Otdfbt

Find the digital key and very simple method to unlock the long digital sequence 16xxxxxx61

Additional clues: Eeryday one clue will be given till problem is solved

Day 1: Bounty

Day 2: My code is much simpler than my twin and my dad’s

Day 3: Our dad’s dear departed siblings a decade older had 3 digit codes.

Day 4: I am fuming mad.I have to get this right out of my chest before I can think with cool head and give you right hints. Close by Cryptic Casino closed their doors forever for us right on the opening night. We are lodging complaint to their madam manager and her assistant Adam. If they don’t reverse their decision, I think we have the right to appeal to the Carlson Casino Commission. I cannot think right now..see you in 12 hrs or so.

Day 4-5: I thought over day 4 post and I have given tons of clues inadvertently. Look at my recent puzzle posts for further clues.Reread the puzzle..none of the answers are equal to my sequence. See you in 12 hrs

Day 5: Divide and conquer is not always the best strategy for all occasions

Day 6: Some of my younger siblings have ridiculously easy codes.

Day 7: None of the answers given upto now are not yet right. Did you have a chance to visit Cryptic and Jack’s Casinos yet. You can have some fun and get some valuable clues too.

Day 8: Last day for any further clues...We have come to the end of the clue road. Since I haven’t seen the right sequence yet, I will leave you with parting words from wise grand dad whom I was very fond of...

Be bold and just don’t go on the beaten path. Make the right decision...take the right path to get the right result. Good luck in your further endeavors!!!

X s shown are for illustrative for missing digits and there are lot more x s than shown. All of the missing digits can be..0,1,2,3,4,5,6,7,8,9 and each of them occur at least twice in the series.

Solution should show the digital code number and the specific method used to construct the sequence using the following clues.

I am middle aged bike rider still in prime shape. I have twin like older brother who is couple of years older than me. We both are cryptographers who love to play with fractions and our digital signatures consisting of long sequences that are closely tied to our age and a tad shorter than our ages. Since my signature is closely tied to the reciprocal of my age, I can recall the entire sequence mentally using the key and simple method.

We are special family of 9 who are in prime shape and none of us have hit the century mark yet.

The numbers $59$ and $61$ are a pair of twin primes ("in prime shape. I have twin like older brother").
$$frac159 = 0.0169491525423728813559322033898305084745762711864406779661016949...$$
This reciprocal ("reciprocal of my age") has a repeating section of $58$ digits ("tad shorter than our ages"), which apart from the leading zero starts with $16$ and ends with $61$. A number $p$ whose reciprocal has a period of $p-1$ is necessarily prime, and called a full reptend prime. The repeated sequence of digits of length $p-1$ is a cyclic number ("bike rider"). There are 9 full reptend primes ("family of 9") under $100$ ("none of us have hit the century mark yet"), namely $7$, $17$, $19$, $23$, $29$, $47$, $59$, $61$, and $97$.

So the answer is

$169491525423728813559322033898305084745762711864406779661$

I am not sure I have found the intended "key" and generating method, but one possible method is the following.

Presumably it has to depend on the fact that the sequence of digits is a cyclic number. This means that if you multiply it by any number between $1$ and $p-1$ you get the same digits, except cyclically rotated. It happens that multiplying by 9 gives you the same sequence starting 5 digits further on. So what you could do is memorise a key of the first five digits, and calculate the rest. Multiplying by 9 is easily done by multiplying by 10 and subtracting once.

The method is then as follows:

Write the key $16949$ twice, the bottom one shifted one digit to the right.

 16949
 16949
 ------ -
 1525... 

You subtract from left to right, and then append the digits you found to continue the subtraction:

 169491525
 169491525
 ---------- -
 152542372... 

Doing the subtraction left to right is a bit unusual but not too hard, as you only need to look ahead one or two digits to check for carries. With a little practice you could write down the answer directly, just once instead of three times like I do here.

Another useful property is that the second half of a cyclic number is the nines complement of the first half (i.e. the two halves add to $999...9$), so once you reach the $983$ you can just recite the complement digits.

Edit:

Quick explanation, mostly covered by Jaap Scherphuis' answer already:


The family of 9 are referring to the full reptend primes under 100, namely $7, 17, 19, 23, 29, 47, 59, 61, 97$, of which the "brothers" are $59$ and $61$, and the "father" is $97$.


The sequence 16xxxx61 is tied to the reciprocal of $1/59$, which has a 58 digit repeating decimal sequence:

$0169491525423728813559322033898305084745762711864406779661$


My construction method:


I found a construction method that doesn't require starting with any particular (long) number. It uses a division trick called Leapfrog division to compute the value of $1/59$:


Start with $1$ ("the digital code"). At each step we will divide the current number by $6$ to get a quotient and a remainder. In step one we have:

$1 / 6$ = $0,1$ (where the 1st value is the quotient, and the 2nd value is the remainder)


Now take the remainder, $1$ and append the quotient $0$ to get the next number to divide: $10$:

$10 / 6$ = $1,4$
$41 / 6$ = $6,5$
$56 / 6$ = $9,2$
$29 / 6$ = $4,5$
$54 / 6$ = $9,0$
$09 / 6$ = $1,3$
$31 / 6$ = $5,1$
$15 / 6$ = $2,3$
$32 / 6$ = $5,2$
$25 / 6$ = $4,1$

etc.


The quotient values form the decimal expansion of $1/59$:
$0.01694915254...$ which is the sequence we are looking for minus the leading zero.


The reason this works is because we are trying to compute $1/59$, and since $59$ ends in a $9$, we can use leapfrog division using $6$ as the divisor instead of $59$.

(Old answer removed, apparently it was confusing the issue)

Jaap Scherpuis found the answer, but the OP wants more explanation and a simple construction...

I am middle aged bike rider still in prime shape.

We're looking for a prime number $p$, somewhere around the 40s or 50s maybe.

I have twin like older brother who is couple of years older than me.

It's the lower of a pair of twin primes $p,p+2$.

We both are cryptographers who love to play with fractions and our digital signatures consisting of long sequences that are closely tied to our age and a tad shorter than our ages.

Something about fractions gives a long sequence of digits related to $p$, a sequence which is just a little shorter than length $p$. That suggests we're looking for a full reptend prime, whose reciprocal $1/p$ has a cyclic decimal expansion consisting of a cyclic number repeated.

Since my signature is closely tied to the reciprocal of my age, I can recall the entire sequence mentally using the key and simple method.

This is the key part (pun intended). Since the signature (cyclic number) times $p$ equals $99999...999$, we can construct each digit of the signature in turn.

Starting from the front of the number ($16...$), we would proceed as follows:

Divide successive powers of $10$ by $p$:

$10^2div59=1.(dots)$,
$10^3div59=16.(dots)$,
$10^4div59=169.(dots)$,
$10^5div59=1,694.(dots)$,
$10^6div59=16,949.(dots)$,
$10^7div59=169,491.(dots)$,
$10^8div59=1,694,915.(dots)$,
$10^9div59=16,949,152.(dots)$,

etc.

Starting from the end of the number ($...61$), we would proceed as follows:

Find the reciprocal of $59$ modulo each power of $10$ and multiply it by $99$.

$59times61equiv99text (mod 10^2)$
$59times661equiv999text (mod 10^3)$
$59times9,661equiv9,999text (mod 10^4)$
$59times79,661equiv99,999text (mod 10^5)$
$59times779,661equiv999,999text (mod 10^6)$
$59times6,779,661equiv9,999,999text (mod 10^7)$
$59times06,779,661equiv99,999,999text (mod 10^8)$
$59times406,779,661equiv999,999,999text (mod 10^9)$

etc.

We are special family of 9 who are in prime shape and none of us have hit the century mark yet.

There are 9 full reptend primes under 100: namely $7, 17, 19, 23, 29, 47, 59, 61, 97$.