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

Revision as of 14:08, 15 December 2019

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

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

$LHS$ $k$ ! = 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

For $k$ = 2: 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

@@ Line 1: / Line 1: @@
+==Problem==
 Find all pairs <math>(k,n)</math> of positive integers such that
 <cmath>k!=(2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1}).</cmath>
+==Solution 1==
+<math>LHS</math> <math>k</math>! = 1(when <math>k</math> = 1), 2 (when <math>k</math> = 2), 6(when <math>k</math> = 3)
+<math>RHS = 1</math>(when <math>n</math> = 1), 6 (when <math>n</math> = 2)
+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.
+For <math>k</math> > 3, <math>n</math> > 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