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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.

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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)

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  • $\begingroup$ @jsi. Although this is nice simplification of Japp’s original suggestion, what I am looking for is the numerical code which used repeatedly to generate the rest of the sequence. Current solution involves generating the first five digits presumably by straightforward division and then applying the algorithm. $\endgroup$ – Uvc May 11 at 10:49
  • $\begingroup$ @Uvc Still not sure what you want, but I came up with a construction method where the starting numeric code is "1". $\endgroup$ – JS1 May 12 at 11:02
  • $\begingroup$ @jsi As I stated, you have to find the code number which used repeatedly generates the next number in the sequence in a very simple way. Beauty of Reptand primes is for some of them , sequence can be written immediately without ever seeing the sequence before..follow me for further clues. $\endgroup$ – Uvc May 12 at 11:08
  • $\begingroup$ @Uvc You must be looking for some specific method. The leapfrog division method in my latest edit is so simple I can do it in my head, but I take it that the one you have in mind is just as simple or even more so. $\endgroup$ – JS1 May 12 at 16:35
  • $\begingroup$ @jsl yes..also your method needs the first 5 digits be computed first and apply leapfrog technique..mine involves only knowing the reptand number, initial digit is given by the rule for all reptand s..then simple operation to be performed for determining digits using the simple numeric code...all the 29 digits can be derived...in conjunction with 9 compliment rule..rest of the digits can be written right. Away..my method involves no initial computation of the seed number and much faster. $\endgroup$ – Uvc May 12 at 19:44
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The numbers $59$ and $61$ are a pair of twin primes ("in prime shape. I have twin like older brother"). $$\frac{1}{59} = 0.0169491525423728813559322033898305084745762711864406779661016949...$$ 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.

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  • $\begingroup$ Answer should include the digital code and the method you employed to generate the whole sequence..that is the key part. I can recall all the digits anywhere even without a calculator. $\endgroup$ – Uvc May 8 at 14:01
  • $\begingroup$ I just edited my answer with another method which does use the special cyclic property, but it seems computationally harder than the direct reciprocal method ... not sure exactly what @Uvc is looking for. $\endgroup$ – Rand al'Thor May 10 at 13:40
  • $\begingroup$ Sorry for my late comment..just came from a long 6 mile walk...as mathematicians i knew most of you will get this far..I will edit my question by adding. “Very simple”..I have already given the first clue today..till the puzzle is solved I will edit to add more clues 1 per day $\endgroup$ – Uvc May 10 at 15:05
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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 $p$:

$10^2\div59=1.(\dots)$,
$10^3\div59=16.(\dots)$,
$10^4\div59=169.(\dots)$,
$10^5\div59=1,694.(\dots)$,
$10^6\div59=16,949.(\dots)$,
$10^7\div59=169,491.(\dots)$,
$10^8\div59=1,694,915.(\dots)$,
$10^9\div59=16,949,152.(\dots)$,
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 $99$.

$59\times61\equiv99\text{ (mod }10^2)$
$59\times661\equiv999\text{ (mod }10^3)$
$59\times9,661\equiv9,999\text{ (mod }10^4)$
$59\times79,661\equiv99,999\text{ (mod }10^5)$
$59\times779,661\equiv999,999\text{ (mod }10^6)$
$59\times6,779,661\equiv9,999,999\text{ (mod }10^7)$
$59\times06,779,661\equiv99,999,999\text{ (mod }10^8)$
$59\times406,779,661\equiv999,999,999\text{ (mod }10^9)$
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 $7, 17, 19, 23, 29, 47, 59, 61, 97$.

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  • $\begingroup$ As nicely explained as this is, it does seem to boil down to "find the reciprocal of p", which as you've shown is trivial and the one used to find any fraction (given a pen and paper, I'd hate to hold that in my head I admit). I felt Uvc's comment below to state the "method employed" implied some mnemonic to remember the digits. $\endgroup$ – Fifth_H0r5eman May 10 at 13:22
  • $\begingroup$ @Fifth_H0r5eman I racked my brains for a neater method ... FWIW, when I found the Wikipedia page for cyclic numbers it described the same construction. Maybe, maybe, we can somehow use modular arithmetic, but I'm not seeing it. $\endgroup$ – Rand al'Thor May 10 at 13:25
  • $\begingroup$ Oh I'm not criticising the method, just that I felt it was the standard way to calculate any of this form so was almost redundant. Finding the actual number wasn't such a problem, finding a way to memorise it without tedious calculation however seems the greater feat. There's a fairly high proportion of double digits, but even then given the length of the number, short of splitting it into segments and base changes, I can't see a simple way... $\endgroup$ – Fifth_H0r5eman May 10 at 13:32
  • $\begingroup$ @Fifth_H0r5eman I found another method which does use the special property of $p$, but it seems computationally harder than the first method. $\endgroup$ – Rand al'Thor May 10 at 13:39
  • $\begingroup$ Oh dear, that really is an unpleasant algorithm to look at. It's really just the inverse of the first, but due to the old left-to-right bias the multiplication instead of division doesn't manage to make it look very friendly! (Especially the 10^8). $\endgroup$ – Fifth_H0r5eman May 10 at 13:45
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Surely not the intended answer

Jaap has of course identified the intended sequence. It seems to me that all the algorithms described so far are strictly more complicated than the following, which is simply another way of formulating the division operation and seems entirely straightforward to do in one's head. However, it can't be the intended answer because supposedly the process is much simpler for this number than for its "twin"'s. (Which the process described below isn't.)

You need to keep track of a single number between 0 and 59 inclusive, and to be able to multiply and divide small numbers. Initially, your number is 10. Now repeat the following process. 1. Multiply the number by 10. 2. Divide it by 59. 3a. Next digit is the quotient. 3b. New value of the number is the remainder. 4. Repeat these steps until the number reaches 1.

I have trouble understanding how any other process could actually be simpler than this. (But perhaps I'll be surprised!)

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  • $\begingroup$ You sure will be surprised. $\endgroup$ – Uvc May 11 at 2:09
  • $\begingroup$ Or perhaps we have different ideas about what's easier than what, of course :-). $\endgroup$ – Gareth McCaughan May 11 at 2:16
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For reptend prime 59, starting digit has to be 1. For 59, code is 6. Operation is multiplication with carry over. Also, you have to start from the right. You will get exact reversed part of the sequence and multiplication is much easier

The hints for all these words were dropped profusely in all the hints. Right is the most frequently used word.

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  • 1
    $\begingroup$ If you're going to self-answer and accept it, you might want to provide a more explicit explanation of the process you're describing. Having read this answer, I have no idea how to apply it to find the expected sequence. (Compare the other answers, which apparently weren't simple enough, but for all their extra complexity are entirely understandable in terms of how to use them to generate the relevant sequence.) $\endgroup$ – Rubio May 21 at 20:05
  • $\begingroup$ Step by step explanation....See@ JSI’s explanation..selection of initial digit 1 and the code..6 are very similar. My sequence is right to left and when fully generated, it will exactly mirror reciprocal period unit. $\endgroup$ – Uvc May 21 at 21:22
  • $\begingroup$ Method..Multiplication from left with carryover as shown: $\endgroup$ – Uvc May 21 at 21:25
  • $\begingroup$ 1. 16...16.36..keep 6 carry 3 to next step..166..6x6+3..keep 9..carry 3...1669..9x6+3..keep 7, carry 5....16697..6x7+5..keep 7 carry 5..now we have generated 6 digits..166977..keep on repeating till you come to middle digit..from then on fillin with compliment to finish the 58 period unit for 59. For 19, key is 2 and the process is even more easy. $\endgroup$ – Uvc May 21 at 21:37
  • $\begingroup$ You should put your explanation in your answer - the answer feels incomplete without it (and comments are ephemeral). In any case, your intended approach is at best slightly simpler (I'm not sure I agree that tracking carries from "multiplication is much easier" than division and remainders), so I'm of the opinion that JS1's answer was an equally valid solution that could extract the solution digits in one's head, using one holdover digit and one operation per digit extracted (that is, the same "complexity" as your solution). You're free to accept as you like, but I'd have given JS1 the nod. $\endgroup$ – Rubio May 23 at 16:17

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