Advanced Math/AMC Contest Questions
Hi, here are some questions from AMC 12 that I couldn't quite figure out:
1. A function f from the integers is defined as follows:
f(n) = n + 3 if n is odd
f(n) = n/2 if n is even
Suppose k is odd and f(f(f(k)))=27. What is the sum of the digits of k?
2. The function f(x) satisfies f(2x+x)=f(2-x) for all real numbers x. If the equation f(x)=0 has exactly 4 distinct real roots, then what is the sum of these roots?
3. What is the sum of all real x such that (2^x-4^3)=(4^x+2^x-6)^3?
4. 3^x^x^x...=9, what is x?
I wasn't able to answer all the questions completely, but hopefully the following info will help you (I posted this on the experts Question Pool but it hasn't been picked up yet).
Prob. 1: f(f(f(n))) = 27, i.e, a function composed with itself twice. Given the rules for the odd and even numbers, let's try working this backwards. Start by labeling 27 = n4 (using an index). Then we have 2 choices;
if the inverse f^-1(n4) = n3 is odd, then f(n3) = n3+3 = n4 = even, but n4 = 27 = odd, so this doesn't work. So try
n3 = even, then n4 = n3/2 = 27 => n3 = 2n4 or n3 = 54, so this works.
Playing the same game, let n2 be even, then n3 = n2/2 = 54 so that n2 = 108. But also let
n2 be odd, then n3 = n2+3 = 54 or n2 = n3-3 = 51 => n2 is odd, which is consistent. So n2 = 108 or 51.
Taking the next step, let n1 = even, then
for n2 = 108, n1 = 2n2 = 216 = even, so this is OK. For n2 = 51, we have n1 = 2n2 = 102, which is also OK;
for n1 = odd, we get for the 2 values of n2, n1(n2=108) = 108-3 = 105 -> OK, and n1(n2=51) = 48, not odd so this solution not OK. Thus
n4 = 27
n3 = 54
n2 = 108 or 51
n1 = 216, 102 or 105 = k, the integer we are looking for. Since k is odd, the sum of its digits is 1+0+5 = 6.
To check, do the forward calculation; for n = n1 = 216, we have f(216) = n2 = 108 -> n3 = 54 -> n4 = 27. The other values also converge to 27.
The function f(x) as defined is non-linear so families of solutions are hard to come by. I'm sure there is a lot of interesting math related to functions of this type but I hope I at least answered the question.
Prob 2: You wrote that f(2x+x) = f(2-x). I'm wondering why the 1st argument is written 2x+x instead of 3x. Is this a typo?
If, for instance, it is really 2x+x^2, then some progress can be made. We are looking for 4 distinct roots. Let r1 be a root so that f(r1) = 0. Set
r1 = 2x+x^2 or x^2+2x-r1 = 0 which gives x = -1 +/- sqrt(1+r1). Given the property f(2x+x^2) = f(2-x), we know solutions for roots on the left-hand side (LHS), i.e., x, will gives roots for the RHS. Plugging these 2 values into the second argument gives
r2 = 2 +1-sqrt(1+r1) and r3 = 2 +1+sqrt(1+r1).
so that r2 + r3 = 6. This partial result shows that the sum of the roots r2 and r3 doesn't depend on the value of the root r1, which is encouraging. It may be that further progress can be made with the arguments of f adjusted, but I hesitate to explore further without knowing if a) the arguments do need to be adjusted and b) what the adjustment might be. Please follow-up to let me know if I am barking up the wrong tree.
Prob. 3: Sum of all real x s.t. (2^x + 4^3) = (4^x + 2^x - 6)^3. Again I'm a little suspicious of there being typos (4^3 = 64; why not write it out?). If (big if) the LHS is (2^x - 4), then
set LHS = RHS = 0, let 2^x = y, then RHS = y^2 + y - 6 -> y = 2 and -3; but since 2^x cannot be negative (for x real), we have just y = 2. This is convenient since 2^2 = 4 so that the LHS also =0.
Again, let me know if the equations were entered properly in the problem.
Prob. 4: 3^x^x^x...= 9. I assume that the ellipsis (3 dots) after the last x means the sequence goes on indefinitely. The first thing one can say is
3^2 = 9 so that x^x^x... = 2. If the ellipsis is not really meant, then we have x^x^x = 2, which can be written x^3 = 2 or x = 2^1/3 = 1.26... If the ellipsis is real, then we have
lim(n->infinity)x^n = 2. Denote "infinity" = inf, then use the definition e = lim(n->inf)(1+1/n)^n where e is the base of the natural logarithm. Then
lim(n->inf)[((2/e)^1/n)(1+1/n)]^n = 2, so that the nth term in the sequence for x is xn = [((2/e)^1/n)(1+1/n)]^n.
Please let me know if this helps and if there are any corrections to the original problems.