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

(Solution 1)
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==Solution 1==
 
==Solution 1==
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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)
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Hence, (1,1), (3,2) satisfy
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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.
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For <math>k</math> > 3, <math>n</math> > 2:
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LHS: Minimum two odd terms other than 1.
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RHS: 1st term odd. No other term will be odd.
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By parity, LHS not equal to RHS.
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Hence, (1,1), (3,2) are the only two pairs that satisfy.
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~flamewavelight and phoenixfire

Revision as of 13:48, 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

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