Anyone willing to take a stab at the high school math exam, Russian style?

[Al B. Sure!]

Sorry, I mistook you for this guy:

I was under the impression that n was not a second variable but represented a constant like how 'e' is a constant.

[Al B. Sure!]

And no, gribbler.

If presented with y superscript x+1 = 352 I have no idea what to do with that. Something with logs. That's what I think but I don't remember how to use logs. Did I know how to convert between exponential and logarithmic forms in the past? Yes. Do I know how to do that now? **** no.

[giblets]

I was under the impression that n was not a second variable but represented a constant like how 'e' is a constant.

If I had that problem I would have said "is 'n' a variable or a constant?" There would have been no reason to bring up superscripted stuff. Either you've failed to communicate clearly yet again, or you're lying.

[giblets]

And no, gribbler.

If presented with y superscript x+1 = 352 I have no idea what to do with that. Something with logs. That's what I think but I don't remember how to use logs. Did I know how to convert between exponential and logarithmic forms in the past? Yes. Do I know how to do that now? **** no.

That is so easy

[Kuciwalker]

n^(k+1) - n! = 5*(30k + 11)

First observation is that there's a factor of n on the LHS, so:

n*(n^k - (n-1)!) = 5*(30k + 11)

This is hint #1 that n is divisible by 5. Let's try setting n=5 and see if a solution emerges:

5^k - 24 = 30k + 11

Eyeballing this gives k = 3.

So (5, 3) is a solution. Are there any others?

Well, if n = c*5 (for natural c) then we've got:
c*(n^k - (n-1)!) = 30k + 11

c can't be even, since 30k + 11 is odd, so c is at least 3. But 15^k - 14! = 30k + 11 looks like it has no solutions, and a bigger c makes this even more difficult. So for n divisible by 5 it looks like (5, 3) is the only solution.

If n isn't divisible by 5 then it must be divisible by a factor of 30k+11. But out of 41 71 101 131 161 the first non-prime is 161 for k = 5 and n = 7 or n = 23. 23 is too big (LHS goes negative) and 7 makes LHS way too large. We're just going to have worse luck with larger k so it looks like (5, 3) is probably the only solution.

[Kuciwalker]

(If I were actually taking the exam I would bother to prove that n is divisible by 5.)

[Al B. Sure!]

This...

n^(k+1) - n! ==> n*(n^k - (n-1)!)

is something I would not have thought to do. It makes sense but I would not have thought of it.

[Al B. Sure!]

What is 'natural c'? You're referring to any other natural (as indicated by the problem) number?

[Kuciwalker]

"Well, if n = c*5 (for natural c)"

What this means is that, given the assumption that n is divisible by 5, I can express n as the product of 5 and some other natural number, which I denote as 'c'.

[Jaguar]

It took me about 10 minutes. n is divisible by 5, but cannot be bigger than 10 and cannot be even.

How did you prove the second of those? (n<10). I observed the other two things you did, and simply guessed 5, but I had no ironclad reason for doing so.

[Al B. Sure!]

"Well, if n = c*5 (for natural c)"

What this means is that, given the assumption that n is divisible by 5, I can express n as the product of 5 and some other natural number, which I denote as 'c'.

I understand. Or it could be n= c*(30K+11) but you explain that away subsequently. That initial transformation of the left hand side to n*(n^k - (n-1)!) makes it a lot easier to understand.


I take away my comments on your inability to teach what you know.

[KrazyHorse]

How did you prove the second of those? (n<10). I observed the other two things you did, and simply guessed 5, but I had no ironclad reason for doing so.

Because the RHS can never be divisible by 25, but for all n multiples of 5 where n > 10 the LHS is divisible by 25

[KrazyHorse]

The last question was the only moderately difficult problem (to actually get the proof; my initial guess that 5,3 was unique took under 30 seconds). In total, the test took me ~30 mins

[KrazyHorse]

A few of the question were at grade 7/grade 8 canadian math competition levels. The last one was either at grade 7 (no proof required) or ~grade 10-11 (proof required). The rest were too trivial to appear on any math competition I've ever taken.

EDIT: I'm aware this is a general high school exam, not a competition. It's still not particularly surprising that any mathematically-inclined 18 year-old could easily get 100% on this with some preparation.

[KrazyHorse]

It was the same way when I took it. Also when I took the GRE's they had two math sections, fap fap fap (I believe they've now replaced one with a "reading" section)

One math one "logic" section. They replaced the logic with "critical writing" or some such nonsense based on some subjective score from 1 to 6.

800Q 800L 760V

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