Big number puzzle

Question

I saw this on rec.puzzles many years ago, but can't find the reference to credit the source.

I am thinking of a large number. If you want to multiply it by a two digit number $ab$ with $a \lt b$, search through the number for the digits $ba$ in order. Write the part starting at $a$, two zeros, and the part ending with $b$, and you have the product. If you represent the original number as $AbaB$ the product is $aB00Ab$. If you want to multiply by $ab$ with $a \gt b$, let $c=a-1$ and look for $ac$. If the number is $AacB$ the product is $cB00Aa$. If you want to multiply it by a multiple of $11$, so $aa$, there are two copies of $aa$ in the number. Do the same as in the previous sentence, but use the one that does not have a $9$ after it in the original number.

What number am I thinking of?

Answer 1

And here is the number you are probably thinking of:

$(10^{108}-1)/109$ = 9174311926605504587155963302752293577981651376146788990825688073394495412844036697247706422018348623853211

It works only for $ab$ where $a \le b$. I suppose that it is a mistake in the problem statement. Others have proven that as it is, the problem is unsolvable.

And here is how I came to that number.

This behaviour of the multiples being a rotation of the original number occurs when the number is the recurrent part of the decimal expansion of some 1/N.

For example for N=7, 1/N = 0.142857142857... . The recurrent part is 142857. The multiples of 142857 are 142857, 285714, 428571, 571428, 714285, 857142 and 999999.

The additional 00 indicate that 1/N actually starts with two zeroes: 0.009174... The zeroes are stripped in the original number, but must be restored in every rotation of the number. The fact that the recurrent part starts with 00 tells me that N is betwen 100 and 999.

The recurrent part of $1/N$ for a large N is of the form $(10^R-1)/N$ where R is the period of the decimal expansion, or the length of the recurrent part. So I checked only numbers of that form.

The number of digits should have been 100 to accomodate with all the combinations of $ab$ once from 10 to 99, plus a second copy of $aa$'s from 11 to 99, plus 1 because the first and last digits can match only once. The actual length is 106. Since cases with $a<b$ are not counted, it actually could be anywhere from 46 and larger. Anyway I tried all lengths up to 120.

So, using a computer, I searched for numbers of the form $(10^R-1)/N$ that contain all combination of 2 digits $ab$. I didn't find any so I ignored the case $b=0$. I found a few candidates. But checking for which $ab$ the multiplication procedure actually works, I noticed that it works only when $a \le b$. Since other have proven that the problem is not possible as stated, I assume it is a mistake, maybe the procedure for $a>b$ is different than for $a<b$.

PS: I have been playing with this problem. You can extend it to $ab$ with $a > b$ with the following rule:

• if a = b+1 then search $bb9$ and split between the b's.
• if a > b+1 then search $b(a-1)$ and split between b and (a-1).

For example, to multiply by 42 don't search for 24 but 23 and split the number between 2 and 3. For the search with 0's you might need to imagine the '00' in front.

Answer 2

I am going to prove that

such a number cannot exist, because for any $a$ there has to be more than one instance of $aa$ not followed by a $9$. The proof relies on the condition that for every digit $a$, there are exactly two instances of $aa$ in the number. Note that the number found by @FlorianF fails to meet this criterion (there is only one $99$).

Indeed

There is $999$ somewhere in $X$, and no other $99$'s. Indeed, we know that there are only two $99$ sequences, and one of them is followed by a $9$.

Now let us multiply $X$ by $99$. Since there is $\color{red}{99}9$ in $X$, $99\times X$ ends with $\color{red}{99}$ and therefore $X$ ends with $01$. This is because the last two digits of $99\times x$ are always equal to $100$ minus the last two digits of $x$.

Now because of this, the product of $X$ by $aa$ for arbitrary $a$ ends with $aa$, and this gives us that for every digit $a$, the sequence $aa$ not followed by a $9$ has to be preceded by yet another $a$, which makes it $aaa(\text{some digit}\neq 9)$, which proves the claim.

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Let me dump here previous thoughts that turned out not to be useful but might be in the future.

First,

$X$ ends with $1$.
Indeed, by Rule #1 the product $X\times(ab)$ always ends with the digit $b$.

Second

$X$ begins with $9$ (followed either by a non-zero digit, or by $09$ and then a non-zero digit).
This is because $11\times X$ has two more digits than $X$. The smallest such number (with any fixed number of digits) is $909090...$ but there can't be more than one instance of $09$ in $X$, hence the precision in parentheses.