A very interesting Putnam problem

A casual yet satisfying Putnam problem.

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#math #putnam #competition #numbertheory

[mcalkis]

Knowing more about modular arithmetic would make this a more fun watch. It'd be cool if you made some videos on it.

[temptate]

Please do more Putnam videos! I would like to use them to study for the Putnam myself thank you!

[JanAgro]

I'm not sure how you justify a^(n +1) = 1 mod 3 leading to a = 1 mod 3 automatically, if n odd, this gives the possibility that a = 2 mod 3 leads to a^(n +1) = 1 mod 3, later on you say n even, but you should establish that before automatically saying a^(n +1) = 1 mod 3 leads to a = 1 mod3

[joseluishablutzelaceijas]

Thank you for the nice problem. My solution was a bit different, maybe worth sharing here: From the equation one can deduce that the LHS is strictly increasing, which can be e.g. derived by subtracting two consecutive LHSs with n and n+1 and noting that this difference cannot be less than one, and that a cannot be larger than 45, which can be concluded by noting that 45^2 = 2025 > 2001 and thus trying 45^2 - 46 = 1979 and 46^2 - 47 = 2069 > 2001 (a > 46 would provide larger LHSs for the smallest possible value of n and thus for all possible values of n). One can then bring the "1" from the LHS to the RHS to obtain that a*(a^n - sum_{k=1, k=n}(nCr(n, k)*a^(k-1))) = 2002 = 2*7*11*13, i.e. a can only result from a combination of 2, 7, 11 and 13, there are 16 possible combinations or, accounting for the observation that a < 46, only 7 possible values, 2, 7, 11, 13, 14, 22 and 26. The case a = 26 cannot work as the first summand in the LHS of 26^(n+1) - 27^n = 2001 ends in 6 and the second one cannot end in 5. The case a = 22 cannot work as 2001 = 3*23*29 and therefore the equation with a = 22 would be equivalent to saying that 22^(n+1) = 23*(87+23^(n-1)), which is not possible as 23 does not divide 22. The cases a = 14, a = 11 and a = 2 can be ruled out by exactly the same way as for a = 22 as 3|15, 3|12 and 3|3, and 3|2001. For the case a = 7 I had to compute a bit and use the strictly increasing property of the LHS, the LHS is namely bigger than 10'000 for n > 3 and n = 1, n = 2 and n = 3 do not work. The only remaining case, a = 13, provides the solution n = 2, which, due to the LHS being strictly increasing, is unique.

[orionspur]

(a+1)^n rapidly dominates a^(n+1) for any fixed a. So one can compute the min and max possible value of n for each small a. All but one pair fails basic congruences.

[kingzenoiii]

putnam problems are mind-boggling for the average stormtrooper, but darth-MATH-vader is above such limitations

[tomasstride]

Mostly I only do problems involving Cauchy's Theorem as I told you. But sometimes I try number theory problems. I tried this one, and indeed I did get to the answer. I used modular arithmetic. Usually I do these problems reasonably quickly but very much doubt I would be able to do them in the few minutes (or even seconds) that is expected in these competitions. I studied mathematics well over 50 tears ago. But that was basically applied mathematics and I knew nothing about number theory at all. Believe me it was entirely a closed book to me. Such number theory that I know I have entirely learnt by watching Michael Penn's Youtube Channel. This has all happened in the last couple of years. This shows that it is possible to learn something new in mathematics even when one is old, decrepit and in decline as I certainly am.

[neilgerace]

1:12 Why can't they both be congruent to 2 mod 3?

[unkown]

I dont know if you include 0 in N (i am european so I do) but in this case a=2002 et n = 0 is also a solution

[angusclark]

I fought a long battle with this problem and got to a = 7 or 13 and if a is 7 then n is odd and if a is 13 then n is even and that allowed me to find the solution a = 13, n = 2. But when I was working mod 3 I didn't realise that (a+1) congruent to 2 mod 3 meant that n is even which would have contradicted my earlier deduction that if a is 7 then n is odd.

[CM_France]

Hi,

"ok, cool" : 0:18, 3:28,

"terribly sorry about that" : 7:37 .

[daveydd]

Hopefully I do not make a fool of myself this time.. but.. Could you please make a video on how to solve:
The integral from 0 to pi of (sin(2nx)cos(x))/sin(x) dx. ? The answer is pi but I do not quite get to know how to solve it

[SkorjOlafsen]

Congrats for finding a number theory problem that didn't require iterating through all positive integers less than some two-digit prime number and checking each case. Those are never satisfying.

[chaosredefined]

Wait, how do we know that 13^4 - 14^3 doesn't work? Or 13^6 - 14^4? etc...

[nicolascamargo]

Porque a no puede 3 mod 4 en 4:54 no entendí esa parte