Difference between revisions of "2013 Indonesia MO Problems/Problem 6"
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a. Take <math>x=2^{2013}-1</math>, notice how <math>x</math> is odd, <math>(2^{2013}-1)^{(2^{2013}-1)2013}+1\equiv -1^{\text{some odd number}}+1\equiv -1+1\equiv 0\mod 2^{2013}</math> so its divisible | a. Take <math>x=2^{2013}-1</math>, notice how <math>x</math> is odd, <math>(2^{2013}-1)^{(2^{2013}-1)2013}+1\equiv -1^{\text{some odd number}}+1\equiv -1+1\equiv 0\mod 2^{2013}</math> so its divisible | ||
− | b. Notice how <math> | + | b. Notice how <math>y</math> is always odd as if it is even then even+odd=odd and cant be divisible by 2^m, and <math>m</math> is always odd as if it is even <math>y^{my}=(y^{ny})^2=1\mod 4</math>, <math>1+1=0\mod 4</math> which is not true because if m=1 then it is odd. By LTE, <math>v_{2}(y^{my}+1^{my})=v_{2}(y+1)\geq m</math> as it is divisible by 2^m, and the smallest <math>y</math> is <math>2^m-1</math> |
==See Also== | ==See Also== |
Latest revision as of 19:54, 26 December 2024
Problem
A positive integer is called "strong" if there exists a positive integer such that is divisible by .
a. Prove that is strong.
b. If is strong, determine the smallest (in terms of ) such that is divisible by .
Solution
a. Take , notice how is odd, so its divisible
b. Notice how is always odd as if it is even then even+odd=odd and cant be divisible by 2^m, and is always odd as if it is even , which is not true because if m=1 then it is odd. By LTE, as it is divisible by 2^m, and the smallest is
See Also
2013 Indonesia MO (Problems) | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 5 |
All Indonesia MO Problems and Solutions |