Apply Binomial Theorem in Number Theory - ISI, CMI Entrance - TOMATO Subjective 26

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Can you show 9 divides 2^{2n} - 3n - 1?

Problem useful for I.S.I B.Stat B.Math Entrance, CMI Entrance and Math Olympiad
  1. Cheenta Academy for Olympiad & Research
  2. Indian Statistical Institute
  3. Chennai Mathematical Institute
  4. Math Olympiad
  5. RMO
  6. IMO
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Комментарии
Автор

How athis much easy question can come in ISI-CMI cant beleive

sarvesh_soni
Автор

This can be done easily by mathematical induction

archismanchattopadhyay
Автор

2^(2n)-3n-1 = (3+1)^n-3n-1 = = 9(nC2+nC3*3+....+3^(n-2) = 9*(an integer), so 9 divides 2^(2n)-3n-1
Again, a gem of a problem!!!! This is why I love TOMATO🙂🙂

arundhatimukherjee
Автор

If n=1, so 2^2n-3-1=2^2*1-3*1-1
=4-3-1
=0,
So the expression is not divisible by 9 for n=1.
If n=2,
2^2n-3n-1=2⁴-6-1
=16-7
=9,
So the expression is divisible by 9 for n=2.

shewlykhanam
Автор

2^2n -3n -1
= 4^n - 3n - 1
= (3 + 1)^n - 3n - 1
= (3^n + nC1 3^n-1 + ... + nC2 3² + 3n + 1) - 3n - 1
All the terms of the expansion except last two terms are divisible by 9.. but the last two terms cancel out..
= 3² (3^(n-2) +... )
= 0 (mod 9)

sharathpr