# Big number puzzle

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?

• Wow, nice puzzle! Can we assume that the digits $ab$ for $a \neq b$ appear only once? – athin Aug 14 '19 at 7:24
• Do digits $b$ and $a$ have to be side by side in the big number? Does the "part starting at $a$" includes $a$? The same for $b$. Are there any restrictions as to what $a$ and $b$ can be? – Ardweaden Aug 14 '19 at 7:59
• Or am I getting this wrong and $ab$ is one specific number? – Ardweaden Aug 14 '19 at 8:14
• Is the essential equation $AbaB \times ab = aB00Ab$ where $A$ and $B$ are some strings of digits of unknown length? – hexomino Aug 14 '19 at 11:25
• @hexomino That's my interpretation. – Arnaud Mortier Aug 14 '19 at 11:33

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 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.

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.

• \geq will produce $\geq$. Also, how did you come up with this number, and with "the" (?) way to fix the statement? – Arnaud Mortier Aug 14 '19 at 12:20
• This does seem to be a very interesting number! It does seem that multiplying it by $ab$ with $a \geq b$ gives you some cyclic permutation of the digits with two 0s included but I don't see how it matches the question description. Could you expand a bit more, please? – hexomino Aug 14 '19 at 12:29
• @ArnaudMortier I've googled around a bit and it looks like this is the periodic part of $\frac{1}{109}$ and is the 10th cyclic number (in decimal). This means that cyclic permutations of the number (including two 0s at the beginning) are multiples of it. – hexomino Aug 14 '19 at 12:37
• @hexomino I was doing the same :) it would still be interesting to see a natural way to come up with this number. – Arnaud Mortier Aug 14 '19 at 12:41
• Multiplying by ab with a<b also gives a rotation of the number, it is only the method to find where to split the number that fails. – Florian F Aug 14 '19 at 13:20

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.

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.

• I arrived at the same conclusion as you by a different route. It seems something is probably wrong with the statement of the problem... – Gareth McCaughan Aug 14 '19 at 12:03
• @GarethMcCaughan Indeed, if this is not the intended answer, since the OP does not have the original reference it will be hard to feel around for the correct statement. – Arnaud Mortier Aug 14 '19 at 12:06
• @ArnaudMortier Apologies my interpretation was that there be just one instance of $ab$ but I'll admit the wording seems to support your interpretation. Does this mean that, a priori, $X$ must contain every possible substring of two digits? – hexomino Aug 14 '19 at 12:13
• @hexomino Yes, that's how I interpret it. – Arnaud Mortier Aug 14 '19 at 12:16