Difference between revisions of "2019 IMO Problems/Problem 4"
(→Solution 1) |
(→Solution 1) |
||
Line 12: | Line 12: | ||
Hence, (1,1), (3,2) satisfy | Hence, (1,1), (3,2) satisfy | ||
− | For <math>k</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. | + | </large>For <math>k</math> = 2:<large> 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: |
Revision as of 13:06, 15 December 2019
Problem
Find all pairs of positive integers such that
Solution 1
! = 1(when = 1), 2 (when = 2), 6(when = 3)
(when = 1), 6 (when = 2)
Hence, (1,1), (3,2) satisfy
</large>For = 2:<large> 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