USAMTS 2014-2015 Round 1

William Zhang 22072 Round 1 Problem 1

1.

William Zhang 22072 Round 1 Problem 2 2. In order for 𝑥, 𝑦, 𝑧, 𝑥 − 𝑦, 𝑦 − 𝑧, and 𝑥 − 𝑧 to all be positive integers, we must have 𝑥 > 𝑦 > 𝑧 > 0. Lemma: The only ordered triplet (𝑎, 𝑏, 𝑐) of consecutive odd numbers that satisfy 𝑎 > 𝑏 > 𝑐 and 𝑎, 𝑏, 𝑐 ∈ {prime numbers} is (𝑎, 𝑏, 𝑐) = (7, 5, 3) Proof for lemma: Any triplet of consecutive positive odd numbers can be expressed in the form (𝑛 + 4, 𝑛 + 2, 𝑛), where 𝑛 is some natural number. Notice that each of 𝑛 + 4, 𝑛 + 2, and 𝑛 all have a unique remainder when divided by three. Thus, at least one of 𝑛 + 4, 𝑛 + 2, and 𝑛 have a remainder of 0 when divided by three, so it is divisible by 3. However, a prime number can only have two positive divisors which are 1 and the number itself. So, the number in the set {𝑛, 𝑛 + 2, 𝑛 + 4} that is divisible by 3 must be equal to 3 in order to be prime. After testing the possibilities 𝑛 = 3, 𝑛 + 2 = 3, and 𝑛 + 4 = 3, only the equation 𝑛 = 3 makes (𝑛 + 4, 𝑛 + 2, 𝑛) a triplet of prime numbers, so the only ordered triplet (𝑎, 𝑏, 𝑐) of consecutive odd numbers that satisfy 𝑎 > 𝑏 > 𝑐 and 𝑎, 𝑏, 𝑐 ∈ {prime numbers} is (𝑎, 𝑏, 𝑐) = (𝑛 + 4, 𝑛 + 2, 𝑛) = (3 + 4, 3 + 2, 3) = (7, 5, 3) We can categorize all possible solutions into exactly one of the following cases:

All of 𝑥, 𝑦, and 𝑧 are odd numbers that satisfy 𝑥, 𝑦, 𝑧, 𝑥 − 𝑦, 𝑦 − 𝑧, and 𝑥 − 𝑧 are all prime. o There are no solutions under this case. If 𝑥, 𝑦, and 𝑧 are all odd, then that means that 𝑥 − 𝑧

and 𝑦 − 𝑧 are both even prime numbers, so 𝑥 − 𝑧 = 2 and 𝑦 − 𝑧 = 2. By the transitive property, we would have 𝑥 − 𝑧 = 𝑦 − 𝑧 ⟹ 𝑥 = 𝑦, which contradicts the strict inequality from earlier, which stated 𝑥 > 𝑦.

Exactly one of 𝑥, 𝑦, and 𝑧 are even while the other two are odd. o We can conclude that 𝑧 must be the even number because 2 is the smallest prime number

and 𝑥 > 𝑦 > 𝑧. This means that 𝑥 and 𝑦 are both odd, so 𝑥 − 𝑦 is even and thus 𝑥 − 𝑦 = 2 ⟹ 𝑦 = 𝑥 − 2. Also, because 𝑦 − 𝑧 is prime, we have 𝑦 − 𝑧 = 𝑦 − 2 = 𝑥 − 4 is prime. Thus we have three prime numbers (𝑥, 𝑦, 𝑦 − 𝑧) = (𝑥, 𝑥 − 2, 𝑥 − 4) which are also consecutive odd numbers. By the lemma, the only solution to this is (𝑥, 𝑦, 𝑦 − 𝑧) = (7, 5, 3) ⟹ (𝑥, 𝑦, 𝑧) = (7, 5, 2)

At least two of 𝑥, 𝑦, and 𝑧 are even. o There are no solutions under this case. If at least two of 𝑥, 𝑦, and 𝑧 are even, then these two

even prime are both equal to 2, and then by the transitive property, would be equal to each other. This contradicts the strict inequality 𝑥 > 𝑦 > 𝑧.

William Zhang 22072 Round 1 Problem 3 3. a) To do this problem, we will use recursion.

In a group with two people with heights 140 and 145, there are two permutations that are in almost-order: the permutation (140, 145) and the permutation (145, 140) In a group of three people with heights 140, 145, and 150, there are three permutations that are in almost-order: the permutations (140, 145, 150), (145, 140, 150), and (140, 150, 145). From the definition of a permutation that is in almost-order, we see that a person of height 135 + 5(𝑚 + 2) cannot be to the left of a person of height 135 + 5𝑚, so each person is changed from their original position (the position they would have if they were in increasing order) by at most 1 place. In a group of 𝑛 people with heights 140, 145, 150,… , and 135 + 5𝑛, any almost-order can be constructed from one of the two following cases:

The person of height 135 + 5𝑛 (tallest person) can be added to the back of a line of length 𝑛 − 1

The person of height 135 + 5𝑛 can be added to the back of a line of length 𝑛 − 2 and then the person of height 130 + 5𝑛 (second-tallest person) can be placed behind the tallest person so that the second-tallest person is at the end of the line and the tallest person is in the second-to-last position in line.

Therefore we have the following recursion: 𝑥𝑛 = 𝑥𝑛−1 + 𝑥𝑛−2, where 𝑥𝑛 is the number of almost-ordered permutations when considering the first 𝑛 shortest people. 2, 3, 5, 8, 13, 21, 34, 55, 89, so our answer is 89 b) We use a very similar recursive strategy.

We define 𝐴𝑛 to be the number of almost-ordered permutations when considering the first 𝑛 shortest

people.

We can use the same Fibonacci-like sequence for the first 9 people of heights 120, 125, 130, 135, 140,

145, 150, 155, and 160. We find that there are 𝐴9 = 55 almost-order permutations, if we only consider

the people with heights between 120 and 160, inclusive.

To construct a permutation that includes all the people of heights between 120 and 164, inclusive, we

can:

Add the person of height 164 to the back of the line that contains the first 9 shortest people, or

Add the person of height 164 to the back of the line that contains the first 8 shortest people then

add the person of height 160 to the back of the new line so that the person of height 264 is second-

to-last.

From here, we can calculate 𝐴10 = 𝐴9 + 𝐴8 = 55 + 34 = 89.

To construct a permutation that includes the first 11 shortest people, we can:

Add the person of height 165 to the back of the line that contains the first 10 shortest people,

Add the person of height 165 to the of the line that contains the first 9 shortest people, and then

place either the person of height 160 or the person of height 164 behind the person of height 165,

or

Add the person of height 165 to the back of the line that contains the first 8 shortest people, and

then place the persons of heights 160 and 164 at the end of the line, in either order.

We calculate that 𝐴11 = 𝐴10 + 2𝐴9 + 2𝐴8 = 89 + 2 × 55 + 2 × 34 = 267

William Zhang 22072 Round 1 Problem 3

To construct a permutation that includes the first 12 shortest people, we can:

Add the person of height 170 to the back of the line that contains the first 11 shortest people,

Add the person of height 170 to the back of the line that contains the first 10 shortest people and

then place either the person of height 165 or the person of height 164 behind the person of height

170, or

Add the person of height 170 to the back of the line that contains the first 9 shortest people and add

both people of heights 164 and 165 behind the person of height 170.

We calculate that 𝐴12 = 𝐴11 + 2𝐴10 + 2𝐴9 = 267 + 2 × 89 + 2 × 55 = 555

Constructing permutation that include the first 13, 14, 15, 16, 17, 18, 19, and 20 shortest people uses

the same Fibonacci-like algorithm as constructing permutations for part a of this problem because the

common difference is now equal to 5.

𝐴13 = 𝐴12 + 𝐴11 = 555 + 267 = 822

𝐴14 = 𝐴13 + 𝐴12 = 822 + 555 = 1377

𝐴15 = 𝐴14 + 𝐴13 = 1377 + 822 = 2199

𝐴16 = 𝐴15 + 𝐴14 = 2199 + 1377 = 3576

𝐴17 = 𝐴16 + 𝐴15 = 3576 + 2199 = 5775

𝐴18 = 𝐴17 + 𝐴16 = 5775 + 3576 = 9351

𝐴19 = 𝐴18 + 𝐴17 = 9351 + 5775 = 15126

𝐴20 = 𝐴19 + 𝐴18 = 15126 + 9351 = 24477

The answer is 24477

William Zhang 22072 Round 1 Problem 4 4. Definition: a polygon is inscribed in a circle iff the vertices of the polygon are also points on the circle. Lemma 1: There exists a regular 𝑛-gon that passes through a set of points on a circle iff the arcs between

any two of the points that lie on the circle can be written in the form 2𝜋𝑚

𝑛, where 𝑚 is some natural number

less than 𝑛. Proof for Lemma:

Let 𝐴 and 𝐵 be two arbitrary points that lie on the circle. We will first show that if there exists a regular 𝑛-gon inscribed in circle 𝑂 that has vertices 𝐴 and 𝐵, then the measure of minor arc 𝐴𝐵 can be written in the

form 2𝜋𝑚

𝑛.

We know that an inscribed regular 𝑛-gon would break the circumference of the circle into 𝑛 arcs of equal measure and that the total circumference has a measure of 2𝜋 radians, so each piece of the circumference

would have measure 2𝜋

𝑛. Since 𝐴 and 𝐵 are vertices of the regular 𝑛-gon, we know that the measure of

arc 𝐴𝐵 can be written as the number of arcs that have a measure of 2𝜋

𝑛 between 𝐴 and 𝐵 times the measure

of each of these arcs, so the product would be 2𝜋𝑚

𝑛, where 𝑚 is the number of these arcs between 𝐴 and 𝐵.

Now we need to prove that if the measure of minor arc 𝐴𝐵 can be written in the form 2𝜋𝑚

𝑛, then there exists

a regular 𝑛-gon inscribed in circle 𝑂 that has vertices at points 𝐴 and 𝐵. Since 𝑚 is an integer, we can place points along the circumference of circle 𝑂 such that the arcs between

any two consecutive points has a measure of 2𝜋

𝑛, there are 𝑛 of these arcs, and arc 𝐴𝐵 is made of 𝑚 of these

arcs. Because these arcs all have equal measures, we can conclude that the endpoints of these arcs are the vertices of the inscribed regular 𝑛-gon. Lemma 2: If two circles of centers 𝑂1 and 𝑂2 that have equal radius intersect at two points, 𝑀 and 𝑁, then the measure of minor arc 𝑀𝑁 of circle 𝑂1 is equal to the measure of minor arc 𝑀𝑁 of circle 𝑂2. Proof for Lemma: We wish to prove that the 2 bolded arcs have equal measures.

William Zhang 22072 Round 1 Problem 4

We know that 𝑀𝑂1 ≅ 𝑁𝑂1 ≅ 𝑀𝑂2 ≅ 𝑁𝑂2 because we are given that the two circles have equal radii, and all the radii of the same circle are congruent. We also have 𝑀𝑁 ≅ 𝑀𝑁 by the reflexive property. Therefore, we can conclude that triangle 𝑀𝑂1𝑁 ≅ 𝑀𝑂2𝑁 by the SSS congruence postulate. Therefore, by CPCTC, we have ∠𝑀𝑂1𝑁 ≅ ∠𝑀𝑂2𝑁 ⇒ 𝑚∡𝑀𝑂1𝑁 = 𝑚∡𝑀𝑂2𝑁 Because the central angles of the two arcs have equal measures, and the measure of an arc is equal to the measure of its central angle, we can conclude that the measure of minor arc 𝑀𝑁 of circle 𝑂1 is equal to the measure of minor arc 𝑀𝑁 of circle 𝑂2. Solution to problem: We are given that circles 𝜔𝑝 and 𝜔𝑞 intersect at points 𝐴 and 𝐵 and that point 𝐵 lies inside circle 𝜔, but

point 𝐴 lies outside circle 𝜔. We are also given that regular polygon 𝑃 and regular polygon 𝑄 are inscribed in circles 𝜔𝑝 and 𝜔𝑞,

respectively, and that points 𝐶 and 𝐷 are vertices of 𝑃 and points 𝐸 and 𝐹 are vertices of 𝑄. By the definition of an inscribed polygon, points 𝐶 and 𝐷 lie on 𝜔𝑝 and points 𝐸 and 𝐹 lie on 𝜔𝑞.

We are also given that 𝜔 intersects all four points 𝐶, 𝐷, 𝐸, and 𝐹.

William Zhang 22072 Round 1 Problem 4 From this information, we can draw the circles to look like this:

We are also given that points 𝐷, 𝐵, and 𝐶 lie on a regular 𝑛-gon that is inscribed in 𝑤𝑝, and

points 𝐸, 𝐵, and 𝐹 lie on a regular 𝑛-gon that is inscribed in 𝑤𝑞. By Lemma 1, we can conclude that

arc 𝐸𝐵𝐹 has a measure that can be written in the form 2𝜋𝑚1

𝑛, where 𝑚1 is positive integer less than 𝑛,

and that arc 𝐷𝐵𝐶 has a measure that can be written in the form 2𝜋𝑚2

𝑛, where 𝑚2 is positive integer less

than 𝑛. By Lemma 2, we know that measure of arc 𝐸𝐵𝐹 = measure of arc 𝐸𝐷𝐹 and that measure of arc 𝐷𝐵𝐶 = measure of arc 𝐷𝐸𝐶, so

we also know that arc 𝐸𝐷𝐹 has a measure that can be written in the form 2𝜋𝑚1

𝑛 and that arc 𝐷𝐸𝐶 has a

measure that can be written in the form 2𝜋𝑚2

𝑛. We again use Lemma 1 to solve the problem.

William Zhang 22072 Round 1 Problem 5 5. Let 𝑓(𝑥) be a function with a domain of all nonnegative integers such that 𝑓(𝑥) = 𝑎𝑥. (In other words, this

just means that 𝑓(𝑥) is a function that returns the 𝑥𝑡ℎ term in the sequence.)

We are given that 𝑓(2) = 5, 𝑓(2014) = 2015, and 𝑓(𝑥) = 𝑓(𝑓(𝑥 − 1)).

We will call the statement 𝑓(𝑥) = 𝑓(𝑓(𝑥 − 1)) Theorem 1

In this solution, we will use the notation 𝑓𝑚(𝑥) = 𝑓(𝑓(𝑓(…𝑓⏟ 𝑚 repeats of 𝑓

(𝑥)… ))), for example,

𝑓4(𝑥) = 𝑓 (𝑓 (𝑓(𝑓(𝑥)))).

A theorem that will be used often in this proof is 𝑎 = 𝑏 ⟹ 𝑔(𝑎) = 𝑔(𝑏), for all functions 𝑔(𝑥). In this proof, this theorem will be denoted as Theorem 2. Lemma 1: 𝑓𝑚(𝑥) = 𝑓𝑚+1(𝑥 − 1) for all positive integers 𝑚 Proof for Lemma 1:

𝑓𝑚(𝑥) = 𝑓𝑚−1(𝑓(𝑥)) = 𝑓𝑚−1 (𝑓(𝑓(𝑥 − 1))) = 𝑓𝑚+1(𝑥 − 1)

Also, repeated use of this lemma results in 𝑓𝑚(𝑥) = 𝑓𝑚+𝑘(𝑥 − 𝑘) where 𝑘 ∈ ℕ and 𝑘 < 𝑛. Lemma 2: 𝑓𝑚(3) = 𝑓𝑚(5).

Proof for Lemma 2: Using the property 𝑓(𝑥) = 𝑓(𝑓(𝑥 − 1)) and 𝑓(2) = 5, we apply Theorem 2 twice to get

𝑓(3) = 𝑓(𝑓(2)) = 𝑓(5) ⟹ 𝑓𝑚−1(𝑓(3)) = 𝑓𝑚−1(𝑓(5)) ⟹ 𝑓𝑚(3) = 𝑓𝑚(5).

From here, we use Lemma 1 to get

𝑓(4) = 𝑓2(3), 𝑓3(4) = 𝑓4(3), 𝑓5(4) = 𝑓6(3), 𝑓7(4) = 𝑓8(3),… and

𝑓2(5) = 𝑓3(4), 𝑓4(5) = 𝑓5(4), 𝑓6(5) = 𝑓7(4), 𝑓8(5) = 𝑓9(4),…. We can use Lemma 2 to get 𝑓2(3) = 𝑓2(5), 𝑓4(3) = 𝑓4(5), 𝑓6(3) = 𝑓6(5), 𝑓8(3) = 𝑓8(5),… And then apply the transitive property on the previous three highlighted lines to get

𝑓(4) = 𝑓2(3) = 𝑓2(5) = 𝑓3(4) = 𝑓4(3) = 𝑓4(5) = 𝑓5(4) = 𝑓6(3) = 𝑓6(5) = ⋯ 𝑓(4) = 𝑓2(5) = 𝑓4(5) = 𝑓6(5) = ⋯.

We can use the statement𝑓𝑚(𝑥) = 𝑓𝑚+𝑘(𝑥 − 𝑘) and the statement on the previous line to get 𝑓1(6) = 𝑓2(5) = 𝑓(4) 𝑓1(8) = 𝑓4(5) = 𝑓(4) 𝑓1(10) = 𝑓6(5) = 𝑓(4) …. Thus, we have 𝑓(4) = 𝑓(6) = 𝑓(8) = 𝑓(10) = ⋯ = 𝑓(2014) = ⋯. By the transitive property, because 𝑓(2014) = 2015, we have 𝑓(4) = 𝑓(6) = 𝑓(8) = 𝑓(10) = ⋯ = 2015. Let the statement on the previous line be called Lemma 3.

We can use the fact 𝑓𝑚(𝑥) = 𝑓𝑚+𝑘(𝑥 − 𝑘) to get 𝑓𝑚(5) = 𝑓𝑚+2(3), so

𝑓(5) = 𝑓3(3), 𝑓3(5) = 𝑓5(3), 𝑓5(5) = 𝑓7(3),…, and then we can use Lemma 2 to get 𝑓(3) = 𝑓(5), 𝑓3(3) = 𝑓3(5), 𝑓5(3) = 𝑓5(5),…. We can now use the equations on the previous two lines as well as the transitive property to get 𝑓(3) = 𝑓(5) = 𝑓3(3) = 𝑓3(5) = 𝑓5(3) = 𝑓5(5) = ⋯ 𝑓(3) = 𝑓(5) = 𝑓3(5) = 𝑓5(5) = 𝑓7(5) = ⋯

William Zhang 22072 Round 1

Then we can use the equation 𝑓𝑚(𝑥) = 𝑓𝑚+𝑘(𝑥 − 𝑘) again as well as the transitive property to get 𝑓1(7) = 𝑓3(5) = 𝑓(3) 𝑓1(9) = 𝑓5(5) = 𝑓(3) 𝑓1(11) = 𝑓7(5) = 𝑓(3) … Now we can use the transitive property to get the equations 𝑓(3) = 𝑓(5) = 𝑓(7) = 𝑓(9) = 𝑓(11) = ⋯ = 𝑓(2015) = ⋯. We can place all possible sequences 𝑎0, 𝑎1, 𝑎2, 𝑎3, 𝑎4, 𝑎5, 𝑎6, 𝑎7, 𝑎8, … under exactly one of the following

three cases:

𝑎1 = 𝑓(1) = 0

o 𝑓(1) = 0 ⟹ 𝑓(𝑓(1)) = 𝑓(0) by Theorem 2, and therefore

𝑓(2) = 𝑓(0) = 5 because 𝑓(2) = 𝑓(𝑓(1)) and 𝑓(2) = 5. We also have

𝑓(1) = 𝑓(𝑓(0)) = 𝑓(5), so

𝑓(1) = 0 ⟹ 𝑓(1) = 𝑓(5) = 0 ⟹ 𝑓(𝑓(5)) = 𝑓(0), where the last statement is true by

Theorem 1.

𝑓(6) = 2015 by Lemma 3, and 𝑓(6) = 𝑓(𝑓(5)) = 𝑓(0) = 5 would imply that 2015 = 5

which is a contradiction. Therefore, there are no possible sequences under this case.

𝑎1 = 𝑓(1) = 1

o 𝑓(1) = 1 ⟹ 𝑓(𝑓(1)) = 𝑓(1) by Theorem 2, but we also have 𝑓(𝑓(1)) = 𝑓(2) = 5, so by

transitive property, we would have 𝑓(1) = 5, which contradicts the statement 𝑓(1) = 1. Therefore, there are no possible sequences under this case.

𝑎1 = 𝑓(1) = 2 o 𝑓(1) = 2 ⟹ 𝑓(𝑓(0)) = 𝑓(1) = 2. We can divide this case in which 𝑎1 = 2 into multiple

sub-cases.

If 𝑓(0) = 0, then 𝑓(𝑓(0)) = 𝑓(0) = 0 by Theorem 1 and the transitive property,

which contradicts the statement 𝑓(𝑓(0)) = 2.

If 𝑓(0) = 1, then our sequence becomes {𝑎𝑖|𝑖 ∈ ℤ≥0} = 1, 2, 5, 𝑓(3), 2015, 𝑓(3), 2015, 𝑓(3), 2015,… From here, we see that 𝑓(3) ∈ {even numbers ≥ 4} ∪ {2015} clearly gives possible

values of 𝑓(3). On the other hand, we see that 𝑓(3) = 2 ⇒ 𝑓(𝑓(3)) = 𝑓(2) = 5,

but 𝑓(4) = 𝑓(𝑓(3)) ⇒ 𝑓(𝑓(3)) = 2015 contradicts this. We also see that

𝑓(3) ∈ {positive odd numbers ≠2015} ⇒ 𝑓(𝑓(3)) = 𝑓(1), 𝑓(3), 𝑓(5), 𝑓(7), …

would give a contradiction to the statement 𝑓(4) = 𝑓(𝑓(3)) ⇒ 𝑓(𝑓(3)) = 2015

because if 𝑓(3) ≠ 2015, then we see that 𝑓(any odd number) ≠ 2015.

If 𝑓(0) = 2, then 𝑓(𝑓(0)) = 𝑓(2) = 5 by Theorem 1 and because 𝑓(2) = 5, but

this contradicts the statement 𝑓(𝑓(0)) = 𝑓(1) = 2.

If 𝑓(0) = 3, 5, 7, 9, …, then 𝑓(𝑓(0)) = 𝑓(3), 𝑓(5), 𝑓(7), 𝑓(9), … = 𝑓(3). We can use

the statement 𝑓(𝑓(0)) = 𝑓(1) = 2 and transitive property to establish that

𝑓(3) = 2 ⟹ 𝑓(𝑓(3)) = 𝑓(2) = 5 by Theorem 2, but by Theorem 1, we

have 𝑓(4) = 𝑓(𝑓(3)), and Lemma 3 gives 𝑓(4) = 2015, so we would

have 𝑓(𝑓(3)) = 2015, which contradicts 𝑓(𝑓(3)) = 5, giving no possible

sequences under this case.

If 𝑓(0) = 4, 6, 8, 10,…, then 𝑓(𝑓(0)) = 𝑓(4), 𝑓(6), 𝑓(8), 𝑓(10), … = 2015, but this

contradicts the statement 𝑓(𝑓(0)) = 𝑓(1) = 2 because 2015 ≠ 2.

𝑎1 = 𝑓(1) = 3, 5, 7, 9, 11,… o 𝑓(1) = 3, 5, 7, 9, 11,… ⟹ 𝑓(𝑓(1)) = 𝑓(3) by Theorem 2. (Remember that 𝑓(3) = 𝑓(5) =

𝑓(7) = ⋯)

𝑓(2) = 𝑓(𝑓(1)) = 5 and 𝑓(𝑓(1)) = 𝑓(3) would imply that 5 = 𝑓(3) = 𝑓(5) = 𝑓(7) = ⋯,

However, we have 𝑓(4) = 2015 and 𝑓(𝑓(3)) = 𝑓(4), so 𝑓(𝑓(3)) = 2015

𝑓(3) = 5 ⟹ 𝑓(𝑓(3)) = 𝑓(5) = 5 by Theorem 2, which is a contradiction

because 2015 ≠ 5. Therefore, there are no possible sequences under this case.

𝑎1 = 𝑓(1) = 4, 6, 8, 10, 12,… o Because 𝑓(𝑓(𝑥 − 1)) = 𝑓(𝑥), we have 𝑓(𝑓(1)) = 𝑓(2) = 5.

Therefore, we know that 𝑓(1) cannot be equal to 4, 6, 8, 10, 12,… because

𝑓(1) = 4, 6, 8, 10, 12,… ⟹ 𝑓(𝑓(1)) = 𝑓(4), 𝑓(6), 𝑓(8), 𝑓(10), 𝑓(12),… = 2015 by

Theorem 2, which gives a contradiction to the statement 𝑓(𝑓(1)) = 𝑓(2) = 5

because 2015 ≠ 5. After examining all possible cases, we conclude that 𝑓(2015) = 𝑓(3) ∈ {even numbers ≥ 4} ∪ {2015}

𝑓(2015) = 𝑓(3) ∈ {even numbers ≥ 4} ∪ {2015}