Find the $26^{th}$ digit of a $50$ digit number divisible by $13$.

$$N$$ is a $$50$$ digit number (in the decimal scale). All digits except the $$26^{th}$$ digit (from the left) are $$1$$. If $$N$$ is divisible by $$13$$, find the $$26^{th}$$ digit.

This question was asked in RMO $$1990$$ and is very similar to this question and the same as this question but it is not solved by the approach used by me whereas I want to verify my approach.

My approach:

Suppose $$N=111\cdots a\cdots111$$ and $$N\equiv 0\pmod {13}$$

Now $$N=10^{49}+10^{48}+\ldots+a10^{24}+\ldots+10+1=(10^{49}+10^{48}\ldots+10+1)+(a-1)10^{24}$$

$$N=\dfrac{10^{50}-1}{9}+(a-1)10^{24}$$

Now $$10^{12}\equiv 1\pmod {13}\Rightarrow 10^{24}\equiv 1\pmod {13}$$ by fermat's little theorem.

Thus $$(a-1)10^{24}\equiv (a-1) \pmod{13}\Rightarrow \dfrac{10^{50}-1}{9}\equiv 1-a\pmod{13}$$ since $$N\equiv 0\pmod{13}$$

$$10^{24}\equiv 1\pmod{13}\Rightarrow 10^{48}\equiv 1\pmod{13}$$ or $$10^{50}-1\equiv -5 \pmod{13}$$

Now $$10^{50}-1\equiv -5\pmod {13}\Rightarrow 9(1-a)\equiv -5\pmod{13}$$

$$a=3$$ clearly satisfies the above conditions

$$\therefore$$ The $$26^{th}$$ digit from the left must be $$3$$.

Please suggest what is incorrect in this solution and advice for alternative solutions.

THANKS

• $111111$ is divislbe by $13$ so only consider 25th and 26th digits Sep 1, 2020 at 9:25
• @Peter Yes, but by then they have already multiplied away the denominator; note how $4$ became $36$. Sep 1, 2020 at 9:26
• @MathLover But we can also consider the residues of $\frac{10^{50}-1}{9}$ and $10^{24}$ mod $13$ to get the solution. Sep 1, 2020 at 9:29
• @Peter absolutely we can and OP's is the right way to do it. I was just making a comment that if we know $111111$ is divisible by $13$, we can ignore first $24$ and last $24$ $1's$. Sep 1, 2020 at 9:32
• @Peter If we want to add $10^{50} + 10^{49} + \cdots + 10 + 1$, then that becomes $\frac{10^{51} - 1}9$. Turns out, however, that that's not what we want to add. So yes, the exponent in the fraction should've been $50$, but only because the original sum was not the one we were after, not because of an arithmetical error. Sep 1, 2020 at 9:32

$$10^{50}$$ is a 51-digit number. And in a 50-digit number, the digit 26th from the left is represented by $$10^{24}$$.

Other than these two mistakes, I find your approach entirely reasonable. And if they were looking for a 51-digit number, with all except the 25th digit from the left being $$1$$, then it would've been correct too.

Edit: After having corrected these two off-by-one errors, the solution looks fine.

• I'm sorry if I am mistaken, but wouldn't the 26th digit frmo the left be represented by $10^{25}$? Sep 1, 2020 at 9:28
• @DevanshKamra In a 51-digit number, yes. Not in a 50-digit number. Sep 1, 2020 at 9:28
• Okay, let me correct the question Sep 1, 2020 at 9:29
• Please check the question now. Is it correct now? Sep 1, 2020 at 9:37
• @DevanshKamra It looks good to me, yes. Sep 1, 2020 at 9:39

Another way is to use the trick from Wikipedia (that doesn't solve your solution)

Taking $$N$$ from the right, and applying the sequence $$(1, −3, −4, −1, 3, 4)$$ as instructed on the page (multiply the digits from the right by the given numbers in sequence), we get

$$0$$ for the 6 first digits from the right ($$1-3-4-1+3+4=0$$), repeating the sequence, $$0$$ up to digit 24 (from right), we still have $$0$$

Then, the next $$6$$ are our $$a$$ and $$5\times 1$$, or $$a-3-4-1+3+4\\=a-1$$

We did $$30$$ digits, $$20$$ to go. The next $$18$$ will give $$0$$, the last $$2$$ give $$1-3$$, thus the whole sum is $$a-1-2=a-3$$ The only digit that would have $$a-3\equiv 0\pmod {13}$$ is $$\bbox[5px,border:2px solid #ba9]{a=3}$$

• not a general approach, but quite a neat one. Thanks for this approach (+1) Sep 1, 2020 at 9:50

The number $$N$$ consists of $$24$$ ones followed by the two digits $$1a$$ (the $$2$$-digit number $$10+a$$) followed by another $$24$$ ones, so with the number $$M$$ consisting of $$24$$ ones, $$M:=\sum_{k=0}^{23}10^k=\frac{10^{24}-1}{9}$$, we have $$N=M\cdot10^{24+2}+(10+a)\cdot10^{24}+M$$ Since $$13$$ is a prime, from Fermat's Little Theorem we know that $$10^{12}\equiv1\pmod{13}$$, and it follows that $$13\mid(10^{12}-1)(10^{12}+1)=10^{24}-1=9M \Rightarrow 13\mid9 \lor 13\mid M$$. Obviously, $$13\nmid 9$$, so $$13\mid M$$.

Now, if $$13\mid N$$, it follows that $$13\mid (10+a)\cdot10^{24}$$, and since $$13\nmid10^{24}$$, it must be $$13\mid10+a$$. Since $$0\le a\le9$$, it must be $$a=3$$.

• That certainly is a good approach. (+1) Sep 1, 2020 at 14:14

There's several tricks you can use but mostly they are similar to yours.

A famous well known trick is that as $$1001 = 13*7*11$$ so your number, $$N$$ is divisible by $$13$$ if and only if the $$N- 1001*10^k$$ is divisible by $$13$$ and so we can remove any pairs of $$1$$s if there are $$3$$ spaces apart. So we can get rid of the $$1$$ and $$4$$ one, the $$2$$nd and $$5$$ one, and the third and $$6$$th ones to get rid of the first $$6$$ ones ($$111111\div 13 = 8547$$ BTW). We can repeat that $$4$$ times to get rid of the first $$24$$ ones, and do it to the end to get rid of the last $$24$$ ones to and up with $$11111...11d111.....11$$ is divisble by $$13$$ if and only if $$1d00000....000= (10+d)\times 10^{24}$$ is.

Now $$1001 = 13*7*11$$ so $$100\equiv -1 \pmod 13$$ so $$10^{24} = 1000^{8}\equiv (-1)^8\equiv 1 \pmod {13}$$. So $$(10+d)\times 10^{24}\equiv (10+d)\times 1\equiv 10+d \pmod {13}$$ so if this is divisble by $$13$$ we must have $$d = 3$$.

That was tedious.....

We could also do, by Fermat's little Theorem $$10^{12} \equiv 1 \pmod {13}$$ so $$10^{12}- 1 =999999999999 \equiv 0 \pmod 13$$ so $$13$$ divides $$999999999999 = 9\times 111111111111$$ and so $$13$$ divides $$9$$ or $$111111111111$$ so $$13|111111111111$$ and we do similar to above to get $$(10+d)\times 10^{24}$$ and as $$10^{12} \equiv 1$$ then $$10^{24} \equiv 1$$ and $$10+d\equiv 0$$ so $$d = 3$$.

.....

Or we could realize the remainder of $$10\div 13$$ is $$10$$. The remainder of $$10^2 \div 13$$ is $$9$$ and so on, and these must eventually cycle through. Just list them all: $$10 \equiv 10; 10^2\equiv 9; 10^3 \equiv 12 \equiv -1$$. SO $$10^4\equiv -10\equiv 3$$ and $$10^{5}\equiv -9\equiv 4$$ and $$10^6\equiv 1$$ and then it repeats. And add them all up. (In groups of $$6$$ whe get $$\sum_{k=0}^5 10^k \equiv 1+10 + 9+(-1)+(-10)+(-9) \equiv 0$$ so $$13|111111$$)

All of theses are more or less the same idea and lead to the conclusion $$d=3$$.