib_i$ are suppose there are two digits $b_j$ and $b_j + 3$ so that $b_j < 9$ and $b_j+3 > 0$.
Then $C = B - 10^i*999 = B -10^i*(1000 - 1) = sum_i= 0^n 10^i c_i$ where $c_j = b_j + 1$ and and $c_j+3 = b_j+3 - 1$ and $c_i = b_i; i ne j, j+3$.
So the sum of the digits of $C$ is the same of the digits of $B$ but that contradicts that $B$ is the lowest multiple of $999$ with digits adding to $18$ or less.
Now $b_n ne 0$ so that means $b_n-3 = 9$ and $18 > b_n + b_n-3 ge 10$ so none of the other digits may equal $9$. Which means if there is an non zero digit $b_j$ it must be that $j < 3$.
This also means the sum of the digits must be exactly $18$.
We don't have many possible choices for $B$. To begin with if $B$ is a multiple of $10$ then $frac B10$ is a smaller multiple of $999$ with the same sums of digits. So $b_0 ne 0$ with means either $b_3 = 0$ or $n =3$.
To spell out the options with have. $B = :$
$9009$ which isn't a multiple of $999$. or
$abc9$ where $a +b+c=9; a> 0$ (easily verified that none of the first nine multiples of $999$ are fo this form. They are all of the form $a99(9-a)$. Also $abc9 - 999 = (a-1)b(c+1)0$ and the sum is less not more.
$a0b9c$ where $a+b+c = 9; c>0; a > 0$. $a0b9c - 999= (a-1)9b9(c+1)$ so the sum of digits is 27. So $a0b9c = wv*999$ for some $wv$. We can verify know such numbers match those forms. (Probably.... It'll involve tedious case checking.)
Final option is $a009bc$ and we can probably verify no $wv*999$ or $wvz*999$ are of that form.
Theres probably a much slicker way to do this.
answered Aug 1 at 18:12
fleablood
60.1k22575
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$9|999$ so the sum of the digits of any multiple of $999$ is a multiple of $9$. So either the sum of the digits is $9$ or $18$ or $ ge 27$.
The sum of the digits of $999 = 27 ge 27$.
Let $k*999$ be the very lowest positive multiple in which the sums of the digits is $le 18$.
Bear with me:
Let $B = 999k = sum_i=0^n 10^ib_i$ are suppose there are two digits $b_j$ and $b_j + 3$ so that $b_j < 9$ and $b_j+3 > 0$.
Then $C = B - 10^i*999 = B -10^i*(1000 - 1) = sum_i= 0^n 10^i c_i$ where $c_j = b_j + 1$ and and $c_j+3 = b_j+3 - 1$ and $c_i = b_i; i ne j, j+3$.
So the sum of the digits of $C$ is the same of the digits of $B$ but that contradicts that $B$ is the lowest multiple of $999$ with digits adding to $18$ or less.
Now $b_n ne 0$ so that means $b_n-3 = 9$ and $18 > b_n + b_n-3 ge 10$ so none of the other digits may equal $9$. Which means if there is an non zero digit $b_j$ it must be that $j < 3$.
This also means the sum of the digits must be exactly $18$.
We don't have many possible choices for $B$. To begin with if $B$ is a multiple of $10$ then $frac B10$ is a smaller multiple of $999$ with the same sums of digits. So $b_0 ne 0$ with means either $b_3 = 0$ or $n =3$.
To spell out the options with have. $B = :$
$9009$ which isn't a multiple of $999$. or
$abc9$ where $a +b+c=9; a> 0$ (easily verified that none of the first nine multiples of $999$ are fo this form. They are all of the form $a99(9-a)$. Also $abc9 - 999 = (a-1)b(c+1)0$ and the sum is less not more.
$a0b9c$ where $a+b+c = 9; c>0; a > 0$. $a0b9c - 999= (a-1)9b9(c+1)$ so the sum of digits is 27. So $a0b9c = wv*999$ for some $wv$. We can verify know such numbers match those forms. (Probably.... It'll involve tedious case checking.)
Final option is $a009bc$ and we can probably verify no $wv*999$ or $wvz*999$ are of that form.
Theres probably a much slicker way to do this.
answered Aug 1 at 18:12
fleablood
60.1k22575
$9|999$ so the sum of the digits of any multiple of $999$ is a multiple of $9$. So either the sum of the digits is $9$ or $18$ or $ ge 27$.
The sum of the digits of $999 = 27 ge 27$.
Let $k*999$ be the very lowest positive multiple in which the sums of the digits is $le 18$.
Bear with me:
Let $B = 999k = sum_i=0^n 10^ib_i$ are suppose there are two digits $b_j$ and $b_j + 3$ so that $b_j < 9$ and $b_j+3 > 0$.
Then $C = B - 10^i*999 = B -10^i*(1000 - 1) = sum_i= 0^n 10^i c_i$ where $c_j = b_j + 1$ and and $c_j+3 = b_j+3 - 1$ and $c_i = b_i; i ne j, j+3$.
So the sum of the digits of $C$ is the same of the digits of $B$ but that contradicts that $B$ is the lowest multiple of $999$ with digits adding to $18$ or less.
Now $b_n ne 0$ so that means $b_n-3 = 9$ and $18 > b_n + b_n-3 ge 10$ so none of the other digits may equal $9$. Which means if there is an non zero digit $b_j$ it must be that $j < 3$.
This also means the sum of the digits must be exactly $18$.
We don't have many possible choices for $B$. To begin with if $B$ is a multiple of $10$ then $frac B10$ is a smaller multiple of $999$ with the same sums of digits. So $b_0 ne 0$ with means either $b_3 = 0$ or $n =3$.
To spell out the options with have. $B = :$
$9009$ which isn't a multiple of $999$. or
$abc9$ where $a +b+c=9; a> 0$ (easily verified that none of the first nine multiples of $999$ are fo this form. They are all of the form $a99(9-a)$. Also $abc9 - 999 = (a-1)b(c+1)0$ and the sum is less not more.
$a0b9c$ where $a+b+c = 9; c>0; a > 0$. $a0b9c - 999= (a-1)9b9(c+1)$ so the sum of digits is 27. So $a0b9c = wv*999$ for some $wv$. We can verify know such numbers match those forms. (Probably.... It'll involve tedious case checking.)
Final option is $a009bc$ and we can probably verify no $wv*999$ or $wvz*999$ are of that form.
Theres probably a much slicker way to do this.
answered Aug 1 at 18:12
fleablood
60.1k22575
answered Aug 1 at 18:12
fleablood
60.1k22575
answered Aug 1 at 18:12
fleablood
60.1k22575
answered Aug 1 at 18:12
fleablood
60.1k22575
60.1k22575
add a comment |Â
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this is how I prove that the digit sum of factorials diverge to infinity, it gives a lower bound that grows laughably slowly
– mercio
Aug 2 at 15:55