1. ## 26th digit

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

2. we can use the facts that
$\displaystyle 10^6 \equiv 1 \mod 13$
so $\displaystyle 10^{6k} \equiv 1 \mod 13$
so $\displaystyle a10^{6k} \equiv a \mod 13$
also
$\displaystyle 10^3 \equiv -1 \mod 13$
then combining this with our other idendity
$\displaystyle 10^{6k+3} \equiv -1 \mod 13$
so $\displaystyle a10^{6k+3} \equiv -a \mod 13$

so our number is
$\displaystyle N = 11,111,111,111,111,111,111,111,11k$$\displaystyle ,111,111,111,111,111,111,111,111 which we can write as.. \displaystyle N = 11*10^{48} + 111*10^{45} + ... +$$\displaystyle 11k*10^{24} + ... + 111*10^3 + 111$

using our idendities from before, we know that for instance
$\displaystyle 11*10^{48} \equiv 11 \mod 13$
since 48 = 7*6 so

$\displaystyle N = 11-111+111-111+111-111+111-111+11k$$\displaystyle -111+111-111+111-111+111-111+111 \mod 13 which simplifies to \displaystyle N \equiv 11k - 100 \mod 13 so if \displaystyle N|13 then \displaystyle 11k-100 \equiv 0 \mod 13 the only value of k where this is true is \displaystyle k=3 thus this is the answer. \displaystyle \mbox{the 26th digit is 3.} 3. Originally Posted by Aradesh we can use the facts that \displaystyle 10^6 \equiv 1 \mod 13 so \displaystyle 10^{6k} \equiv 1 \mod 13 so \displaystyle a10^{6k} \equiv a \mod 13 also \displaystyle 10^3 \equiv -1 \mod 13 then combining this with our other idendity \displaystyle 10^{6k+3} \equiv -1 \mod 13 so \displaystyle a10^{6k+3} \equiv -a \mod 13 so our number is \displaystyle N = 11,111,111,111,111,111,111,111,11k$$\displaystyle ,111,111,111,111,111,111,111,111$

which we can write as..
$\displaystyle N = 11*10^{48} + 111*10^{45} + ... +$$\displaystyle 11k*10^{24} + ... + 111*10^3 + 111 using our idendities from before, we know that for instance \displaystyle 11*10^{48} \equiv 11 \mod 13 since 48 = 7*6 so You made amistake. 48 = 6*8 not 7*6 never mind. I could follow your solution. Are there any more methods? 4. Originally Posted by malaygoel You made amistake. 48 = 6*8 not 7*6 never mind. I could follow your solution. Are there any more methods? my finger must have slipped. this is the only way i know how to do this problem. 5. Originally Posted by Aradesh my finger must have slipped. this is the only way i know how to do this problem. I was just wondering could the given number be just divided by 13 and get the answer. 6. oh, good point. i just checked it on mathematica (a computer program that does big sums) and it gives: 854700854700854700854701008547008547008547008547 while other values for our 26th digits don't give a whole number. 7. Originally Posted by Aradesh oh, good point. i just checked it on mathematica (a computer program that does big sums) and it gives: 854700854700854700854701008547008547008547008547 while other values for our 26th digits don't give a whole number. Of course this is not a proof as M. could have a bug in its BigNum arithmetic RonL 8. Originally Posted by CaptainBlack Of course this is not a proof as M. could have a bug in its BigNum arithmetic RonL Mathematica have a BUG?? It's not possible! - 9. Originally Posted by malaygoel I was just wondering could the given number be just divided by 13 and get the answer. What I wanted to say(although I have not tried) is that we could substitute the 26th digit as 1 and perform divison(which will be easy due to repeating 1's) and then find find the value of k by guessing the remainder that the 26th digit should give while performing division. 10. Originally Posted by malaygoel N is a 50digit nomber(in the decimal scale). All digits except the 26th digit(from the left) are 1. If N is divisible by 13, find the 26th digit. I think I have found it. By performing simple division, I found that 111111 is divisible by 13 Now, \displaystyle N=11,111,111,111,111,111,111,111,11k,$$\displaystyle 111,111,111,111,111,111,111,111,$
there are 25 1's before k and 24 1's after k.
24 1's before k and 24 1's after k are divisible by 13
We are left with $\displaystyle 1k$ and hence the answer 3.

11. Excellent solution, Malay!

I've known for years that $\displaystyle 7 \times 11 \times 13 \:=\:1001$ **

. . hence: $\displaystyle 111,111\:=\:111 \times 1001 \:=\:111\times 7 \times 11 \times 13$

I'm embarrassed that none of this occured to me.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

More trivia . . .

. . $\displaystyle 10,001 \;=\;73\cdot137$
. .$\displaystyle 100,001 \;= \;11\cdot9091$
.$\displaystyle 1,000,001\;=\;101\cdot9901$
$\displaystyle 10,000,001\;=\;11\cdot909091$

. . . .$\displaystyle 111\;=\;3\cdot37$
. . .$\displaystyle 1,111 \;=\;11\cdot101$
. . $\displaystyle 11,111\;=\;41\cdot271$
. .$\displaystyle 111,111\;=\;111\cdot1001\;=\;3\cdot7\cdot11\cdot13 \cdot37$
.$\displaystyle 1,111,111\;=\;239\cdot4649$
$\displaystyle 11,111,111\;=\;10001\cdot1111\;=\;11\cdot73\cdot10 1\cdot137$

12. Originally Posted by Soroban
Excellent solution, Malay!

I've known for years that $\displaystyle 7 \times 11 \times 13 \:=\:1001$ **

. . hence: $\displaystyle 111,111\:=\:111 \times 1001 \:=\:111\times 7 \times 11 \times 13$

I'm embarrassed that none of this occured to me.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

More trivia . . .

. . $\displaystyle 10,001 \;=\;73\cdot137$
. .$\displaystyle 100,001 \;= \;11\cdot9091$
.$\displaystyle 1,000,001\;=\;101\cdot9901$
$\displaystyle 10,000,001\;=\;11\cdot909091$

. . . .$\displaystyle 111\;=\;3\cdot37$
. . .$\displaystyle 1,111 \;=\;11\cdot101$
. . $\displaystyle 11,111\;=\;41\cdot271$
. .$\displaystyle 111,111\;=\;111\cdot1001\;=\;3\cdot7\cdot11\cdot13 \cdot37$
.$\displaystyle 1,111,111\;=\;239\cdot4649$
$\displaystyle 11,111,111\;=\;10001\cdot1111\;=\;11\cdot73\cdot10 1\cdot137$

It seems that 1001,10001,.... are all composite. Can it be proved that $\displaystyle 10^k + 1(k\geq4)$ is composite?
Can it be proved that 111,1111,.......$\displaystyle \frac{10^k - 1}{9}(k\geq1)$is composite?

13. Originally Posted by malaygoel
It seems that 1001,10001,.... are all composite. Can it be proved that $\displaystyle 10^k + 1(k\geq4)$ is composite?
Can it be proved that 111,1111,.......$\displaystyle \frac{10^k - 1}{9}(k\geq1)$is composite?
No, because it is not true.

These numbers are called "repunits" we can however show if a repunit is prime then the number of ones must also be prime.

14. Hello, ThePerfectHacker!

Good call!

These numbers are called "repunits".
We can however show: if a repunit is prime, then the number of ones must also be prime.

"Repunits" is an abbreviation for repeated units.

Even if the numbers of 1's is a prime number, there is no dependable rule.

Let $\displaystyle 1^{(n)}$ represent a number composed of n 1's.

$\displaystyle 1^{(3)} \;= (3)(37)$
$\displaystyle 1^{(5)} \;= (41)(271)$
$\displaystyle 1^{(7)} \;= (239)(4649)$
$\displaystyle 1^{(11)} = (21,\!649)(313,\!239)$
$\displaystyle 1^{(13)} = (53)(79)(265,\!371,\!653)$
$\displaystyle 1^{(17)} = (2,\!071,\!723)(5,\!363,\!222,\!857)$
$\displaystyle 1^{(19)} = prime$
$\displaystyle 1^{(23)} = prime$
$\displaystyle 1^{(29)} = (3191)(16,\!763)(43,\!037)(62,\!003)(77,\!843,\!83 9,\!397)$
$\displaystyle 1^{(31)} = composite \text{ (divisible by }2791)$

15. Originally Posted by ThePerfectHacker
No, because it is not true.

These numbers are called "repunits" we can however show if a repunit is prime then the number of ones must also be prime.
I would be happy if you post here the proof of the statement in the quoted work.But the question still remains:Is there any criteria by which we can say that "$\displaystyle \frac{10^k - 1}{9}$ is prime or composite depends on the value of k"?
Similarly, what can you say about the numbers $\displaystyle 10^k + 1(k\geq4)$.
Keep Smiling
Malay

Page 1 of 2 12 Last