Difference between revisions of "2019 IMO Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
LHS = <math>k</math>! = 1(when <math>k</math> = 1), 2 (when <math>k</math> = 2), 6(when <math>k</math> = 3) | LHS = <math>k</math>! = 1(when <math>k</math> = 1), 2 (when <math>k</math> = 2), 6(when <math>k</math> = 3) | ||
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RHS = 1(when <math>n</math> = 1), 6 (when <math>n</math> = 2) | RHS = 1(when <math>n</math> = 1), 6 (when <math>n</math> = 2) | ||
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For <math>k</math> = 2: | For <math>k</math> = 2: | ||
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RHS is strictly increasing, and will never satisfy <math>k</math> = 2 for integer n since RHS = 6 when <math>n</math> = 2. | RHS is strictly increasing, and will never satisfy <math>k</math> = 2 for integer n since RHS = 6 when <math>n</math> = 2. | ||
For <math>k</math> > 3, <math>n</math> > 2: | For <math>k</math> > 3, <math>n</math> > 2: | ||
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LHS: Minimum two odd terms other than 1. | LHS: Minimum two odd terms other than 1. | ||
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RHS: 1st term odd. No other term will be odd. | RHS: 1st term odd. No other term will be odd. | ||
By parity, LHS not equal to RHS. | By parity, LHS not equal to RHS. | ||
Revision as of 12:49, 15 December 2019
Problem
Find all pairs 
of positive integers such that
![\[k!=(2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1}).\]](http://latex.artofproblemsolving.com/7/c/a/7ca521b92d40def7f320702893c01502439174e8.png)
Solution 1
LHS = 
! = 1(when = 1), 2 (when = 2), 6(when = 3)
RHS = 1(when 
= 1), 6 (when = 2)
Hence, (1,1), (3,2) satisfy
For = 2:
RHS is strictly increasing, and will never satisfy = 2 for integer n since RHS = 6 when = 2.
For > 3, > 2:
LHS: Minimum two odd terms other than 1.
RHS: 1st term odd. No other term will be odd. By parity, LHS not equal to RHS.
Hence, (1,1), (3,2) are the only two pairs that satisfy.
~flamewavelight and phoenixfire
