math puzzle

[Orpz]

this is actually a thinly veiled request for help but i'll disguise it as a puzzle so it's not moved to like circlejerk

1)n is a natural number
2)f(n) is a natural number
3)f(n) < f(n+1)
4)f(f(n)) = 3n

what is f(10)

[Nazzhoul]

Okay.
Natural number is positive whole numbers including zero.

f1<f2<f3<f4<f5<f6<f7<f8<f9<f10<f11<f12

0<f1< 3
0<f2<6
f3<9
f4<12
f5<15
f6<18
f7<21
f8<24
0<f9<27
0<f10< 30
0<f11< 33

I'm sure that there's a way to continue from there, but I can't figure it out at the moment. And I'm really tired...

[Gerik]

what is f(10)

I'm going to go with 20.

[Espozito]

Well, the idea is simple here. We must find out how to raise n into 3n making 2 same actions. The bad thing is that second time that action is made not on n, but on f(n), which makes it harder to find out. Let me think of it.

[Gerik]

Well, the idea is simple here. We must find out how to raise n into 3n making 2 same actions. The bad thing is that second time that action is made not on n, but on f(n), which makes it harder to find out. Let me think of it.

I've already solved it.

I'm going to go with 20.

See?

[Orpz]

It's not 20 close though

I have the answer, but there's no explanation with the answer. I suspect it has to do with a different base notation.

[creedkingsx]

30

[vornksr]

This was fun; thanks!

Gerik was indeed close (if he wasn't just guessing), but he got a step wrong.


The trick is that f acts differently depending on the number's factorization. For what highest value z is 2^z a factor of n? If z is odd, f(n)=(3/2)*n. If z is even, f(n)=2*n.

Thus, for example, 3 is not divisible by 2, so z=0. Therefore f(3)=6. Likewise 4 is divisible by 2^2, so z is even, and f(4) = 8. On the other hand, f(2)=3 and f(8 )=12, since z=1 for n=2 and z=3 for n=8.

When n=10, z=1 (since 10=2*5). Therefore f(10)=15. (As a check, z=0 for 15, so f(15)=30 like it should).

So your answer is 15.

[vornksr]

A simpler way of phrasing what the function f does:

If n is odd, double it.
If n is even (but NOT divisible by 4), increase it by 150%. (That is, add half of its value to it, or multiply by 3/2. So 2 becomes 3, 10 becomes 15, and so on.)
If n is divisible by 4, cry a little bit. Then use my formula above.

[Orpz]

The answer is 19

Using logic and what we are given, we have

f(1) = 2
f(2) = 3
f(3) = 6
f(6) = 9
f(9) = 18
f(10) = ?

Converting each number into base 3 gets you
f(1) = 2
f(2) = 10
f(10) = 20
f(20) = 100
f(100) = 200
f(101) = ??

The pattern is that if the original number in base 3, n, starts with 1, then f(n)'s starting digit is 2, and the rest of the number is the same. If the original number in base 3, n, starts with 2, then f(n)'s starting digit is 1 and gets an additional 0 at the end of the number.

Therefore f(101) = 201

201 converted back to base 10 is 19.

I welcome any and more math puzzles

[Mateo]

1 cross + 3 nails = 4given

[louiswill]

Orpz, why?
"Converting each number into base 3 gets you
f(1) = 2
f(2) = 10
f(10) = 20
f(20) = 100
f(100) = 200
f(101) = ??"

why f(1) = 2, f(2) = 10?

f(2) = f(f(1)) = 10? not 3?

what happen to f(f(n)) = 3 n???

[louiswill]

I can explain more clearly why is it 19. I should get some reputation after all this....

consider

domain of f(n) is all N.

For obvious reason,
f(f(1)) = 3
f(f(2)) = 6
f(f(3)) = 9
f(f(4)) = 12
let g(n) = f(f(n)) = 3n

Therefore,
f(1) must be define with a nice number,
otherwise it will be a problem

i.e. let f(1) = 3, then f(f(1)) = f(3) = 3, contradict with f(f(3)) = 9

similarly, if let f(1) = 3k, k is a natural number, then f(f(1)) = f(3k) = 3, contradict with all f(f(3k)) >> f(f(3)) = 9.

Thus, the only way to define f(x) with its domain as all Natural number is

f(1) = 1 or 2, but if f(1) = 1, it will contradict with f(f(1)) = 3

Therefore,

Define f(1) as f(1) = 2

The rest is much easier:

f(1) = 2
f(2) = g(1) = 3
f(3) = g(2) = 6
...

f(1) = 2
f(2) = 3
f(3) = 6
f(6) = 9
f(9) = 18
f(10) = ?

Notice 4,5,7,8 are also open to define before it comes to 10

Just in case, lets go over it.
g(4) = 12
thus, define 3<f(4)<6 as 5, for f(4) can not be 4, contradicting to g(4)

f(4) = 5

g(5) = 15 thus, define 5<f(5)<9 as 7 or 8? lets hold it there.

g(7) = 21 g( = 24 thus, define 9<f(7)<f(<18? HOLD


we can see we basically are filling the gap of all 2,3,6,9,18.
f(1) = 2
f(2) = 3
f(3) = 6
f(6) = 9
f(9) = 18

That mean, f(10) -f(17) will all need define.

add f(1 = f(f(9)) = 27 in to list, 27 will be the maximum range of our choice for f(17).

after f(10), there are still 7 numbers need to be define nicely,

so, 27-7 = 20, 20 will be the range for f(10)

f(10) < 20

I guess this is why someone is giving 20 as the answer, but f(10) can not be 20.

Then,

How about 19?

Errr......why not? but why not 11?

f(10) = 11, then f(11) = 30 > f(1 = 27, Nope.

So f(10) > 17, because if f(10) = any number between 11 and 17 will be a define problem for corresponding f(x)

That is, f(10) = 19, because f(1 is already defined.

I do not know how to remove those emotion icon?

those are "8" and ")"

[WHITE BOY SLIM AYO]

This is a pretty old thread.

Let me try a few math puzzles that I have heard a few years ago:

1.) Two men enter a class reunion: Joe and John.
A conversation ensues.

Joe: Hi John.
John: Hey Joe.
Joe: Have you heard that I have 3 kids now?
John: Oh, how old are they?
Joe: Since you were always so good in math, let me put it this way: The product of their age is 36 and the sum of their age is your favorite number.
John: I don't think that's enough information to answer your question.
Joe: Well the oldest has blue eyes.

What are the ages of Joe's kids? What is John's favorite number?

2.) 1-2+3-4....=1/4

How is this possible?