Interesting Pattern - Adding or Multiplying Sequential Integers and Then Reducing












2












$begingroup$


I noticed that if you sequentially add integers together, or even multiply them, and then reduce the result, an interesting pattern emerges.



(+1) 1  >> 1
(+2) 3 >> 3
(+3) 6 >> 6
(+4) 10 >> 1
(+5) 15 >> 6
(+6) 21 >> 3
(+7) 28 >> 1
(+8) 36 >> 9
(+9) 45 >> 9


This pattern, 1, 3, 6, 1, 6, 3, 1, 9, 9, repeats itself without end. Obviously this is due at least in part to the fact that we use a base 10 number system.




  1. Are there any other interesting patterns that I'm missing?


  2. Is there a logical reason for this?



If you double the numbers, 1, 2, 4, 8, 16, 32, 64, etc, and then reduce them, you get a different pattern, 1, 2, 4, 8, 7, 5 that repeats itself endlessly.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you mean by "reduce the result?"
    $endgroup$
    – Eevee Trainer
    Jan 2 at 4:34






  • 1




    $begingroup$
    @EeveeTrainer i mean like 10 = 1 + 0 = 1 or 561 = 5 + 6 + 1 = 12 = 1 + 2 = 3
    $endgroup$
    – Anthony
    Jan 2 at 4:36










  • $begingroup$
    Ah, so taking the sum of the numbers' digits repeatedly until you get a one-digit number, I see. Might be appropriate to explain that more explicitly in the OP (unless it's actually a commonly-used term, still a novice with this but I've never heard of that).
    $endgroup$
    – Eevee Trainer
    Jan 2 at 4:38






  • 3




    $begingroup$
    The repetition, and basic patterns, have to do with the "reduced" value having the same remainder as the original value when it's divided by $9$, and the sum of the integers from $1$ to $n$, inclusive, being $frac{left(nright)left(n + 1right)}{2}$.
    $endgroup$
    – John Omielan
    Jan 2 at 4:39












  • $begingroup$
    @EeveeTrainer i believe that is the term. it's the only term i've heard when people describe this. there could be more terms, but i've never heard of them
    $endgroup$
    – Anthony
    Jan 2 at 4:41
















2












$begingroup$


I noticed that if you sequentially add integers together, or even multiply them, and then reduce the result, an interesting pattern emerges.



(+1) 1  >> 1
(+2) 3 >> 3
(+3) 6 >> 6
(+4) 10 >> 1
(+5) 15 >> 6
(+6) 21 >> 3
(+7) 28 >> 1
(+8) 36 >> 9
(+9) 45 >> 9


This pattern, 1, 3, 6, 1, 6, 3, 1, 9, 9, repeats itself without end. Obviously this is due at least in part to the fact that we use a base 10 number system.




  1. Are there any other interesting patterns that I'm missing?


  2. Is there a logical reason for this?



If you double the numbers, 1, 2, 4, 8, 16, 32, 64, etc, and then reduce them, you get a different pattern, 1, 2, 4, 8, 7, 5 that repeats itself endlessly.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you mean by "reduce the result?"
    $endgroup$
    – Eevee Trainer
    Jan 2 at 4:34






  • 1




    $begingroup$
    @EeveeTrainer i mean like 10 = 1 + 0 = 1 or 561 = 5 + 6 + 1 = 12 = 1 + 2 = 3
    $endgroup$
    – Anthony
    Jan 2 at 4:36










  • $begingroup$
    Ah, so taking the sum of the numbers' digits repeatedly until you get a one-digit number, I see. Might be appropriate to explain that more explicitly in the OP (unless it's actually a commonly-used term, still a novice with this but I've never heard of that).
    $endgroup$
    – Eevee Trainer
    Jan 2 at 4:38






  • 3




    $begingroup$
    The repetition, and basic patterns, have to do with the "reduced" value having the same remainder as the original value when it's divided by $9$, and the sum of the integers from $1$ to $n$, inclusive, being $frac{left(nright)left(n + 1right)}{2}$.
    $endgroup$
    – John Omielan
    Jan 2 at 4:39












  • $begingroup$
    @EeveeTrainer i believe that is the term. it's the only term i've heard when people describe this. there could be more terms, but i've never heard of them
    $endgroup$
    – Anthony
    Jan 2 at 4:41














2












2








2





$begingroup$


I noticed that if you sequentially add integers together, or even multiply them, and then reduce the result, an interesting pattern emerges.



(+1) 1  >> 1
(+2) 3 >> 3
(+3) 6 >> 6
(+4) 10 >> 1
(+5) 15 >> 6
(+6) 21 >> 3
(+7) 28 >> 1
(+8) 36 >> 9
(+9) 45 >> 9


This pattern, 1, 3, 6, 1, 6, 3, 1, 9, 9, repeats itself without end. Obviously this is due at least in part to the fact that we use a base 10 number system.




  1. Are there any other interesting patterns that I'm missing?


  2. Is there a logical reason for this?



If you double the numbers, 1, 2, 4, 8, 16, 32, 64, etc, and then reduce them, you get a different pattern, 1, 2, 4, 8, 7, 5 that repeats itself endlessly.










share|cite|improve this question











$endgroup$




I noticed that if you sequentially add integers together, or even multiply them, and then reduce the result, an interesting pattern emerges.



(+1) 1  >> 1
(+2) 3 >> 3
(+3) 6 >> 6
(+4) 10 >> 1
(+5) 15 >> 6
(+6) 21 >> 3
(+7) 28 >> 1
(+8) 36 >> 9
(+9) 45 >> 9


This pattern, 1, 3, 6, 1, 6, 3, 1, 9, 9, repeats itself without end. Obviously this is due at least in part to the fact that we use a base 10 number system.




  1. Are there any other interesting patterns that I'm missing?


  2. Is there a logical reason for this?



If you double the numbers, 1, 2, 4, 8, 16, 32, 64, etc, and then reduce them, you get a different pattern, 1, 2, 4, 8, 7, 5 that repeats itself endlessly.







sequences-and-series analysis arithmetic pattern-recognition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 5:06







Anthony

















asked Jan 2 at 4:30









AnthonyAnthony

11617




11617








  • 1




    $begingroup$
    What do you mean by "reduce the result?"
    $endgroup$
    – Eevee Trainer
    Jan 2 at 4:34






  • 1




    $begingroup$
    @EeveeTrainer i mean like 10 = 1 + 0 = 1 or 561 = 5 + 6 + 1 = 12 = 1 + 2 = 3
    $endgroup$
    – Anthony
    Jan 2 at 4:36










  • $begingroup$
    Ah, so taking the sum of the numbers' digits repeatedly until you get a one-digit number, I see. Might be appropriate to explain that more explicitly in the OP (unless it's actually a commonly-used term, still a novice with this but I've never heard of that).
    $endgroup$
    – Eevee Trainer
    Jan 2 at 4:38






  • 3




    $begingroup$
    The repetition, and basic patterns, have to do with the "reduced" value having the same remainder as the original value when it's divided by $9$, and the sum of the integers from $1$ to $n$, inclusive, being $frac{left(nright)left(n + 1right)}{2}$.
    $endgroup$
    – John Omielan
    Jan 2 at 4:39












  • $begingroup$
    @EeveeTrainer i believe that is the term. it's the only term i've heard when people describe this. there could be more terms, but i've never heard of them
    $endgroup$
    – Anthony
    Jan 2 at 4:41














  • 1




    $begingroup$
    What do you mean by "reduce the result?"
    $endgroup$
    – Eevee Trainer
    Jan 2 at 4:34






  • 1




    $begingroup$
    @EeveeTrainer i mean like 10 = 1 + 0 = 1 or 561 = 5 + 6 + 1 = 12 = 1 + 2 = 3
    $endgroup$
    – Anthony
    Jan 2 at 4:36










  • $begingroup$
    Ah, so taking the sum of the numbers' digits repeatedly until you get a one-digit number, I see. Might be appropriate to explain that more explicitly in the OP (unless it's actually a commonly-used term, still a novice with this but I've never heard of that).
    $endgroup$
    – Eevee Trainer
    Jan 2 at 4:38






  • 3




    $begingroup$
    The repetition, and basic patterns, have to do with the "reduced" value having the same remainder as the original value when it's divided by $9$, and the sum of the integers from $1$ to $n$, inclusive, being $frac{left(nright)left(n + 1right)}{2}$.
    $endgroup$
    – John Omielan
    Jan 2 at 4:39












  • $begingroup$
    @EeveeTrainer i believe that is the term. it's the only term i've heard when people describe this. there could be more terms, but i've never heard of them
    $endgroup$
    – Anthony
    Jan 2 at 4:41








1




1




$begingroup$
What do you mean by "reduce the result?"
$endgroup$
– Eevee Trainer
Jan 2 at 4:34




$begingroup$
What do you mean by "reduce the result?"
$endgroup$
– Eevee Trainer
Jan 2 at 4:34




1




1




$begingroup$
@EeveeTrainer i mean like 10 = 1 + 0 = 1 or 561 = 5 + 6 + 1 = 12 = 1 + 2 = 3
$endgroup$
– Anthony
Jan 2 at 4:36




$begingroup$
@EeveeTrainer i mean like 10 = 1 + 0 = 1 or 561 = 5 + 6 + 1 = 12 = 1 + 2 = 3
$endgroup$
– Anthony
Jan 2 at 4:36












$begingroup$
Ah, so taking the sum of the numbers' digits repeatedly until you get a one-digit number, I see. Might be appropriate to explain that more explicitly in the OP (unless it's actually a commonly-used term, still a novice with this but I've never heard of that).
$endgroup$
– Eevee Trainer
Jan 2 at 4:38




$begingroup$
Ah, so taking the sum of the numbers' digits repeatedly until you get a one-digit number, I see. Might be appropriate to explain that more explicitly in the OP (unless it's actually a commonly-used term, still a novice with this but I've never heard of that).
$endgroup$
– Eevee Trainer
Jan 2 at 4:38




3




3




$begingroup$
The repetition, and basic patterns, have to do with the "reduced" value having the same remainder as the original value when it's divided by $9$, and the sum of the integers from $1$ to $n$, inclusive, being $frac{left(nright)left(n + 1right)}{2}$.
$endgroup$
– John Omielan
Jan 2 at 4:39






$begingroup$
The repetition, and basic patterns, have to do with the "reduced" value having the same remainder as the original value when it's divided by $9$, and the sum of the integers from $1$ to $n$, inclusive, being $frac{left(nright)left(n + 1right)}{2}$.
$endgroup$
– John Omielan
Jan 2 at 4:39














$begingroup$
@EeveeTrainer i believe that is the term. it's the only term i've heard when people describe this. there could be more terms, but i've never heard of them
$endgroup$
– Anthony
Jan 2 at 4:41




$begingroup$
@EeveeTrainer i believe that is the term. it's the only term i've heard when people describe this. there could be more terms, but i've never heard of them
$endgroup$
– Anthony
Jan 2 at 4:41










1 Answer
1






active

oldest

votes


















4












$begingroup$

The answer to all of these question lies in what actually happens when you "reduce" a number by repeatedly summing its digits. In particular, the following is true




Let $n$ be a positive integer. If $n$ is divisible by $9$, its reduction will be $9$. Otherwise, its reduction will be the remainder of $n$ when divided by $9$.




Here, for formal purposes, "remainder" means the number $0leq r<9$ such that you can write
$$n=9q+r$$
For instance, the remainder of $23$ divided by $9$ is $5$ since $9times 2 + 5 = 23$. Note that $2+3=5$ as well.



The basic thing to notice in order to prove this is that the difference between an integer $n$ and the sum of its digits $s$ is always a multiple of $9$. For instance, $23-(2+3)=18=9times 2$. You can prove this by looking at the expanded form of decimal representation; in particular if $n$ has decimal expansion $d_kd_{k-1}ldots d_1d_0$ for some sequence of digits $d_i$, we have
$$n=10^kd_k+10^{k-1}d_{k-1}+ldots+10^1cdot d_1 + 10^0cdot d_0.$$
Then, the sum of the digits is
$$s=d_k+d_{k-1}+ldots+d_1+d_0.$$
The difference between these two values is
$$n-s=(10^k-1)d_k+(10^{k-1}-1)d_{k-1}+ldots+(10^1-1)cdot d_1 + (10^0-1)cdot d_0.$$
However, note that $$10^k-1=underbrace{99ldots 99}_{ktext{ digits}}=9cdot underbrace{11ldots 11}_{ktext{ digits}}.$$
Thus, every term in the difference $n-s$ is $9$ times something, so the whole thing is $9$ times something - so $n-s$ is divisible by $9$.



You can repeat this logic as many times as necessary to see that if $r$ was the reduction of $n$, then $n-r$ is divisible by $9$ - but then you find that there are only actually $9$ positive single digits numbers - and $r$ is definitely one of them - so it's only a matter of checking that there is only one positive single digit number $r$ so that $n-r$ is a multiple of $9$.





Okay, so why does this help us? Well, it reveals another surprising fact:




Let $n$ and $n'$ be positive integers with reductions $r$ and $r'$ respectively. The reduction of $n+n'$ and $ncdot n'$ equal the reductions of $r+r'$ or $rcdot r'$ respectively.




If you want to prove this, you basically write $n=9q+r$ and $n'=9q+r'$ and then note that $n+n'=9(q+q')+(r+r')$ then $nn'=9(9qq'+qr'+q'r)+rr'$ and use the last theorem, which gets rid of all those multiples of $9$ being added, since they can't change the result.



The significance is that you get a multiplication table and addition table where, given the two reductions, you can look up the reduction of the sum and product without knowing the original two numbers. Those tables look like this:



Addition Table



Multiplication Table



This framework basically sets up the answer to your questions: the reductions of the doubling sequence is easiest to explain. To find that sequence, you start at $1$, then, using the table above, multiply that by $2$. You repeat that for a while - and eventually you get back to $1$. The sequence is bound to repeat from there, since you've entered a loop. Note that this method completely forgets about what the number your reducing was - it just cares about the reduction!



For the addition, note that as we add the integers together sequentially, we'll be adding reductions where we start at $1$, add $2$, and so on - until we add $9$. Then instead of adding $10$, we add its reduction of $1$, then instead of $11$ we add $2$ - and so on. So, we are adding an infinitely cycling sequence of $1,2,3,4,5,6,7,8,9$ instead of thinking about adding the integers up sequentially. However, we note that after we add $1+2+3+4+5+6+7+8+9$, we get $9$ - which has a special property; in the above table, we always have $9+r=r$. Thus, when we jump back to adding a reduction of $1$, we get back to a sum with reduction $1$ - and the cycle repeats.



In more generality, this sort of reasoning is known as modular arithmetic. There's a lot of interesting things that happen when one develops this theory further.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    A very detailed & well done explanation. There is just one small thing I saw to correct. About a fifth of the way down, you have an equation starting with "$10^k = 99ldots 99$" with "$k$ digits" below it. You should have a "$- 1$" after $10^k$.
    $endgroup$
    – John Omielan
    Jan 2 at 6:17








  • 1




    $begingroup$
    @Anthony For the first comment: It's the remainder since, whenever you sum the digits of a number, you subtract a multiple of $9$ from it (e.g. $92$ becomes $11$ by subtracting $81=9times 9$). Repeating this, you get that to get from $n$ to $r$, you subtract some multiple of $9$, so $n-9q=r$ for some $q$. Then, rearranging gives $n=9q+r$. Since we know $r$ is a single digit number, this turns out to be the remainder (and this is the only possibility since if we successively subtract $9$ from a number, we'll only once end up between $1$ and $9$). Basically, it's the only possibility.
    $endgroup$
    – Milo Brandt
    Jan 3 at 16:59






  • 1




    $begingroup$
    ...Not all the resulting totals are divisible by $3$ - for instance, $1$ and $1+2+3+4=10$ are not. In general, every third total will not be. There's a formula for the sum of $1+2+ldots+n=frac{ncdot (n+1)}2$. From this, you can see that $3$ divides the sum whenever $3$ divides $n$ or $n+1$ - which happens $2/3$ of the time. (It's also worth noting that almost everything in this post works by replacing "reduction" with "remainder when divided by $k$" for any $k$ - so if you work things out with $k=3$, you can get an addition table for those remainders and see it there)
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:01






  • 1




    $begingroup$
    ...Nothing in this post is specific to base $10$ - essentially, everything works if you replace $9$ by $10-1$ in whatever base you're in (or, otherwise said, $b-1$ in base $b$). You'll get a (eventually) repeating pattern of reductions in any base, though the length of the pattern might change. It's worth looking at.
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:03






  • 1




    $begingroup$
    @Anthony Yes - modular arithmetic, more generally, deals with remainders (so, this whole post talks about "mod $9$" structure)
    $endgroup$
    – Milo Brandt
    Jan 3 at 23:28












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









4












$begingroup$

The answer to all of these question lies in what actually happens when you "reduce" a number by repeatedly summing its digits. In particular, the following is true




Let $n$ be a positive integer. If $n$ is divisible by $9$, its reduction will be $9$. Otherwise, its reduction will be the remainder of $n$ when divided by $9$.




Here, for formal purposes, "remainder" means the number $0leq r<9$ such that you can write
$$n=9q+r$$
For instance, the remainder of $23$ divided by $9$ is $5$ since $9times 2 + 5 = 23$. Note that $2+3=5$ as well.



The basic thing to notice in order to prove this is that the difference between an integer $n$ and the sum of its digits $s$ is always a multiple of $9$. For instance, $23-(2+3)=18=9times 2$. You can prove this by looking at the expanded form of decimal representation; in particular if $n$ has decimal expansion $d_kd_{k-1}ldots d_1d_0$ for some sequence of digits $d_i$, we have
$$n=10^kd_k+10^{k-1}d_{k-1}+ldots+10^1cdot d_1 + 10^0cdot d_0.$$
Then, the sum of the digits is
$$s=d_k+d_{k-1}+ldots+d_1+d_0.$$
The difference between these two values is
$$n-s=(10^k-1)d_k+(10^{k-1}-1)d_{k-1}+ldots+(10^1-1)cdot d_1 + (10^0-1)cdot d_0.$$
However, note that $$10^k-1=underbrace{99ldots 99}_{ktext{ digits}}=9cdot underbrace{11ldots 11}_{ktext{ digits}}.$$
Thus, every term in the difference $n-s$ is $9$ times something, so the whole thing is $9$ times something - so $n-s$ is divisible by $9$.



You can repeat this logic as many times as necessary to see that if $r$ was the reduction of $n$, then $n-r$ is divisible by $9$ - but then you find that there are only actually $9$ positive single digits numbers - and $r$ is definitely one of them - so it's only a matter of checking that there is only one positive single digit number $r$ so that $n-r$ is a multiple of $9$.





Okay, so why does this help us? Well, it reveals another surprising fact:




Let $n$ and $n'$ be positive integers with reductions $r$ and $r'$ respectively. The reduction of $n+n'$ and $ncdot n'$ equal the reductions of $r+r'$ or $rcdot r'$ respectively.




If you want to prove this, you basically write $n=9q+r$ and $n'=9q+r'$ and then note that $n+n'=9(q+q')+(r+r')$ then $nn'=9(9qq'+qr'+q'r)+rr'$ and use the last theorem, which gets rid of all those multiples of $9$ being added, since they can't change the result.



The significance is that you get a multiplication table and addition table where, given the two reductions, you can look up the reduction of the sum and product without knowing the original two numbers. Those tables look like this:



Addition Table



Multiplication Table



This framework basically sets up the answer to your questions: the reductions of the doubling sequence is easiest to explain. To find that sequence, you start at $1$, then, using the table above, multiply that by $2$. You repeat that for a while - and eventually you get back to $1$. The sequence is bound to repeat from there, since you've entered a loop. Note that this method completely forgets about what the number your reducing was - it just cares about the reduction!



For the addition, note that as we add the integers together sequentially, we'll be adding reductions where we start at $1$, add $2$, and so on - until we add $9$. Then instead of adding $10$, we add its reduction of $1$, then instead of $11$ we add $2$ - and so on. So, we are adding an infinitely cycling sequence of $1,2,3,4,5,6,7,8,9$ instead of thinking about adding the integers up sequentially. However, we note that after we add $1+2+3+4+5+6+7+8+9$, we get $9$ - which has a special property; in the above table, we always have $9+r=r$. Thus, when we jump back to adding a reduction of $1$, we get back to a sum with reduction $1$ - and the cycle repeats.



In more generality, this sort of reasoning is known as modular arithmetic. There's a lot of interesting things that happen when one develops this theory further.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    A very detailed & well done explanation. There is just one small thing I saw to correct. About a fifth of the way down, you have an equation starting with "$10^k = 99ldots 99$" with "$k$ digits" below it. You should have a "$- 1$" after $10^k$.
    $endgroup$
    – John Omielan
    Jan 2 at 6:17








  • 1




    $begingroup$
    @Anthony For the first comment: It's the remainder since, whenever you sum the digits of a number, you subtract a multiple of $9$ from it (e.g. $92$ becomes $11$ by subtracting $81=9times 9$). Repeating this, you get that to get from $n$ to $r$, you subtract some multiple of $9$, so $n-9q=r$ for some $q$. Then, rearranging gives $n=9q+r$. Since we know $r$ is a single digit number, this turns out to be the remainder (and this is the only possibility since if we successively subtract $9$ from a number, we'll only once end up between $1$ and $9$). Basically, it's the only possibility.
    $endgroup$
    – Milo Brandt
    Jan 3 at 16:59






  • 1




    $begingroup$
    ...Not all the resulting totals are divisible by $3$ - for instance, $1$ and $1+2+3+4=10$ are not. In general, every third total will not be. There's a formula for the sum of $1+2+ldots+n=frac{ncdot (n+1)}2$. From this, you can see that $3$ divides the sum whenever $3$ divides $n$ or $n+1$ - which happens $2/3$ of the time. (It's also worth noting that almost everything in this post works by replacing "reduction" with "remainder when divided by $k$" for any $k$ - so if you work things out with $k=3$, you can get an addition table for those remainders and see it there)
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:01






  • 1




    $begingroup$
    ...Nothing in this post is specific to base $10$ - essentially, everything works if you replace $9$ by $10-1$ in whatever base you're in (or, otherwise said, $b-1$ in base $b$). You'll get a (eventually) repeating pattern of reductions in any base, though the length of the pattern might change. It's worth looking at.
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:03






  • 1




    $begingroup$
    @Anthony Yes - modular arithmetic, more generally, deals with remainders (so, this whole post talks about "mod $9$" structure)
    $endgroup$
    – Milo Brandt
    Jan 3 at 23:28
















4












$begingroup$

The answer to all of these question lies in what actually happens when you "reduce" a number by repeatedly summing its digits. In particular, the following is true




Let $n$ be a positive integer. If $n$ is divisible by $9$, its reduction will be $9$. Otherwise, its reduction will be the remainder of $n$ when divided by $9$.




Here, for formal purposes, "remainder" means the number $0leq r<9$ such that you can write
$$n=9q+r$$
For instance, the remainder of $23$ divided by $9$ is $5$ since $9times 2 + 5 = 23$. Note that $2+3=5$ as well.



The basic thing to notice in order to prove this is that the difference between an integer $n$ and the sum of its digits $s$ is always a multiple of $9$. For instance, $23-(2+3)=18=9times 2$. You can prove this by looking at the expanded form of decimal representation; in particular if $n$ has decimal expansion $d_kd_{k-1}ldots d_1d_0$ for some sequence of digits $d_i$, we have
$$n=10^kd_k+10^{k-1}d_{k-1}+ldots+10^1cdot d_1 + 10^0cdot d_0.$$
Then, the sum of the digits is
$$s=d_k+d_{k-1}+ldots+d_1+d_0.$$
The difference between these two values is
$$n-s=(10^k-1)d_k+(10^{k-1}-1)d_{k-1}+ldots+(10^1-1)cdot d_1 + (10^0-1)cdot d_0.$$
However, note that $$10^k-1=underbrace{99ldots 99}_{ktext{ digits}}=9cdot underbrace{11ldots 11}_{ktext{ digits}}.$$
Thus, every term in the difference $n-s$ is $9$ times something, so the whole thing is $9$ times something - so $n-s$ is divisible by $9$.



You can repeat this logic as many times as necessary to see that if $r$ was the reduction of $n$, then $n-r$ is divisible by $9$ - but then you find that there are only actually $9$ positive single digits numbers - and $r$ is definitely one of them - so it's only a matter of checking that there is only one positive single digit number $r$ so that $n-r$ is a multiple of $9$.





Okay, so why does this help us? Well, it reveals another surprising fact:




Let $n$ and $n'$ be positive integers with reductions $r$ and $r'$ respectively. The reduction of $n+n'$ and $ncdot n'$ equal the reductions of $r+r'$ or $rcdot r'$ respectively.




If you want to prove this, you basically write $n=9q+r$ and $n'=9q+r'$ and then note that $n+n'=9(q+q')+(r+r')$ then $nn'=9(9qq'+qr'+q'r)+rr'$ and use the last theorem, which gets rid of all those multiples of $9$ being added, since they can't change the result.



The significance is that you get a multiplication table and addition table where, given the two reductions, you can look up the reduction of the sum and product without knowing the original two numbers. Those tables look like this:



Addition Table



Multiplication Table



This framework basically sets up the answer to your questions: the reductions of the doubling sequence is easiest to explain. To find that sequence, you start at $1$, then, using the table above, multiply that by $2$. You repeat that for a while - and eventually you get back to $1$. The sequence is bound to repeat from there, since you've entered a loop. Note that this method completely forgets about what the number your reducing was - it just cares about the reduction!



For the addition, note that as we add the integers together sequentially, we'll be adding reductions where we start at $1$, add $2$, and so on - until we add $9$. Then instead of adding $10$, we add its reduction of $1$, then instead of $11$ we add $2$ - and so on. So, we are adding an infinitely cycling sequence of $1,2,3,4,5,6,7,8,9$ instead of thinking about adding the integers up sequentially. However, we note that after we add $1+2+3+4+5+6+7+8+9$, we get $9$ - which has a special property; in the above table, we always have $9+r=r$. Thus, when we jump back to adding a reduction of $1$, we get back to a sum with reduction $1$ - and the cycle repeats.



In more generality, this sort of reasoning is known as modular arithmetic. There's a lot of interesting things that happen when one develops this theory further.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    A very detailed & well done explanation. There is just one small thing I saw to correct. About a fifth of the way down, you have an equation starting with "$10^k = 99ldots 99$" with "$k$ digits" below it. You should have a "$- 1$" after $10^k$.
    $endgroup$
    – John Omielan
    Jan 2 at 6:17








  • 1




    $begingroup$
    @Anthony For the first comment: It's the remainder since, whenever you sum the digits of a number, you subtract a multiple of $9$ from it (e.g. $92$ becomes $11$ by subtracting $81=9times 9$). Repeating this, you get that to get from $n$ to $r$, you subtract some multiple of $9$, so $n-9q=r$ for some $q$. Then, rearranging gives $n=9q+r$. Since we know $r$ is a single digit number, this turns out to be the remainder (and this is the only possibility since if we successively subtract $9$ from a number, we'll only once end up between $1$ and $9$). Basically, it's the only possibility.
    $endgroup$
    – Milo Brandt
    Jan 3 at 16:59






  • 1




    $begingroup$
    ...Not all the resulting totals are divisible by $3$ - for instance, $1$ and $1+2+3+4=10$ are not. In general, every third total will not be. There's a formula for the sum of $1+2+ldots+n=frac{ncdot (n+1)}2$. From this, you can see that $3$ divides the sum whenever $3$ divides $n$ or $n+1$ - which happens $2/3$ of the time. (It's also worth noting that almost everything in this post works by replacing "reduction" with "remainder when divided by $k$" for any $k$ - so if you work things out with $k=3$, you can get an addition table for those remainders and see it there)
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:01






  • 1




    $begingroup$
    ...Nothing in this post is specific to base $10$ - essentially, everything works if you replace $9$ by $10-1$ in whatever base you're in (or, otherwise said, $b-1$ in base $b$). You'll get a (eventually) repeating pattern of reductions in any base, though the length of the pattern might change. It's worth looking at.
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:03






  • 1




    $begingroup$
    @Anthony Yes - modular arithmetic, more generally, deals with remainders (so, this whole post talks about "mod $9$" structure)
    $endgroup$
    – Milo Brandt
    Jan 3 at 23:28














4












4








4





$begingroup$

The answer to all of these question lies in what actually happens when you "reduce" a number by repeatedly summing its digits. In particular, the following is true




Let $n$ be a positive integer. If $n$ is divisible by $9$, its reduction will be $9$. Otherwise, its reduction will be the remainder of $n$ when divided by $9$.




Here, for formal purposes, "remainder" means the number $0leq r<9$ such that you can write
$$n=9q+r$$
For instance, the remainder of $23$ divided by $9$ is $5$ since $9times 2 + 5 = 23$. Note that $2+3=5$ as well.



The basic thing to notice in order to prove this is that the difference between an integer $n$ and the sum of its digits $s$ is always a multiple of $9$. For instance, $23-(2+3)=18=9times 2$. You can prove this by looking at the expanded form of decimal representation; in particular if $n$ has decimal expansion $d_kd_{k-1}ldots d_1d_0$ for some sequence of digits $d_i$, we have
$$n=10^kd_k+10^{k-1}d_{k-1}+ldots+10^1cdot d_1 + 10^0cdot d_0.$$
Then, the sum of the digits is
$$s=d_k+d_{k-1}+ldots+d_1+d_0.$$
The difference between these two values is
$$n-s=(10^k-1)d_k+(10^{k-1}-1)d_{k-1}+ldots+(10^1-1)cdot d_1 + (10^0-1)cdot d_0.$$
However, note that $$10^k-1=underbrace{99ldots 99}_{ktext{ digits}}=9cdot underbrace{11ldots 11}_{ktext{ digits}}.$$
Thus, every term in the difference $n-s$ is $9$ times something, so the whole thing is $9$ times something - so $n-s$ is divisible by $9$.



You can repeat this logic as many times as necessary to see that if $r$ was the reduction of $n$, then $n-r$ is divisible by $9$ - but then you find that there are only actually $9$ positive single digits numbers - and $r$ is definitely one of them - so it's only a matter of checking that there is only one positive single digit number $r$ so that $n-r$ is a multiple of $9$.





Okay, so why does this help us? Well, it reveals another surprising fact:




Let $n$ and $n'$ be positive integers with reductions $r$ and $r'$ respectively. The reduction of $n+n'$ and $ncdot n'$ equal the reductions of $r+r'$ or $rcdot r'$ respectively.




If you want to prove this, you basically write $n=9q+r$ and $n'=9q+r'$ and then note that $n+n'=9(q+q')+(r+r')$ then $nn'=9(9qq'+qr'+q'r)+rr'$ and use the last theorem, which gets rid of all those multiples of $9$ being added, since they can't change the result.



The significance is that you get a multiplication table and addition table where, given the two reductions, you can look up the reduction of the sum and product without knowing the original two numbers. Those tables look like this:



Addition Table



Multiplication Table



This framework basically sets up the answer to your questions: the reductions of the doubling sequence is easiest to explain. To find that sequence, you start at $1$, then, using the table above, multiply that by $2$. You repeat that for a while - and eventually you get back to $1$. The sequence is bound to repeat from there, since you've entered a loop. Note that this method completely forgets about what the number your reducing was - it just cares about the reduction!



For the addition, note that as we add the integers together sequentially, we'll be adding reductions where we start at $1$, add $2$, and so on - until we add $9$. Then instead of adding $10$, we add its reduction of $1$, then instead of $11$ we add $2$ - and so on. So, we are adding an infinitely cycling sequence of $1,2,3,4,5,6,7,8,9$ instead of thinking about adding the integers up sequentially. However, we note that after we add $1+2+3+4+5+6+7+8+9$, we get $9$ - which has a special property; in the above table, we always have $9+r=r$. Thus, when we jump back to adding a reduction of $1$, we get back to a sum with reduction $1$ - and the cycle repeats.



In more generality, this sort of reasoning is known as modular arithmetic. There's a lot of interesting things that happen when one develops this theory further.






share|cite|improve this answer











$endgroup$



The answer to all of these question lies in what actually happens when you "reduce" a number by repeatedly summing its digits. In particular, the following is true




Let $n$ be a positive integer. If $n$ is divisible by $9$, its reduction will be $9$. Otherwise, its reduction will be the remainder of $n$ when divided by $9$.




Here, for formal purposes, "remainder" means the number $0leq r<9$ such that you can write
$$n=9q+r$$
For instance, the remainder of $23$ divided by $9$ is $5$ since $9times 2 + 5 = 23$. Note that $2+3=5$ as well.



The basic thing to notice in order to prove this is that the difference between an integer $n$ and the sum of its digits $s$ is always a multiple of $9$. For instance, $23-(2+3)=18=9times 2$. You can prove this by looking at the expanded form of decimal representation; in particular if $n$ has decimal expansion $d_kd_{k-1}ldots d_1d_0$ for some sequence of digits $d_i$, we have
$$n=10^kd_k+10^{k-1}d_{k-1}+ldots+10^1cdot d_1 + 10^0cdot d_0.$$
Then, the sum of the digits is
$$s=d_k+d_{k-1}+ldots+d_1+d_0.$$
The difference between these two values is
$$n-s=(10^k-1)d_k+(10^{k-1}-1)d_{k-1}+ldots+(10^1-1)cdot d_1 + (10^0-1)cdot d_0.$$
However, note that $$10^k-1=underbrace{99ldots 99}_{ktext{ digits}}=9cdot underbrace{11ldots 11}_{ktext{ digits}}.$$
Thus, every term in the difference $n-s$ is $9$ times something, so the whole thing is $9$ times something - so $n-s$ is divisible by $9$.



You can repeat this logic as many times as necessary to see that if $r$ was the reduction of $n$, then $n-r$ is divisible by $9$ - but then you find that there are only actually $9$ positive single digits numbers - and $r$ is definitely one of them - so it's only a matter of checking that there is only one positive single digit number $r$ so that $n-r$ is a multiple of $9$.





Okay, so why does this help us? Well, it reveals another surprising fact:




Let $n$ and $n'$ be positive integers with reductions $r$ and $r'$ respectively. The reduction of $n+n'$ and $ncdot n'$ equal the reductions of $r+r'$ or $rcdot r'$ respectively.




If you want to prove this, you basically write $n=9q+r$ and $n'=9q+r'$ and then note that $n+n'=9(q+q')+(r+r')$ then $nn'=9(9qq'+qr'+q'r)+rr'$ and use the last theorem, which gets rid of all those multiples of $9$ being added, since they can't change the result.



The significance is that you get a multiplication table and addition table where, given the two reductions, you can look up the reduction of the sum and product without knowing the original two numbers. Those tables look like this:



Addition Table



Multiplication Table



This framework basically sets up the answer to your questions: the reductions of the doubling sequence is easiest to explain. To find that sequence, you start at $1$, then, using the table above, multiply that by $2$. You repeat that for a while - and eventually you get back to $1$. The sequence is bound to repeat from there, since you've entered a loop. Note that this method completely forgets about what the number your reducing was - it just cares about the reduction!



For the addition, note that as we add the integers together sequentially, we'll be adding reductions where we start at $1$, add $2$, and so on - until we add $9$. Then instead of adding $10$, we add its reduction of $1$, then instead of $11$ we add $2$ - and so on. So, we are adding an infinitely cycling sequence of $1,2,3,4,5,6,7,8,9$ instead of thinking about adding the integers up sequentially. However, we note that after we add $1+2+3+4+5+6+7+8+9$, we get $9$ - which has a special property; in the above table, we always have $9+r=r$. Thus, when we jump back to adding a reduction of $1$, we get back to a sum with reduction $1$ - and the cycle repeats.



In more generality, this sort of reasoning is known as modular arithmetic. There's a lot of interesting things that happen when one develops this theory further.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 16:36

























answered Jan 2 at 5:17









Milo BrandtMilo Brandt

40k476140




40k476140












  • $begingroup$
    A very detailed & well done explanation. There is just one small thing I saw to correct. About a fifth of the way down, you have an equation starting with "$10^k = 99ldots 99$" with "$k$ digits" below it. You should have a "$- 1$" after $10^k$.
    $endgroup$
    – John Omielan
    Jan 2 at 6:17








  • 1




    $begingroup$
    @Anthony For the first comment: It's the remainder since, whenever you sum the digits of a number, you subtract a multiple of $9$ from it (e.g. $92$ becomes $11$ by subtracting $81=9times 9$). Repeating this, you get that to get from $n$ to $r$, you subtract some multiple of $9$, so $n-9q=r$ for some $q$. Then, rearranging gives $n=9q+r$. Since we know $r$ is a single digit number, this turns out to be the remainder (and this is the only possibility since if we successively subtract $9$ from a number, we'll only once end up between $1$ and $9$). Basically, it's the only possibility.
    $endgroup$
    – Milo Brandt
    Jan 3 at 16:59






  • 1




    $begingroup$
    ...Not all the resulting totals are divisible by $3$ - for instance, $1$ and $1+2+3+4=10$ are not. In general, every third total will not be. There's a formula for the sum of $1+2+ldots+n=frac{ncdot (n+1)}2$. From this, you can see that $3$ divides the sum whenever $3$ divides $n$ or $n+1$ - which happens $2/3$ of the time. (It's also worth noting that almost everything in this post works by replacing "reduction" with "remainder when divided by $k$" for any $k$ - so if you work things out with $k=3$, you can get an addition table for those remainders and see it there)
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:01






  • 1




    $begingroup$
    ...Nothing in this post is specific to base $10$ - essentially, everything works if you replace $9$ by $10-1$ in whatever base you're in (or, otherwise said, $b-1$ in base $b$). You'll get a (eventually) repeating pattern of reductions in any base, though the length of the pattern might change. It's worth looking at.
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:03






  • 1




    $begingroup$
    @Anthony Yes - modular arithmetic, more generally, deals with remainders (so, this whole post talks about "mod $9$" structure)
    $endgroup$
    – Milo Brandt
    Jan 3 at 23:28


















  • $begingroup$
    A very detailed & well done explanation. There is just one small thing I saw to correct. About a fifth of the way down, you have an equation starting with "$10^k = 99ldots 99$" with "$k$ digits" below it. You should have a "$- 1$" after $10^k$.
    $endgroup$
    – John Omielan
    Jan 2 at 6:17








  • 1




    $begingroup$
    @Anthony For the first comment: It's the remainder since, whenever you sum the digits of a number, you subtract a multiple of $9$ from it (e.g. $92$ becomes $11$ by subtracting $81=9times 9$). Repeating this, you get that to get from $n$ to $r$, you subtract some multiple of $9$, so $n-9q=r$ for some $q$. Then, rearranging gives $n=9q+r$. Since we know $r$ is a single digit number, this turns out to be the remainder (and this is the only possibility since if we successively subtract $9$ from a number, we'll only once end up between $1$ and $9$). Basically, it's the only possibility.
    $endgroup$
    – Milo Brandt
    Jan 3 at 16:59






  • 1




    $begingroup$
    ...Not all the resulting totals are divisible by $3$ - for instance, $1$ and $1+2+3+4=10$ are not. In general, every third total will not be. There's a formula for the sum of $1+2+ldots+n=frac{ncdot (n+1)}2$. From this, you can see that $3$ divides the sum whenever $3$ divides $n$ or $n+1$ - which happens $2/3$ of the time. (It's also worth noting that almost everything in this post works by replacing "reduction" with "remainder when divided by $k$" for any $k$ - so if you work things out with $k=3$, you can get an addition table for those remainders and see it there)
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:01






  • 1




    $begingroup$
    ...Nothing in this post is specific to base $10$ - essentially, everything works if you replace $9$ by $10-1$ in whatever base you're in (or, otherwise said, $b-1$ in base $b$). You'll get a (eventually) repeating pattern of reductions in any base, though the length of the pattern might change. It's worth looking at.
    $endgroup$
    – Milo Brandt
    Jan 3 at 17:03






  • 1




    $begingroup$
    @Anthony Yes - modular arithmetic, more generally, deals with remainders (so, this whole post talks about "mod $9$" structure)
    $endgroup$
    – Milo Brandt
    Jan 3 at 23:28
















$begingroup$
A very detailed & well done explanation. There is just one small thing I saw to correct. About a fifth of the way down, you have an equation starting with "$10^k = 99ldots 99$" with "$k$ digits" below it. You should have a "$- 1$" after $10^k$.
$endgroup$
– John Omielan
Jan 2 at 6:17






$begingroup$
A very detailed & well done explanation. There is just one small thing I saw to correct. About a fifth of the way down, you have an equation starting with "$10^k = 99ldots 99$" with "$k$ digits" below it. You should have a "$- 1$" after $10^k$.
$endgroup$
– John Omielan
Jan 2 at 6:17






1




1




$begingroup$
@Anthony For the first comment: It's the remainder since, whenever you sum the digits of a number, you subtract a multiple of $9$ from it (e.g. $92$ becomes $11$ by subtracting $81=9times 9$). Repeating this, you get that to get from $n$ to $r$, you subtract some multiple of $9$, so $n-9q=r$ for some $q$. Then, rearranging gives $n=9q+r$. Since we know $r$ is a single digit number, this turns out to be the remainder (and this is the only possibility since if we successively subtract $9$ from a number, we'll only once end up between $1$ and $9$). Basically, it's the only possibility.
$endgroup$
– Milo Brandt
Jan 3 at 16:59




$begingroup$
@Anthony For the first comment: It's the remainder since, whenever you sum the digits of a number, you subtract a multiple of $9$ from it (e.g. $92$ becomes $11$ by subtracting $81=9times 9$). Repeating this, you get that to get from $n$ to $r$, you subtract some multiple of $9$, so $n-9q=r$ for some $q$. Then, rearranging gives $n=9q+r$. Since we know $r$ is a single digit number, this turns out to be the remainder (and this is the only possibility since if we successively subtract $9$ from a number, we'll only once end up between $1$ and $9$). Basically, it's the only possibility.
$endgroup$
– Milo Brandt
Jan 3 at 16:59




1




1




$begingroup$
...Not all the resulting totals are divisible by $3$ - for instance, $1$ and $1+2+3+4=10$ are not. In general, every third total will not be. There's a formula for the sum of $1+2+ldots+n=frac{ncdot (n+1)}2$. From this, you can see that $3$ divides the sum whenever $3$ divides $n$ or $n+1$ - which happens $2/3$ of the time. (It's also worth noting that almost everything in this post works by replacing "reduction" with "remainder when divided by $k$" for any $k$ - so if you work things out with $k=3$, you can get an addition table for those remainders and see it there)
$endgroup$
– Milo Brandt
Jan 3 at 17:01




$begingroup$
...Not all the resulting totals are divisible by $3$ - for instance, $1$ and $1+2+3+4=10$ are not. In general, every third total will not be. There's a formula for the sum of $1+2+ldots+n=frac{ncdot (n+1)}2$. From this, you can see that $3$ divides the sum whenever $3$ divides $n$ or $n+1$ - which happens $2/3$ of the time. (It's also worth noting that almost everything in this post works by replacing "reduction" with "remainder when divided by $k$" for any $k$ - so if you work things out with $k=3$, you can get an addition table for those remainders and see it there)
$endgroup$
– Milo Brandt
Jan 3 at 17:01




1




1




$begingroup$
...Nothing in this post is specific to base $10$ - essentially, everything works if you replace $9$ by $10-1$ in whatever base you're in (or, otherwise said, $b-1$ in base $b$). You'll get a (eventually) repeating pattern of reductions in any base, though the length of the pattern might change. It's worth looking at.
$endgroup$
– Milo Brandt
Jan 3 at 17:03




$begingroup$
...Nothing in this post is specific to base $10$ - essentially, everything works if you replace $9$ by $10-1$ in whatever base you're in (or, otherwise said, $b-1$ in base $b$). You'll get a (eventually) repeating pattern of reductions in any base, though the length of the pattern might change. It's worth looking at.
$endgroup$
– Milo Brandt
Jan 3 at 17:03




1




1




$begingroup$
@Anthony Yes - modular arithmetic, more generally, deals with remainders (so, this whole post talks about "mod $9$" structure)
$endgroup$
– Milo Brandt
Jan 3 at 23:28




$begingroup$
@Anthony Yes - modular arithmetic, more generally, deals with remainders (so, this whole post talks about "mod $9$" structure)
$endgroup$
– Milo Brandt
Jan 3 at 23:28


















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