Difference between revisions of "2019 IMO Problems/Problem 4"

(Solution 1)
(Solution 1)
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</Large>
 
</Large>
 
For <math>k</math> = 2:
 
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.
+
<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 14:07, 15 December 2019

Problem

Find all pairs $(k,n)$ of positive integers such that

\[k!=(2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1}).\]

Solution 1

$LHS$ $k$! = 1(when $k$ = 1), 2 (when $k$ = 2), 6(when $k$ = 3)

$RHS = 1$(when $n$ = 1), 6 (when $n$ = 2)

Hence, (1,1), (3,2) satisfy

</Large> For $k$ = 2: <Large> RHS is strictly increasing, and will never satisfy $k$ = 2 for integer n since RHS = 6 when $n$ = 2.

For $k$ > 3, $n$ > 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