# Math Help - 26th digit

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
$10^6 \equiv 1 \mod 13$
so $10^{6k} \equiv 1 \mod 13$
so $a10^{6k} \equiv a \mod 13$
also
$10^3 \equiv -1 \mod 13$
then combining this with our other idendity
$10^{6k+3} \equiv -1 \mod 13$
so $a10^{6k+3} \equiv -a \mod 13$

so our number is
$N = 11,111,111,111,111,111,111,111,11k$ $,111,111,111,111,111,111,111,111$

which we can write as..
$N = 11*10^{48} + 111*10^{45} + ... +$ $11k*10^{24} + ... + 111*10^3 + 111$

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

$N = 11-111+111-111+111-111+111-111+11k$ $-111+111-111+111-111+111-111+111 \mod 13$

which simplifies to
$N \equiv 11k - 100 \mod 13$
so if $N|13$ then $11k-100 \equiv 0 \mod 13$
the only value of k where this is true is $k=3$
thus this is the answer. $\mbox{the 26th digit is 3.}$

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

so our number is
$N = 11,111,111,111,111,111,111,111,11k$ $,111,111,111,111,111,111,111,111$

which we can write as..
$N = 11*10^{48} + 111*10^{45} + ... +$ $11k*10^{24} + ... + 111*10^3 + 111$

using our idendities from before, we know that for instance
$11*10^{48} \equiv 11 \mod 13$
since 48 = 7*6 so
48 = 6*8 not 7*6
never mind.
Are there any more methods?

4. Originally Posted by malaygoel
48 = 6*8 not 7*6
never mind.
Are there any more methods?
my finger must have slipped.

this is the only way i know how to do this problem.

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.

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,
$N=11,111,111,111,111,111,111,111,11k,$ $111,111,111,111,111,111,111,111,$
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 $1k$ and hence the answer 3.

11. Excellent solution, Malay!

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

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

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

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

**

More trivia . . .

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

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

12. Originally Posted by Soroban
Excellent solution, Malay!

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

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

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

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

**

More trivia . . .

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

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

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

13. Originally Posted by malaygoel
It seems that 1001,10001,.... are all composite. Can it be proved that $10^k + 1(k\geq4)$ is composite?
Can it be proved that 111,1111,....... $\frac{10^k - 1}{9}(k\geq1)$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 $1^{(n)}$ represent a number composed of n 1's.

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

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 " $\frac{10^k - 1}{9}$ is prime or composite depends on the value of k"?
Similarly, what can you say about the numbers $10^k + 1(k\geq4)$.
Keep Smiling
Malay

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