The set $A=\{1,2,3, \ldots, 2044,2045\}$ contains 2045 elements. A subset $S$ of $A$ is called triple-free if no element of $S$ equals three times another element of $S$. For example, $\{1,2,4,5,10,2043\}$ is triple-free, but $\{1,2,4,5,10,681,2043\}$ is not triplefree. The triple-free subsets of $A$ that contain the largest number of elements contain exactly 1535 elements. There are $n$ triple-free subsets of $A$ that contain exactly 1535 elements. The integer $n$ can be written in the form $p^{a} q^{b}$, where $p$ and $q$ are distinct prime numbers and $a$ and $b$ are positive integers. If $N=p^{2}+q^{2}+a^{2}+b^{2}$, then the last three digits of $N$ are
(A) 202
(B) 102
(C) 302
(D) 402
(E) 502
We begin by partitioning the set $A$ into disjoint subsets of the form

$$
P_{b}=\left\{b, 3 b, 9 b, \ldots, 3^{k} b\right\}
$$

where $b$ is a positive integer with $1 \leq b \leq 2045$ and $k$ is the largest non-negative integer with $3^{k} b \leq 2045$.

The first two of these sets are $P_{1}=\{1,3,9,27,81,243,729\}$ and $P_{2}=\{2,6,18,54,162,486,1458\}$. These sets have $b=1$ and $b=2$, and thus $k=6$.

We will see that each element of $A$ is an element of exactly one of these sets.

Since $3^{7}=2187$, then $3^{7} b \geq 2187>2045$ for every positive integer $b$, so $k=6$ is the largest possible value of $k$ that we can have.

For each $k$ with $0 \leq k \leq 6$, we determine the range of values of $b$ that give that $k$ :

- $k=6: 1 \leq b \leq 2$ since $2 \cdot 3^{6}=14582045$
- $k=5: 3 \leq b \leq 8$ since $8 \cdot 3^{5}=19442045$
- $k=4: 9 \leq b \leq 25$ since $25 \cdot 3^{4}=20252045$
- $k=3: 26 \leq b \leq 75$ since $75 \cdot 3^{3}=20252045$
- $k=2: 76 \leq b \leq 227$ since $227 \cdot 3^{2}=20432045$
- $k=1: 228 \leq b \leq 681$ since $681 \cdot 3^{1}=20432045$
- $k=0: 682 \leq b \leq 2045$ since $2045 \cdot 3^{0}=2045$ and $2046 \cdot 3^{0}=2046>2045$

Since we want to create disjoint sets $P_{b}$ whose union is the set $A=\{1,2,3, \ldots, 2044,2045\}$, then we exclude all $b$ 's that are multiples of 3. (If $b$ were a multiple of 3 , it would appear as an element of a set $P_{r}$ with $r \leq b$.)

For each of the ranges above, we count the number of multiples of 3 that we need to exclude:

| $k$ | Range of values of $b$ | \# of multiples of 3 | \# of remaining $b$ | \# of elements in each $P_{b}$ |
| :---: | :---: | :---: | :---: | :---: |
| 6 | $1 \leq b \leq 2$ | 0 | $2-0=2$ | 7 |
| 5 | $3 \leq b \leq 8$ | 2 | $6-2=4$ | 6 |
| 4 | $9 \leq b \leq 25$ | 6 | $17-6=11$ | 5 |
| 3 | $26 \leq b \leq 75$ | 17 | $50-17=33$ | 4 |
| 2 | $76 \leq b \leq 227$ | 50 | $152-50=102$ | 3 |
| 1 | $228 \leq b \leq 681$ | 152 | $454-152=302$ | 2 |
| 0 | $682 \leq b \leq 2045$ | 454 | $1364-454=910$ | 1 |

Note for example that the range $26 \leq b \leq 75$ contains $75-25=50$ integers and includes the multiples of 3 from $9 \cdot 3$ to $25 \cdot 3$, inclusive, so includes $25-8=17$ multiples of 3 , and the range $228 \leq b \leq 681$ contains $681-227=454$ integers and includes the multiples of 3 from $76 \cdot 3$ to $227 \cdot 3$, inclusive, so includes $227-75=152$ multiples of 3 .

An integer $b$ in $A$ that is not a multiple of 3 generates a set $P_{b}$ and cannot appear in a different set $P_{r}$. Any multiple of 3 , say $m$, will appear in exactly one $P_{b}$ where the value of $b$ (which is itself not a multiple of 3 ) is obtained by dividing out all of factors of 3 from $m$.

We note that $2 \cdot 7+4 \cdot 6+11 \cdot 5+33 \cdot 4+102 \cdot 3+302 \cdot 2+910 \cdot 1=2045$ so the union of these sets $P_{b}$ includes enough elements to be the entire set $A$. No element of $A$ appears in more than one $P_{b}$, so the union of these sets $P_{b}$ is the entire set $A$.

Why are these sets useful in the given problem?

A subset of $A$ is triple-free if it does not contain two elements, one of which is three times the other.

This means that a subset of $A$ is triple-free exactly when it does not contain two consecutive elements of any $P_{b}$.

A triple-free subset $T$ of $A$ that contains as many elements as possible will contain as many elements as possible from each of the $P_{b}$ defined above.

Since no two consecutive elements of any $P_{b}$ can appear in $T$, then a $P_{b}$ with 7 elements can contribute at most 4 elements to $T$ (every other element starting with the first), a $P_{b}$ with 6 elements can contribute at most 3 elements to $T$ (it cannot contain 4 elements without having two consecutive elements), and so on.

For each $k$ from $k=7$ to $k=0$, each $P_{b}$ above can contribute at most the following number of elements to $T$ :

| $k$ | \# of elements in each $P_{b}$ | Number of elements that can be chosen |
| :---: | :---: | :---: |
| 6 | 7 | 4 |
| 5 | 6 | 3 |
| 4 | 5 | 3 |
| 3 | 4 | 2 |
| 2 | 3 | 2 |
| 1 | 2 | 1 |
| 0 | 1 | 1 |

This means that $T$ can contain at most $2 \cdot 4+4 \cdot 3+11 \cdot 3+33 \cdot 2+102 \cdot 2+302 \cdot 1+910 \cdot 1=1535$ elements. This agrees with the information in the problem statement.

Where is there choice in creating $T$ ?

If a $P_{b}$ contains 1 element, then this element must be chosen for $T$.

If a $P_{b}$ contains 3 elements, then there is 1 way to choose 2 elements for $T$, since in this case the 1st and 3rd elements (in increasing order) must be chosen.

If a $P_{b}$ contains 5 elements, then there is 1 way to choose 3 elements for $T$, since in this case the 1st, 3rd and 5th elements (in increasing order) must be chosen.

If a $P_{b}$ contains 7 elements, then there is 1 way to choose 4 elements for $T$, since in this case the 1st, 3rd, 5th and 7th (in increasing order) elements must be chosen.

If a $P_{b}$ contains 2 elements, then there are 2 ways to choose 1 element for $T$.

If a $P_{b}$ contains 4 elements, then there are 3 ways to choose 2 elements for $T$. (If we want to choose 2 elements from $\{A, B, C, D\}$ without choosing consecutive elements, we can choose $A$ and $C$, or $A$ and $D$, or $B$ and $D$.)

If a $P_{b}$ contains 6 elements, then there are 4 ways to choose 3 elements for $T$. (If we want to choose 3 elements from $\{A, B, C, D, E, F\}$ without choosing consecutive elements, we can choose $A$ and $C$ and $E$, or $A$ and $C$ and $F$, or $A$ and $D$ and $F$, or $B$ and $D$ and $F$.)

This means that the number of ways of choosing a triple-free subset $T$ of $A$ with as many elements as possible is $1^{2} \cdot 4^{4} \cdot 1^{11} \cdot 3^{33} \cdot 1^{102} \cdot 2^{302} \cdot 1^{910}$.

This is because there are 2 sets $P_{b}$ with $k=6$ and 1 way of choosing 4 elements from each of them (giving $1^{2}$ choices in total), 4 set $P_{b}$ with $k=5$ and 4 ways of choosing 3 elements from each of them (giving $4^{4}$ choices in total), and so on.

The seven factors in this product give, from $k=6$ to $k=0$, the number of ways of choosing the maximum number of elements from a set $P_{b}$ corresponding to that value of $k$ raised to the power of the number of such sets $P_{b}$.

Therefore, the number of ways equals $4^{4} \cdot 3^{33} \cdot 2^{302}=2^{310} \cdot 3^{33}$.

From the statement of the problem, we thus have $N=2^{2}+3^{2}+310^{2}+33^{2}=97202$.

The final three digits of this integer are 202.