NO CALCULATORS !! | Moscow Mathematical Olympiad 1949

[David_W_Wood]

One interesting point is that the proof doesn't depend on the power being 8. The same proof would show that 26460 divides 27195^n - 10887^n + 10152^n for any other positive integer value of n.

[themathsgeek]

This is a fairly easy question bar the computations maybe if one is familiar with congruence theory. Nice one!

[satyapalsingh]

Nice problem and solved nicely .Keep it up!!!

[jjpower]

I am unsure what the question is asking me to do.
Could it have been phrased differently so that people for whom English is their first language might understand what is being asked?

[f-th]

Modding by 4 and 5 is easy. Modding by 49 is also easy, but not obvious. To do it, turn every 100 into a 2.
For example, to reduce 27195 mod 49, 27100 = 271×100 becomes 271×2. So 27195 is congruent to 271×2 + 95, which is congruent to (2×2+71)×2+95 = 75×2 + 95 = 150 + 95, congruent to 52 + 95 = 147, congruent to 2 + 47= 49.

[davidblauyoutube]

In fact, this number is divisible by 2^5 * 3^10 * 5 * 7^2 = 680, 244, 480. :)

[mtaur]

I was wondering if there were any shorter shortcuts than this, but this works and it isn't unreasonably long. A bit of old fashioned work and the ideas of modular arithmetic are of use.

[simonrobertson]

I noticed that 735 is a factor of the divisor and the first term (is actually the difference between them), and is also the difference between terms 2 and 3 so can be removed by completing the square three times -a^8 + b^8= (b-a)(a+b)(a^2+b^2)(a^4+b^4). I still had to do the modular arithmetic for other factors, 4 and 9 though

[jongyonp]

I made it easier by removing all the 3's from each side. Right hand side is divisible by 3^8 so the LHS can lose all its 3's, then the rest of the prime components cannot divide the 3's left on the RHS so it must divide whatever's left.

[rslitman]

For the mod 4, would 3 instead of (-1) also work? If so, why did you use (-1) here?

[bhoomib]

how did we got 6 on dividing I tried by congruence as well