Is it too late for April Fools?

Algebra Level 5

Let N \mathbb{N} be the set of nonnegative integers. Define a function f : N 3 N f : \mathbb{N}^3 \to \mathbb{N} satisfying the following for any a , b , c N { 0 } a,b,c \in \mathbb{N} \setminus \{0\} :

{ f ( 0 , 0 , 0 ) = 2015 f ( a , 0 , 0 ) = f ( a 1 , 1 , 0 ) f ( 0 , b , 0 ) = f ( 0 , b 1 , 1 ) f ( 0 , 0 , c ) = f ( c 1 , 0 , 0 ) f ( a , b , 0 ) = f ( a 1 , b 1 , 2 ) f ( a , 0 , c ) = f ( c 1 , 1 , a 1 ) f ( 0 , b , c ) = f ( 0 , b 1 , c + 1 ) f ( a , b , c ) = f ( b , c , a 1 ) \begin{cases} f(0,0,0) &= 2015 \\ f(a,0,0) &= f(a-1,1,0) \\ f(0,b,0) &= f(0,b-1,1) \\ f(0,0,c) &= f(c-1,0,0) \\ f(a,b,0) &= f(a-1,b-1,2) \\ f(a,0,c) &= f(c-1,1,a-1) \\ f(0,b,c) &= f(0,b-1,c+1) \\ f(a,b,c) &= f(b,c,a-1) \\ \end{cases}

A happinization of a positive integer n n is an infinite sequence of integers, where the first term of the sequence is n n , and every term afterwards is the sum of the squares of the digits of the previous term. For example, the happinization of 2015 2015 is 2015 , 30 , 9 , 81 , 65 , 61 , 37 , 58 , 89 , 145 , 20 , 4 , 16 , 37 , 2015, 30, 9, 81, 65, 61, 37, 58, 89, 145, 20, 4, 16, 37, \ldots . A happy number is a positive integer whose happinization ends with an infinite number of 1 1 s; all other positive integers are sad numbers . For example, 2015 2015 above is sad, but 1 1 (with happinization 1 , 1 , 1 , 1 , 1, 1, 1, 1, \ldots ) is happy.

For any positive integer n n , define p n p_n as the product of the first n ! n! happy numbers. Let A = p 1 , F = p 2 , I = p 3 , L = p 4 , O = p 5 , P = p 6 , R = p 7 , S = p 8 A = p_1, F = p_2, I = p_3, L = p_4, O = p_5, P = p_6, R = p_7, S = p_8 . Compute the value of f ( A P R I L , F O O L S , 2015 ) f(APRIL,FOOLS,2015) .


The answer is 2015.

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3 solutions

Ivan Koswara
Apr 1, 2015

Finding the answer is trivial: the only possible value that f f can take is clearly 2015 \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 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 f(0,0,0) = 2015 .

Suppose we want to compute f ( a , b , c ) f(a,b,c) . Define three sequences ( a i ) , ( b i ) , ( c i ) (a_i), (b_i), (c_i) indexed starting from 0 0 , such that a 0 = a , b 0 = b , c 0 = c 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 ) f(a_i, b_i, c_i) = f(x,y,z) , set a i + 1 = x , b i + 1 = y , c i + 1 = z 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:

f ( 2 , 1 , 0 ) = f ( 1 , 0 , 2 ) (Def 5) = f ( 0 , 1 , 1 ) (Def 6) = f ( 0 , 0 , 2 ) (Def 7) = f ( 1 , 0 , 0 ) (Def 4) = f ( 0 , 1 , 0 ) (Def 2) = f ( 0 , 0 , 1 ) (Def 3) = f ( 0 , 0 , 0 ) (Def 4) = 2015 (Def 1) \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 ) (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 0 ).

We claim that:

  • a i + b i + c i a_i + b_i + c_i is non-increasing, and
  • whenever a i + b i + c i a_i + b_i + c_i is constant, 2 a i + b i 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 ) (0,0,0) as otherwise we can apply some other Definition.

Suppose the sequences don't terminate. Thus a i + b i + c i a_i + b_i + c_i , being a sequence of natural numbers that is non-increasing, must eventually be constant. But then 2 a i + b i 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 ) f(a,b,c) will eventually become f ( 0 , 0 , 0 ) = 2015 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 2 a + b 2a+b :

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

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

Now we need to prove that p 1 , p 2 , , p 8 p_1, p_2, \ldots, p_8 , used for computing A P R I L APRIL and F O O L S FOOLS , are defined; they can probably be not defined if there are only finitely many happy numbers , for example. But this is simple; 1 0 n 10^n is trivially happy for all natural number n n as its happinization is 1 0 n , 1 , 1 , 1 , 10^n, 1, 1, 1, \ldots , thus we can compute p n p_n by simply inspecting all numbers not greater than 1 0 n ! 10^{n!} . We are guaranteed to find at least n ! n! happy numbers ( 10 , 1 0 2 , 1 0 3 , , 1 0 n ! 10, 10^2, 10^3, \ldots, 10^{n!} ), so p n p_n is defined. Additionally, happy numbers are positive integers by definition, so p n p_n is a positive integer for all n n . This proves that our expression f ( A P R I L , F O O L S , 2015 ) f(APRIL, FOOLS, 2015) is indeed a defined expression, and our result above shows that f ( A P R I L , F O O L S , 2015 ) = 2015 f(APRIL, FOOLS, 2015) = \boxed{2015} .

John Wang
Apr 9, 2015

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

Stewart Gordon
Apr 2, 2015

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

These decrease the value of a + b + c 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 ) 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 ) 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 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 ) 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 f will eventually terminate. Since there is only one terminating condition, with a constant value, we know that f f is just a constant function.

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