Moscow Middle School Math Olympiad Competition | Prove 2222^5555 +5555^2222 is divisible by 7.

How to prove this huge number is divisible by 7? This video will complete the proof by applying two basic algebraic results. This is a question in number theory from Moscow Middle School Math Olympiad Competition.

[imonkalyanbarua]

Didn't know about those 2 results. It was amazing! Thank you. 👏👏👏😇

[john]

Interesting question, brilliant solution!

[ravinderrathod]

Sir, it is humble request to make solutions of Mathematical Holiday solutions of Russian Olympiad which held on 18.02.2024

[actions-speak]

A more boring method: Note that 3^6 = 4^3 = 1 (mod 7). Then 2222^5555 + 5555^2222 = 3^5555 + 4^2222 = 3^5 + 4^2 = 5 + 2 = 0 (mod 7).

[absolutezero]

Thank you for your solutions
I have learnt a lot

Is my solution correct?

2222/7 = 317.4 (1 dp)
317 x 7 = 2219

5555/7 = 793.6 (1 dp)
794 x 7 = 5558

2222⁵⁵⁵⁵ + 5555²²²²
= (2219 + 3)⁵⁵⁵⁵ + (5558 - 3)²²²²
= [(2219 + 3)⁵]¹¹¹¹ + [(5558 - 3)²]¹¹¹¹

In the binomial expansion of (2219 + 3)⁵,
All the terms except 3⁵ are divisible by 7

In the binomial expansion of (5558 - 3)²,
All the terms except (-3)² are divisible by 7

3⁵/7
= 243/7
= 34.7 (1 d.p)

35 x 7 = 245

Hence,
3⁵ is 2 less than 245,
Hence,
3⁵ gives a remainder of -2 when divided by 7

The last term in the binomial expansion of
[(2219 + 3)⁵]¹¹¹¹
= (3⁵)¹¹¹¹

Hence,
Remainder when (3⁵)¹¹¹¹ is divided by 7
= (-2)¹¹¹¹

(Of course,
The quotient is a lot smaller than “actual”)

(-3)²/7
= 9/7
= 1 r 2

(-3)² is 2 more than 7
Hence,
3² gives a remainder of 2 when divided by 7

Hence,
Remainder when (3²)¹¹¹¹ is divided by 7
= 2¹¹¹¹

(Again, the quotient is a lot smaller than “actual”)

The expansion of
[(2219 + 3)⁵]¹¹¹¹ + [(5558 - 3)²]¹¹¹¹
yields this expression:
……. + (3⁵)¹¹¹¹ + [(-3)²]¹¹¹¹

All the terms in front are divisible by 7
Hence,
The sum of all these terms gives a remainder of 0 when divided by 7

Hence,
The remainder when [(3⁵)¹¹¹¹ - (3²)¹¹¹¹]
is divided by 7
= (-2)¹¹¹¹ + 2¹¹¹¹
= -2¹¹¹¹ + 2¹¹¹¹
= 0

Hence,
The overall remainder is 0
Hence,
2222⁵⁵⁵⁵ + 5555²²²² is divisible by 7 (proved)