For each positive integer n , we define f ( n ) such that f ( n ) is a positive integer, f ( n + 1 ) > f ( n ) and ( f ( f ( n ) ) = 3 n . What is the value of f ( 1 0 ) ?
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The best way to solve this is with a number line (natural nos. 1–21). Simply try to fill up the number line according to the rules of function f . Let us call the ordering rule "Rule O" ( f ( n + 1 ) > f ( n ) ) and the multiple rule "Rule M" ( f ( f ( n ) ) = 3 n ). Your working might progress as follows:
1. Find f ( 1 ) .
If f ( 1 ) = 1 , then f ( f ( 1 ) ) = f ( 1 ) = 1 = 3 . So f ( 1 ) = 1 , by Rule M. In fact, f ( 1 ) > 1 , and so f ( n ) > n for all n , by Rule O.
If f ( 1 ) ≥ 3 , then f ( f ( 1 ) ) = 3 ≤ f ( 1 ) . So f ( 1 ) < 3 , as above.
Therefore:
f ( 1 ) = 2 .
Now, applying Rule M, we have the following progression by f :
1 → 2 → 3 → 6 → 9 → 1 8 → 5 4 …
2. Find the f -trails starting at 4 and 5 .
By Rule O, we must have that 6 = f ( 3 ) < f ( 4 ) < f ( 5 ) < f ( 6 ) = 9 , so immediately we have the following:
f ( 4 ) = 7 ; f ( 5 ) = 8 .
Applying Rule M, we have the following progressions by f :
4 → 7 → 1 2 → 2 1 … ; 5 → 8 → 1 5 → 2 4 …
3. Find f ( 1 0 ) .
Again applying Rule O, we have that 1 8 = f ( 9 ) < f ( 1 0 ) < f ( 1 1 ) < f ( 1 2 ) = 2 1 . So, immediately:
f ( 1 0 ) = 1 9 .
Of course, we could carry on! But at this point, the question is answered.
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Wow that's an awesome solution @Raven Herd
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@Devansh Sharma – Thanks! I wish I'd worked it through properly before I entered the wrong answer... ;)
We will prove via induction on all non-negative integers n a claim P n that:
First, we prove that f ( 1 ) = 2 .
If f ( 1 ) = 1 , we have f ( f ( 1 ) ) = 1 = 3 , a contradiction, hence f ( 1 ) > 1 and with the condition f ( n + 1 ) > f ( n ) , we have f ( n ) > n for all positive integers n .
If f ( 1 ) ≥ 3 , we'd have f ( f ( 1 ) ) ≥ f ( 3 ) > 3 , also a contradiction. Hence we must have f ( 1 ) = 2 . Also, we have f ( 2 ) = f ( f ( 1 ) ) = 3 and f ( 3 ) = f ( f ( 2 ) ) = 6 .
We note that we have our claim P n be true for n = 0 , given our values of f ( 1 ) , f ( 2 ) , f ( 3 ) . Now assume for some non-negative integer k , we have P k be true, i.e. that:
We then consider the claim P k + 1 :
For 3 k + 1 ≤ x ≤ 2 ⋅ 3 k + 1 , consider the x where x = 3 y for some integer 3 k ≤ y ≤ 2 ⋅ 3 k . We then have x = f ( f ( y ) ) = f ( y + 3 k ) , and consequently f ( x ) = f ( f ( y + 3 k ) ) ) = 3 ( y + 3 k ) = 3 y + 3 k + 1 = x + 3 k + 1 .
For 3 k + 1 ≤ x ≤ 2 ⋅ 3 k + 1 , there also exist x , x + 1 where 3 y < x < x + 1 < 3 ( y + 1 ) for some ( 3 k ≤ y ≤ 2 ⋅ 3 k − 1 . We can infer that x = 3 y + 1 , x + 1 = 3 y + 2 . Then using our results that f ( 3 y ) = 3 y + 3 k + 1 , f ( 3 y + 3 ) = 3 y + 3 + 3 k + 1 from just now, we have:
3 y + 3 k + 1 = f ( 3 y ) < f ( x ) < f ( x + 1 ) < f ( 3 y + 3 ) = 3 y + 3 + 3 k + 1
We thus have f ( x ) = 3 y + 1 + 3 k + 1 = x + 3 k + 1 , f ( x + 1 ) = 3 y + 2 + 3 k + 1 = x + 1 + 3 k + 1 , hence we have it that for all 3 k + 1 ≤ x ≤ 2 ⋅ 3 k + 1 , we have f ( x ) = x + 3 k + 1
For 2 ⋅ 3 k + 1 ≤ x ≤ 3 k + 2 , we note that 3 k + 1 ≤ x − 3 k + 1 ≤ 2 ⋅ 3 k + 1 and from our previous result we have f ( x − 3 k + 1 ) = x − 3 k + 1 + 3 k + 1 = x . Consequently, we have f ( x ) = f ( f ( x − 3 k + 1 ) ) = 3 x − 3 k + 2 .
Thus we have proven that with P 0 being true and P k being true ⇒ P k + 1 being true, we have proven our claim that P n is true for all non-negative integers n.
Using the above formula, we have f ( 1 0 ) = 1 9 .
Typo on your 7th line: you say 'f(1)=3', but it equals 2.
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It is very easy to see that: f ( f ( 1 ) ) = 3 → f ( 1 ) = 2 Because, if f(1)=1 then f(1)=3 (this is a confusion). f ( f ( 1 ) ) = f ( 2 ) = 3 , f ( f ( 2 ) ) = f ( 3 ) = 6 , f ( f ( 3 ) ) = f ( 6 ) = 9 , → f ( 4 ) = 7 , f ( 5 ) = 8 f ( f ( 4 ) ) = f ( 7 ) = 1 2 , f ( f ( 5 ) ) = f ( 8 ) = 1 5 , f ( f ( 6 ) ) = f ( 9 ) = 1 8 , f ( f ( 7 ) ) = f ( 1 2 ) = 2 1 → f ( 1 0 ) = 1 9