# Find "a" so that 99*a=1a1

(This is my very first question here! :-) ) So I encountered this fun problem yesterday and am feeling like sharing it with you. I almost solved it with the help of my computer but I guess there is a way to solve it without.

So the question is: find all the integer $$a$$ so that $$a*99$$ is equal to $$1a1$$. (The same number "$$a$$" between two 1)

This is what I've done so far (spoiler alert if you want to solve it all by yourself):

I quickly converted this equation into this one: $$89*a -1 =10^n$$ with $$n$$ equals to the number of numbers in $$a$$ plus one.

Then comes $$a=(10^n +1)/89$$

So now the problem is equivalent to finding the n so that $$10^n +1$$ can be divided by $$89$$. One can notice maybe that $$89$$ is a prime number. Fermat theorem gives us : $$10^{88} -1 =0 [mod 89]$$ but this doesn't seem to help :/

Conceptually speaking, I find difficult to convert those two ideas mathematically: -How searching integers solutions differs from searching any solutions. -How to use the fact that n is actually the number of numbers in $$a$$ plus one.

Hope you find the problem interesting and that some very bright solutions will come to your mind :-)

NB: using a computer, I have found some n that verify the equation but this is cheating somehow...

• Welcome to the site ! Commented Sep 8, 2019 at 8:39
• What does $1a1$ mean here? Commented Sep 8, 2019 at 8:39
• I explained between parenthesis what it means but maybe I was not clear enough. If "a" is 333 for instance, 1a1 would be 13331. Commented Sep 8, 2019 at 8:44
• Really nice first question :) You can make your math formulas more readable with MathJax. Here is a tutorial: math.meta.stackexchange.com/questions/5020/… Commented Sep 8, 2019 at 8:51
• You need to find the smallest $n$ such that $10^n=1\mod 89$ such $n\mid 88$(n divides $88$). If such $n$ is divisible by $2$ then $10^{n/2}=-1\mod 89$ and $n/2$ is the smallest such $n$. otherwise the congruence $10^k=-1\mod 89$ has no solutions. Commented Sep 8, 2019 at 8:55

The observation that $$10^{88}\equiv1\pmod{89}$$ is very useful.

If $$a$$ is an $$n$$-digit number, the equation can be rewritten as $$99a=10^n+10a+1$$ and so $$a=\frac{10^n+1}{89}$$ so we need $$10^n\equiv-1\pmod{89}$$.

By Euler-Fermat, $$10^{88}\equiv 1\pmod{89}$$. Let's try and find the order of $$10$$ modulo $$89$$. We have $$10^2\equiv11\pmod{89}$$, $$10^4\equiv32\pmod{89}$$ and $$10^{11}\equiv55\pmod{89}$$, so $$10^{44}\equiv1\pmod{89}$$, but $$10^{22}\equiv88\equiv-1\pmod{89}$$. There's no need to look at $$10^8$$.

So the order is $$44$$ and we have also found a solution to $$10^n\equiv-1\pmod{89}$$, namely $$n=22$$.

Suppose $$10^m\equiv-1\pmod{89}$$; then $$10^m\equiv10^{22}$$, so $$10^{m-22}\equiv1\pmod{89}$$ and therefore $$m\equiv22\pmod{44}$$.

The solutions are the positive integers of the form $$22+44k$$.

The smallest solution is $$\frac{10^{22}+1}{89}=112359550561797752809$$ The next solution is $$\frac{10^{66}+1}{89}= 11235955056179775280898876404494382022471910112359550561797752809$$

• Interestingly, the solutions are very close to sums $\sigma_n:=\sum_{k=1}^{n}10^{n-k}F_n$, where $F_n$ is the $n$-th Fibonacci number: $$\tfrac1{89}\left(10^{22}+1\right)-\sigma_{21}=2113\qquad\tfrac1{89}\left(10^{66}+1\right)-\sigma_{65}=3314006522814$$ both tiny relative errors. Even better, it appears that $$\tfrac1{89}\left(10^{22}+1\right)=\left\lceil\frac{\sigma_{21+p}}{10^p}\right\rceil\qquad \tfrac1{89}\left(10^{66}+1\right)=\left\lceil\frac{\sigma_{65+q}}{10^q}\right\rceil$$ for $p\geq 5$ and $q\geq 17$. I feel like I'm missing an obvious connection.
– Blue
Commented Sep 8, 2019 at 10:53
• @Blue Quite interesting indeed. Commented Sep 8, 2019 at 11:02
• Whoops ... My definition of $\sigma_n$ should use "$F_k$", not "$F_n$".
– Blue
Commented Sep 8, 2019 at 11:08
• With the related sum $\sigma_\star:=\sum_{k=1}^\infty 10^{-k}F_k = .11235955\ldots$ (omitting the leading $0$ on purpose), the recurrence $F_n+F_{n+1}=F_{n+2}$ makes clear why $99\sigma_\star=100\sigma_\star - 1\sigma_\star=11.1235955\ldots$ would have a digit string (without a decimal point) that is the concatenation of $1$ with $\sigma_\star$'s digit string (without a decimal point). So, maybe it's not surprising that the Fibonacci sums $\sigma_n$ enter into the "finitized" problem.
– Blue
Commented Sep 8, 2019 at 11:36
• Thanks for your answer :) I don't understand why there is no need to look at 10^8 and why 10^¹¹ =55 imply 10^44=1 (I don't understand the "so"). Could you explain ? :-) Commented Sep 9, 2019 at 12:46