Christmas Streak 13/88: Much exponential

Number Theory Level 3

Find the maximum value of the positive integer $n$ that satisfies

$\large 2^n \Bigg|\left(2017^{2^{100}}-2015^{2^{100}}\right).$

This problem is a part of <Christmas Streak 2017> series .

The answer is 106.

1 solution

Boi (보이)
Oct 11, 2017

Note that $2017^{2^k}+2015^{2^k}\equiv 1^{2^k}+(-1)^{2^k}\equiv 2~\text{(mod 4)}$

Then $2017^{2^{100}}-2015^{2^{100}}=(2017-2015)(2017+2015)(2017^2+2015^2)(2017^{2^2}+2015^{2^2})\cdots(2017^{2^{99}}+2015^{2^{99}}).$

And since $2017+2015=4032=2^6\times 3^2\times 7,$

$n=1+6+99=\boxed{106}.$

How did you get from the second-last step to the last step?

Siva Budaraju - 3 years, 8 months ago

As a factor $2017 - 2015$ has 1 two's, $2017 + 2015$ has 6 two's, $2017^2+2015^2, ~\cdots,~ 2017^{2^{99}}+2015^{2^{99}}$ has 1 two's each.

So the whole number has 1 + 6 + 99 two's in it.

Boi (보이) - 3 years, 8 months ago

Oh, I see. Thanks!

Siva Budaraju - 3 years, 8 months ago

@Siva Budaraju – No problem! ^^

Boi (보이) - 3 years, 8 months ago

I solved this using the binomial theorem, but your solution looks a 100 times simpler. Only problem is, I have NO IDEA what it means. I've never seen this "mod" notation. Could you suggest where I can learn what it is and the pre requisites I would need to be able solve this using it?

Soumil Sahu - 3 years, 7 months ago

Mod is essentially finding the remainder when it is divided. For example 5 mod 4 = the remainder of 5/4 = 1.

Siva Budaraju - 3 years, 7 months ago

That's simple, but the solution seems to have used multiple identities or formulae directly. What I mean is, even after you telling me the definition, I don't understand what he first step is trying to say.

Soumil Sahu - 3 years, 7 months ago

Christmas Streak 13/88: Much exponential

The answer is 106.

1 solution

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