Yep, can confirm, such system existed in other large Soviet cites as well. I also graduated from one such maths and physics oriented school. Compared to regular public schools the education level was much higher, and not only in those subjects. The studying there was extremely intense, 6-day week (which was standard in USSR at that time), long hours, and tons of homework to keep you busy late at nights and weekends (read: Sunday). Best graduates were expected to go further study in MSU or other top universities. The rest of us who went to local universities found themselves bored in the first years of their studies.

Now, living in a western European country I found in a hard way that the key to a better education for your talented kid hides in your bank account.

Don't get me wrong, I in no way support Soviet system, but this was one of the things they did good.

Well, 2 is prime, and so is 57, so it makes sense those would be math focused schools. ;) [0]

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[0]: https://hsm.stackexchange.com/questions/6358/story-of-grothe...

> And indeed 57 looks a bit prime for some psychological reason

It tricked me, until I read the link.

No one taught me this in school, but if you add up the digits (5+7=12) and the result is divisible by 3, then so is the original number. It works recursively.

My maths teacher made me prove that, my first proof by induction. Try it. You know you want to procrastinate with a nice proof by induction now.

Induction seems like an odd way to prove this. Do you need to preform induction over all the numbers that are not multiples of three?

The converse claim (every multiple of 3 has a digit sum that is a multiple of 3) is a more natural one for induction, though that's not the most standard proof there either.

If induction is the only tool you have then it’s ok. What you really want is modular arithmetic. If you don’t have modular arithmetic then pricing it directly is hard:

  n = Sum(d_k 10^k)
  Let S = sum of digits = Sum(d_k)
  n - S = Sum(d_k (10^k - 1))
  But (10^k - 1) = ((9 + 1)^k - 1) = ((1 + k*9 + (2 choose k)*9^2 + ... + 9^k) - 1)
  by binomial expansion so is divisible by 9.
  Therefore n - S is divisible by 9, so n is divisible by 9 iff S is.
  The same is true when you replace divisibility by 9 with divisibility by 3.

I remember being asked this in an interview for a place at university and I solved it by induction (which was all I had) and was then shown this direct proof.

Odd if you're looking for the simplest way to prove that. Less odd, perhaps, if you're looking for something that can be proved simply using induction, in order to make students do their first proof by induction.

I don't remember precisely what it was we were asked to prove. It may have been the converse.

Yes, a simple consequence of e.g. 3-digit number being:

100x + 10y + z modulo 3

Removing 9y and 99x gives equivalence: x + y + z modulo 3

Now what’s left is induction with proper base (1-digit).

As dan-robertson mentioned in another branch, you don't even need induction - you can sidestep it by writing X = sum_i x_i 10^i and noticing that all the 10^i are 1 modulo 9 (and therefore modulo 3).

"What is Mathematics" by Courant and Robbins teaches some more divisibility tricks, might be useful sometimes. btw, we learned divisibility by 3 in 5th or 6th grade (not the proof, of course)

Works for 9 as well (or any b such that 10 mod b = 1 .. so 3 and 9).

> and I didn't go because was too lazy to commute 40mins one way when my own school was a 3-minute walk

Hah, that mirrors my wife’s experience in Riga - she came second and first in national physics competitions for two years, and was twice invited to enroll in a different school - but she stayed put, because she didn’t want to have to change trams twice, rather than a direct trolley bus. She’s happy with her choice.