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. 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== | ||
Revision as of 19:49, 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. 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 | ||
