An Interesting Series Sequel

Number Theory Level 4

In the infinite series $\dfrac{1}{9}+\dfrac{1}{99}+\cdots + \dfrac{1}{10^n-1}+\cdots$ at what place after the decimal point does the second $1$ occur?

$\text{Details and Assumptions}$

The first $1$ occurs at the first place after the decimal point.

The answer is 576.

2 solutions

Jason Gross
May 31, 2016

The way to solve this is by factorization. The terms in the sequence (not the series) are as follows:

1/9 = .1111111...

1/99 = .01010101...

1/999 = .001001001001....

As you can see, when I compute $\frac{1}{10^n-1}$ , I get a 1 in any decimal place divisible by n. For 1/9, n=1, for 1/99, n=2, and for 1/999, n=3. As logically follows, when summing these terms, I should therefore add 1 to a decimal place if that place is divisible by n.

For example, summing up the first three terms, I get .122213122213... Taking the first place, only 1 goes into 1, and so I only add 1 once. For the second place, both 1 and 2 go into 2, so I add 1 twice. For the third place, 1 and 3 go into 3, so I add 1 twice. So on, and so forth.

You will notice, of course, that I get 3 in places where 1, 2, and 3 all go into the place evenly. For example, in the sixth place, one can see that 1, 2, and 3 all go evenly into 6, and so I add 1 three times.

Thus, the value of a place p in the infinite sum, where n-values encompass all whole numbers, is equivalent to the number of whole numbers I can divide evenly into p. Put another way:

The value in place p is equivalent to the number of whole factors of p.

Now, obviously none of these values but the first one can be 1. However, they need not be: if the value at p is, say, 12, then we have carry-over: the value at p-1 increases by 1, and p becomes 2. For example, the factorization of 48 is 1, 2, 3, 4, 6, 8, 12, 16, 24, 48, or 10 numbers. Thus, the value at 48 becomes 0, and the value at 47 increases from 2 (1, 47) to 3.

So how do we use this knowledge to find the first place p with a value of 1? Well, we are looking for a place where the value of p plus the carry-over of the next term is a number ending in a 1.

(Note: One can prove, by looking at the convergences of the infinite series minus the partial sums, that carry-overs do not matter beyond a certain point, but this proof is annoying and complicated and you don't actually need it for the brute force method I employ. Instead, intuition is enough to tell you that you are unlikely to receive multiple numbers with sizable carry-over in a row, at least with low enough numbers)

I could not figure out a formula for this, so instead I simply wrote a program to cycle through every place p and inform me if the number of its factors, plus the carry-over of the next term, was a number ending in a 1. If so, it would print that number. The output was:

576

944

1024

1424

...

The further down the list, the less likely they were to work, but after this I simply printed out the factors of each number and examined all the spots around 576. They are:

576: 21

577: 2

578: 6

...

As one can see, there is no carry-over from the following terms, and so 576 carries over into just 1, and therefore it works.

(There may be a way to solve this without using a program, but I can't think of one.)

Jack D'Aurizio
Apr 9, 2014

Due to the fact that the decimal representation of $\frac{1}{10^n-1}$ has a one in every position that is a multiple of $n$ , the second one in the decimal representation of the series occurs in position $m>1$ where $d(m)\equiv 1\pmod{10}$ . The smallest of such $m$ s is just $576=2^6\cdot 3^2$ . Additional care is needed in order to prove that carriages do not affect the answer.

How would you show that carry-over doesn't affect the answer?

How do you know that there isn't a smaller value where $d(m) \equiv 0 \pmod{10}$ , and we carried over 1, which would then give us a value of 1? This could easily get complicated with multiple carry-overs, like if $d(m) = 9, d (m+1) = 19, d(m+2) = 20$ .

Calvin Lin Staff - 7 years, 2 months ago

Seeing that this has to do with the distribution of the number of factors of a number, which is pretty random, the best way I could think of to showing that carryovers yield a smaller solution is by guessing and checking.

Daniel Liu - 7 years, 2 months ago

Yes! I want to know that (not just checking...), 576 was going to be my first attempt but I was thinking "I'm sure it's going to be a*b^4 and then I'll have to carry a 1..."

Richard Polak - 7 years, 2 months ago

I hope this may help:

Search for the smallest number having value of σ0(n) (mod 10)=1 from the sequence A000005.

where, σ0(n) is divisor function which gives the number of divisors of n.

http://en.wikipedia.org/wiki/Divisor_function

So, σ0(n)=1,11,21,31, and ...so on. But σ0(n)=1 corresponds to the 1st decimal place.

Writing prime factorisation of the rest,

11=11, so the smallest number with σ0(n) of 11 is n = $2^{10}$ = 1024

Taking base 2 will generate the smallest number required. If there are more prime factors,

then we take 3,5,7,11 and so on in order.

21= 3 $\times$ 7, so n = $2^{6}$ $\times$ $3^{2}$ = 576 <1024

31=31, n = $2^{30}$ > 576

41=41, n = $2^{40}$ > 576

51=3 $\times$ 17, no.= $2^{16}$ $\times$ $3^{2}$ > 576

As we can see that this goes on increasing... so the smallest no. is 576.

Now, to check the carry: 577 is prime no. and so at 577th decimal place, there'll be digit 2 and so no carry is generated here. Since σ0(575)=6, so there'll be digit 6+2=8 at 575th position. 2 is added because of carry of 2 from the 576th place.

Hence the required number is 576.

Bhaskar Sukulbrahman - 7 years, 1 month ago

i dont understand your explanation from "if it is 11 so n is 1024" i think 21and 51 has different method so that is why it is confused me.

actually i still dont understand how this methods work :P

please anyone..

Hafizh Ahsan Permana - 6 years, 11 months ago

I was thinking that number at k position will be number of factors of k. So the number with 11 factors which is not followed by number with more than 10 factors would be the answer. So the obvious answer according to me was 1024 and then I also tried a^4*b type numbers. Plz explain where I went wrong.

Murlidhar Sharma - 7 years, 1 month ago

You checked $\phi(n)=11$ giving $n=2^{10}$ as the smallest $n$ . But you forgot to check $\phi(n)=21=7*3$ giving $n=2^6 3^2=576$ .

Ron van den Burg - 3 years, 5 months ago

i'm just wondering why it wouldn't appear in the 72nd position as i think 72 is the first with 11 factors?? just wanted to know what i was doing wrong

Toby Chamberlain - 5 years, 7 months ago

$72$ has $12$ factors.

Shourya Pandey - 5 years, 1 month ago

$72=2^3 3^2$ giving $(3+1)(2+1)=12$ factors ( $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$ ).

Ron van den Burg - 3 years, 5 months ago

What is (mod10)?

Vishal Ajwani - 7 years, 2 months ago

An Interesting Series Sequel

The answer is 576.

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