Daisy Zeng’s Post

Hi. Sometimes simple things need simple solutions.

It's an easy DIY fix.

Every time you talk about a new feature.

Do you think free service have advantage

Easy and quick fix😜

Figure out what you can control and what you can't. Then decide if what you can't is ok with you.

(5,3), (2,6),(3,4) A different approach to solve this problem using inequality If a>b=> (1+1/a)(1+2/b)<(1+1/b)(1+2/b) =>2b^2<(b^2+3b+2) => b^2-3b-2<0 =>( 3-sqrt(13))/2 <b <(3+sqrt(13))/2 => 0<b<=3 (using b is positive integer) => b=1,2,3 Only for b=3 a=5, so one solution is (5,3) No a possible now if b>=a then replacing b with a in case a>b, we get possible value of a as 1,2 3 So only solution here is (2,6)(3,4)

Try: Observe: f(8) - f(4) = 1 f(4) - f(2) = 1 f(2) - f(1) = 1 Thus let P(x) = f(2x) - f(x) Note if we let x = 1, 2, and 4 P(x) - 1 = 0 Thus P(x) is a cubic function So let P(x) = a(x-1)(x-2)(x-4) + 1 Now let x = 0 f(0) - f(0) = a(-1)(-2)(-4) + 1 a = 1/8 P(x) = f(2x) - f(x) = (1/8)(x-1)(x-2)(x-4) + 1 Now let x = 8 f(16) - f(8) = (1/8)(7)(6)(4) + 1 f(16) - 3 = 21+1 f(16) = 25 Therefore, f(2⁴) = 25

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