Is it too late for April Fools?

Finding the answer is trivial: the only possible value that $f$ can take is clearly $\boxed{2015}$ , and for the problem to have a single answer, this must be the answer. The meat of the problem lies on whether $f$ is always defined.

We will refer to the eight definitions as Definition 1 through Definition 8 as they are listed. For example, Definition 1 is $f(0,0,0) = 2015$ .

Suppose we want to compute $f(a,b,c)$ . Define three sequences $(a_i), (b_i), (c_i)$ indexed starting from $0$ , such that $a_0 = a, b_0 = b, c_0 = c$ , and whenever we apply a Definition in the form $f(a_i, b_i, c_i) = f(x,y,z)$ , set $a_{i+1} = x, b_{i+1} = y, c_{i+1} = z$ . Essentially, the three sequences keep track of the current state of the arguments.

For example, we can perform this computation:

$\begin{aligned} f(2,1,0) &= f(1,0,2) & \text{(Def 5)} \\ &= f(0,1,1) & \text{(Def 6)} \\ &= f(0,0,2) & \text{(Def 7)} \\ &= f(1,0,0) & \text{(Def 4)} \\ &= f(0,1,0) & \text{(Def 2)} \\ &= f(0,0,1) & \text{(Def 3)} \\ &= f(0,0,0) & \text{(Def 4)} \\ &= 2015 & \text{(Def 1)} \end{aligned}$

Thus, by reading the above downwards, we obtain the sequences:

$(a_i) = (2, 1, 0, 0, 1, 0, 0, 0) \\ (b_i) = (1, 0, 1, 0, 0, 1, 0, 0) \\ (c_i) = (0, 2, 1, 2, 0, 0, 1, 0)$

Note that if a sequence of natural numbers is strictly decreasing, by well-ordering principle on the natural numbers, this sequence must eventually terminate. Similarly, if a sequence of natural numbers is non-increasing, it must eventually be constant (although not necessarily constant $0$ ).

We claim that:

• $a_i + b_i + c_i$ is non-increasing, and
• whenever $a_i + b_i + c_i$ is constant, $2a_i + b_i$ is strictly decreasing.

First, we will show that the above is enough to show that the sequences will terminate, necessarily at $(0,0,0)$ as otherwise we can apply some other Definition.

Suppose the sequences don't terminate. Thus $a_i + b_i + c_i$ , being a sequence of natural numbers that is non-increasing, must eventually be constant. But then $2a_i + b_i$ must be decreasing forever, while it is still a sequence of natural numbers, contradicting the well-ordering principle.

With the above, we can show that $f(a,b,c)$ will eventually become $f(0,0,0) = 2015$ . Now we will prove both of the above claims.

First, we can observe that no Definition increases the sum of its arguments, proving the first claim immediately.

Now, we can ignore Definitions 4, 6, and 8, as they decrease the sum; the second promise doesn't claim anything when the sum decreases. We can also ignore Definition 1 as it terminates the sequence. However, we can verify that the remaining four Definitions indeed decrease the value of $2a+b$ :

• For Definition 2, $2 \cdot a + 0 > 2 \cdot (a-1) + 1$ , since it simplifies to $1 > 0$ .
• For Definition 3, $2 \cdot 0 + b > 2 \cdot 0 + (b-1)$ , since it simplifies to $1 > 0$ .
• For Definition 5, $2 \cdot a + b > 2 \cdot (a-1) + (b-1)$ , since it simplifies to $3 > 0$ .
• For Definition 7, $2 \cdot 0 + b > 2 \cdot 0 + (b-1)$ , since it simplifies to $1 > 0$ .

This proves both of our claims, showing that $f$ is a constant function, returning $2015$ no matter what the arguments are.

Now we need to prove that $p_1, p_2, \ldots, p_8$ , used for computing $APRIL$ and $FOOLS$ , are defined; they can probably be not defined if there are only finitely many happy numbers , for example. But this is simple; $10^n$ is trivially happy for all natural number $n$ as its happinization is $10^n, 1, 1, 1, \ldots$ , thus we can compute $p_n$ by simply inspecting all numbers not greater than $10^{n!}$ . We are guaranteed to find at least $n!$ happy numbers ( $10, 10^2, 10^3, \ldots, 10^{n!}$ ), so $p_n$ is defined. Additionally, happy numbers are positive integers by definition, so $p_n$ is a positive integer for all $n$ . This proves that our expression $f(APRIL, FOOLS, 2015)$ is indeed a defined expression, and our result above shows that $f(APRIL, FOOLS, 2015) = \boxed{2015}$ .

OMG OMG OMG LOL I GUESSED IT OP!!!! LOLOLOL CUZ IT WAS 2015 I GUESSED IT XDDDD 4% yeah!!!

None of the definitions of $f$ increase the value of $a+b+c$ .

These decrease the value of $a+b+c$ :

$f(0, 0, c) = f(c-1, 0, 0) \\ f(a, 0, c) = f(c-1, 1, a-1) \\ f(a, b, c) = f(b, c, a-1)$

These, on the other hand, keep it the same:

$f(a, 0, 0) = f(a-1, 1, 0) \\ f(0, b, 0) = f(0, b-1, 1) \\ f(a, b, 0) = f(a-1, b-1, 2) \\ f(0, b, c) = f(0, b-1, c+1)$

However, we can see that each of these eventually leads to $a+b+c$ being decreased, as follows:

$f(1, 0, 0) = f(0, 1, 0) = f(0, 0, 1) \\ f(2, 0, 0) = f(1, 1, 0) = f(0, 0, 2) = f(1, 0, 0) \\ f(a+2, 0, 0) = f(a+1, 1, 0) = f(a, 0, 2) = f(1, 1, a-1) \\ f(0, b, 0) = f(0, b-1, 1) = f(0, 0, b) = f(b-1, 0, 0) \\ f(a+1, b+1, 0) = f(a, b, 2) = f(b, 2, a-1) \\ f(1, b+1, 0) = f(0, b, 2) = f(0, 0, b+2) = f(b+1, 0, 0) \\ f(a+1, 1, 0) = f(a, 0, 2) = f(1, 1, a-1) \\ f(1, 1, 0) = f(0, 0, 2) = f(1, 0, 0) \\ f(0, b, c) = f(0, 0, b+c) = f(b+c-1, 0, 0)$

Consequently, we know that the evaluation of $f$ will eventually terminate. Since there is only one terminating condition, with a constant value, we know that $f$ is just a constant function.