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.
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$$\displaystyle -111+111-111+111-111+111-111+111 \mod 13$
which simplifies to
$\displaystyle N \equiv 11k - 100 \mod 13$
so if $\displaystyle N|13$ then $\displaystyle 11k-100 \equiv 0 \mod 13$
the only value of k where this is true is $\displaystyle k=3 $
thus this is the answer. $\displaystyle \mbox{the 26th digit is 3.}$
What I wanted to say(although I have not tried) is that we could substitute the 26th digit as 1 and perform divison(which will be easy due to repeating 1's) and then find find the value of k by guessing the remainder that the 26th digit should give while performing division.Originally Posted by malaygoel
I think I have found it.Originally Posted by malaygoel
By performing simple division, I found that 111111 is divisible by 13
Now,
$\displaystyle N=11,111,111,111,111,111,111,111,11k,$$\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.
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.
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More trivia . . .
. . $\displaystyle 10,001 \;=\;73\cdot137$
. .$\displaystyle 100,001 \;= \;11\cdot9091$
.$\displaystyle 1,000,001\;=\;101\cdot9901$
$\displaystyle 10,000,001\;=\;11\cdot909091$
. . . .$\displaystyle 111\;=\;3\cdot37$
. . .$\displaystyle 1,111 \;=\;11\cdot101$
. . $\displaystyle 11,111\;=\;41\cdot271$
. .$\displaystyle 111,111\;=\;111\cdot1001\;=\;3\cdot7\cdot11\cdot13 \cdot37$
.$\displaystyle 1,111,111\;=\;239\cdot4649$
$\displaystyle 11,111,111\;=\;10001\cdot1111\;=\;11\cdot73\cdot10 1\cdot137$
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.
$\displaystyle 1^{(3)} \;= (3)(37)$
$\displaystyle 1^{(5)} \;= (41)(271)$
$\displaystyle 1^{(7)} \;= (239)(4649)$
$\displaystyle 1^{(11)} = (21,\!649)(313,\!239)$
$\displaystyle 1^{(13)} = (53)(79)(265,\!371,\!653)$
$\displaystyle 1^{(17)} = (2,\!071,\!723)(5,\!363,\!222,\!857)$
$\displaystyle 1^{(19)} = prime$
$\displaystyle 1^{(23)} = prime$
$\displaystyle 1^{(29)} = (3191)(16,\!763)(43,\!037)(62,\!003)(77,\!843,\!83 9,\!397)$
$\displaystyle 1^{(31)} = composite \text{ (divisible by }2791)$
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"?Originally Posted by ThePerfectHacker
Similarly, what can you say about the numbers $\displaystyle 10^k + 1(k\geq4)$.
Keep Smiling
Malay