A Recurring Decimal Problem

[@leofun01]

First time seeing this notation. Thanks for the explanation how it should be used.

[@nathanisbored]

I’ve actually never seen the two dots notation before this video

[@Drayiss]

Yo heya there Nathan! Didn’t know you’d be here! :0

[@lollol-tt3fx]

neither have i, bucko (im cool beacuse i said bucko by the way, or as the kids are saying it these days, btw 😎😎😎😎), stay geeking

[@aoay]

In New Zealand I was taught the dot notation. I started to notice the overline notation online in the last decade or so.

[@ronbannon]

Notations may vary, and that's why it is essential to clarify symbols that are not seen elsewhere. The bar is perhaps easier to write on a blackboard, which may be why it is used more often. I have noticed some notations used result from the typesetter's inability to type certain symbols, especially in old textbooks. LaTeX has made it easy to type loads of math, but pre-LaTeX, a lot of math needed to typeset using what they had. We've come a long way; nowadays, just about anyone can create high-quality math using LaTeX. One day, KZbin will allow us to comment using LaTeX.

[@honeybee9455]

Thought this was fancy lambda calculus from the thumbnail. I think i have officially gone crazy.

[@ronbannon]

Nice problem. I still need to look over your work, but I did work the problem out and I was surprised how it proceeded. Thanks for sharing.

[@ngc-fo5te]

I have seen the two dots used this way in an introductory number theory course at uni in the 1950s. The bar notation was more common even back then if I recall correctly. Edit: I just checked and probably the most famous Number Theory text of all by Hardy and Wright uses the two dots notation.

[@mrphlip]

When I solved this, I wasn't expecting a=0 to be an option (not used to leading zeros being allowed in these sorts of problems) but it's not hard to retrofit it into my solution... Much like in the video, I converted the recurring decimals to fractions, but earlier in the process, and in a slightly different way... for instance, the LHS of [ab.c̄ab̄] is just 100*[0.ābc̄] = 100*[abc]/999 = [abc00]/999. Similarly for the other fractions. This gives us: [abc00]/999 = [pqp0]/999 + [q0]/9 Multiplying through by 999, and then also dividing through by 10, gives us: [abc0] = [pqp] + [qqq] which is now a nice equation over integers, much easier to work with. The units digit gives us p + q = 0 (mod 10), the only options, are 0 or 10. p + q = 0 quickly gives us p = q = 0, then the entire RHS is 0, so the entire LHS is 0, all the digits are 0, hooray. p + q = 10 means that the p+q in the hundreds place will also carry and give us a = 1 if q < 5 then the addition in the tens digit won't carry, and we have b = 0, c = 2q + 1 if q >= 5 then the addition in the tens digit will carry, and we have b = 1, c = 2q - 9 All of the possible values for q (except zero) here give solutions. This gives us our full set of solutions: 00.000000... = 0.000000... + 0.000... 10.310310... = 9.199199... + 1.111... 10.510510... = 8.288288... + 2.222... 10.710710... = 7.377377... + 3.333... 10.910910... = 6.466466... + 4.444... 11.111111... = 5.555555... + 5.555... 11.311311... = 4.644644... + 6.666... 11.511511... = 3.733733... + 7.777... 11.711711... = 2.822822... + 8.888... 11.911911... = 1.911911... + 9.999... Of these, only two satisfy the (not in force here, but usually implied for these sorts of problems) additional restriction that each letter represents a _different_ digit: 10.510510... = 8.288288... + 2.222... 10.910910... = 6.466466... + 4.444...

[@aoay]

I made the same 'mistake' of not expecting that valid solutions could incorporate leading zeros.

[@RGP_Maths]

@mrphlip Yes, I supposed, based on familiar letters-for-digits puzzles, that each letter had to represent a different digit. That eliminates a lot of possibilities from the outset, and especially if leading zeroes are forbidden, so that a can only be 1. This leads pretty quickly from the thumbnail to the two solutions you highlighted, so I watched the video to find out how it could possibly take 18.5 minutes!

[@barutjeh]

In the a=1 case, p=q=0 isn't a trivial solution, since the left hand side is at least 10. It's just plain impossible.

[@staircase72]

It seems to be easier to just expand it out and treat the entire expression algebraically as long as possible rather than starting making restrictions early. If you do you get 10a+b+(100c+10a+b)/999 = p + (100q+10p+p)/999+q+q/9 We then multiply through by 999 and combine like terms to get 10000a+1000b+100c = 1010p+1210q We can clearly see that everything is a multiple of 10 so divide by 10 to get 1000a+100b+10c = 101p+121q The left hand side again is a multiple of 10 so the right hand side must be to therefore p + q = 0 mod 10 As p and q are digits, 0

[@connerpartaker5149]

I did the same thing and started going crazy when he immediately broke into cases, thank you for beating me to the punch here!

[@barutjeh]

Your suggestion of proving p+q=10 works well actually! From the integer part, we know 10a+b = p+q+d, where d is either 0 or 1. From the last two digits of the recurring digita, we get that 10a+b ≡ 11(p+q) + e (mod 100), where e is another carry, so 0 or 1. Combining the two, you get 10(p+q) ≡ d-e (mod 100). Since d and e are either 0 or 1 and the left hand side is a multiple of 10, d-e=0, so 10(p+q) ≡ 0 (mod 100), therefore p+q is a multiple of 10. Since p and q are 9 or less, either p+q=0 (leading to the trivial solution) or p+q=10 (leading to the other solutions).

[@tmowlee]

You could simplify the problem straight off the bat by dividing everything by 10 to get a.b'cab' = .p'qpp' + .q'q' . Here, single quotes are the repeating part. then noting that the rightmost non-repeating digit is the same as the right most repeating digit in each term, you can write this as: a.'bca' = .'pqp' + .'q'

[@NoNameAtAll2]

what a wonderful notation much easier to write than the standard "line all the way over" I feel it's gonna be ignored just as much as "three vertical dots" as a standin for "divisible by" a|b swaping sides to mean "b/a is integer" is straight up idiotic xD

[@gunhasirac]

I think if you start you analysis by splitting case using integer part and second decimal place, you can reach the finer cases fast which matches up with your suspicion. This is kind of backward engineering so i think your approach looks more natural. Integer: b = p+q, p+q+1 Second decimal place: a = p+q, p+q+1 (mod 10) Second equation is especially useful because there are only two possibilities of a.

[@michaelslack8900]

Fun problem, but I occasionally was getting confused by occurrences of 9, a, p and q, because writing on whiteboards is hard

[@vetbaitednv]

what dry erase marker is that?

[@holyshit922]

Ths recurring decimals can be treated as sum of infinite geometric sequence

[@iMíccoli]

Never seen that notation.

[@bscutajar]

8:45 but this is wrong, p+q could be 0 mod 10

[@peterstromberg9614]

I have some problems with that video. First, using p and q, while handling with a lot of 9s is a bit confusing and not appealing at all. Second, the dot notation is not explained enough. Why not using the bar notation for all? Isn't it two notations that do the same thing in this video? It should be clear (or stated) that we are working with a digit representation. Third, it's really hard to write a comments using both (or even one the) notations above. :-)

[@stenzenneznets]

Acab