WakeUp wrote: Find all the three-digit numbers $\overline{abc}$ such that the $6003$-digit number $\overline{abcabc\ldots abc}$ is divisible by $91$. solutions are: 182, 273, ... all threedigit multiples of 91 such that <1000

Let the $6003$-digit number be $N$ and let $x= \overline{abc} $. Then \[N=x(1+10^3+10^6+\cdots+10^{6000})=x\cdot\frac{10^{6003}-1}{10^3-1}\] It can be proven by Fermat's little theorem that $10^{6003}\equiv 6\pmod 7$ and $10^{6003}\equiv 12\pmod 7$ and hence, \[13\times 7|N\Longrightarrow 91|x\] All the three digit multiples of $91$ are the required answer.

gouthamphilomath wrote: Let the $6003$-digit number be $N$ and let $x= \overline{abc} $. Then \[N=x(1+10^3+10^6+\cdots+10^{6000})=x\cdot\frac{10^{6003}-1}{10^3-1}\] It can be proven by Fermat's little theorem that $10^{6003}\equiv 6\pmod 7$ and $10^{6003}\equiv 12\pmod 7$ and hence, \[13\times 7|N\Longrightarrow 91|x\] All the three digit multiples of $91$ are the required answer. $10^{6003}\equiv 6\pmod 7$ and $10^3\equiv 6\pmod 7$, so a mistake with FLT. $91|10^3+1$ and so you find $91|abc$

SCP wrote: gouthamphilomath wrote: Let the $6003$-digit number be $N$ and let $x= \overline{abc} $. Then \[N=x(1+10^3+10^6+\cdots+10^{6000})=x\cdot\frac{10^{6003}-1}{10^3-1}\] It can be proven by Fermat's little theorem that $10^{6003}\equiv 6\pmod 7$ and $10^{6003}\equiv 12\pmod 7$ and hence, \[13\times 7|N\Longrightarrow 91|x\] All the three digit multiples of $91$ are the required answer. $10^{6003}\equiv 6\pmod 7$ and $10^3\equiv 6\pmod 7$, so a mistake with FLT. Sorry but what is wrong there?

gouthamphilomath wrote: SCP wrote: gouthamphilomath wrote: Let the $6003$-digit number be $N$ and let $x= \overline{abc} $. Then \[N=x(1+10^3+10^6+\cdots+10^{6000})=x\cdot\frac{10^{6003}-1}{10^3-1}\] It can be proven by Fermat's little theorem that $10^{6003}\equiv 6\pmod 7$ and $10^{6003}\equiv 12\pmod 7$ and hence, \[13\times 7|N\Longrightarrow 91|x\] All the three digit multiples of $91$ are the required answer. $10^{6003}\equiv 6\pmod 7$ and $10^3\equiv 6\pmod 7$, so a mistake with FLT. Sorry but what is wrong there? You wrote the same different. So I thougt you meant what I wrote, but maybe the idea was $10^{6003}\equiv 6\pmod 7$ and $10^{6003}\equiv 12\pmod [b]13[/b]$ But how get you the result in that case?

Maybe I've missed something, but I don't see how the above proves these are the $only$ values of abc for which 91 is a divisor. Couldn't there be a sequence of, say, $n$ abc's (where n is a divisor of 2001) such that 91 divides this sequence, but 91 does not divide abc by itself? I can show a proof of this if need be, but would like to know first whether it's already been shown in your earlier posts. Merlin

Tons of little typos, never clearly corrected. Since $10^3 + 1 = 1001 = 7\cdot 11\cdot 13 = 11\cdot 91$, it follows that $10^{6003} - 1 = (10^3)^{2001} - 1 \equiv (-1)^{2001} - 1 = -2 \pmod{91}$, hence $\gcd(10^{6003} - 1,91) = 1$. Therefore we need $91 \mid \overline{abc}$.

Note that $\overline{abcabc\ldots abc} = abc(10^{6000}+10^{6000}+...+10^6+10^3+10^0) = abc \left( \frac{10^{6003}-1}{10^3-1} \right) = abc \left( \frac{10^{6003}-1}{999} \right)$. $\gcd(999,91) = 1$. Now, we simply want to ensure that $10^{6003} \mod 13$ is not $1$ and $10^{6003} \mod 7$ isn't $1$. We start with the former, $$10^0 \mod 13 \equiv 1$$$$10^1 \mod 13 \equiv 10$$$$10^2 \mod 13 \equiv 9$$$$10^3 \mod 13 \equiv 12$$$$10^4 \mod 13 \equiv 3$$$$10^5 \mod 13 \equiv 4$$$$10^6 \mod 13 \equiv 1$$Continuing this trend, we see that $10^{6003}-1$ is not divisible by $13$. We do a similar thing, this time $\mod 7$. $$10^0 \mod 7 \equiv 1$$$$10^1 \mod 7 \equiv 3$$$$10^2 \mod 7 \equiv 2$$$$10^3 \mod 7 \equiv 6$$$$10^4 \mod 7 \equiv 4$$$$10^5 \mod 7 \equiv 5$$$$10^6 \mod 7 \equiv 1$$Continuing this trend, we see that $10^{6003}-1$ is not divisible by $7$. Hence, in order for $\overline{abcabc\ldots abc}$ to be divisible by $91$, $\overline{abc}$ must be divisible by $91$.

we have $91\mid 1001$ and in particular, $91\mid \overline{abcabc}$, so since we have $6003$ digits, we must have the remaining digits $\overline{abc}$ divisible by $91$ which are multiples of $91$ greater than or equal to $100$ but smaller than $1000$.

Cool problem. Use the shorthand $\overline{abc}=abc$. Claim: $91|abc$ Proof: Write $$abcabc...=abc*100100100...001=abc\sum_{n=0}^{2000}10^{3n}$$Suffices to show that the summation is $0\pmod{91}$. Lemma: For odd n, $10^{3n}=-1$ in the 91 residue system Proof: Use induction. True for $n=1$. To get from $2k+1$ to $2k+3$, we multiply by $10^6$, which we find as $1$ mod 91, so our induction is complete. $\boxed{}$ From our lemma, we can square to get that for even $n$, $10^{3n}=1\pmod{91}$. So the sum is essentially $1-1+1-1+1...$. Since there is an odd number of terms, the summation is $\ne 0\pmod{91}$, so we’re done.$\blacksquare$ 3000 posts