Set Theory is difficult, isn't it?

From 1) and 2) we get that every number greater than 2 that is 2 mod 5 is also in S.

Using 3) once tells us that 6 is in S since 2 * 3 = 6. This combined with 2) tells us that every number greater than 6 that is 1 mod 5 is also in S.

Using 3) again tells us that 18 is in S since 6 * 3 = 18. This combined with 2) tells us that every number greater than 18 that is 3 mod 5 is also in S.

Using 3) once more tells us that 54 is in S since 18 * 3 = 54. This combined with 2) tells us that every number greater than 54 that is 4 mod 5 is also in S.

One more application of 3) tells us that 162 is in S since 54 * 3 = 162. We already knew this, however, because 162 is greater than 2 and is 2 mod 5. Repeated applications of 3) from here yield no new elements of S.

This process tells us that every number that is greater than 54 that is either 1, 2, 3, or 4 mod 5 is in S. Therefore, our answer must be 0 mod 5.

It can be shown that no number that is 0 mod 5 is in S. Via Siddhartha Srivastava :

Let the smallest member of S divisible by 5 be $x$ . We either get $x$ from $x - 5$ or $\frac{x}{3}$ . But both of these are smaller than $x$ and are divisible by 5, which is a contradiction, since $x$ is the smallest member divisible by 5. Therefore no member of S is divisible by 5.

This means that 2005 is our answer since it is the largest number in the given set of possible answers that is 0 mod 5.

Every element of $S$ can be expressed as $a (i,j) = 2\dot{} 3^i + 5j$ , where $i,j = 0,1,2,....$

Even for $a(m,n) = 3a(i,j) = 3(2\dot{} 3^i + 5j)$ $\quad \quad \quad \quad \quad \quad \space = 2\dot{} 3^{i+1} + 5(3j) = a(i+1,3j)$ .

We note that $a (i,j) \equiv 2\dot{} 3^i + 5j \equiv (5+1)3^{i-1} + 5j \equiv 3^{i-1} \pmod{5}$ for $i>0$ . Since $3^{i-1} \ne 0$ , this means that multiples of $5$ are not members of $S\quad \Rightarrow \overline{S}_{5|} = \{ 5,10,15,...,2005 \}$

We know that integer with units digit being $2$ are $\in S$ , because, for $k=0,1,2,...$ , we have:

$n_2 = 10k + 2 = 2\dot{}3^0 + 5(2k) \in S \text{ for all }k\quad \Rightarrow \overline{S}_{2} = \{ \space \}$

Similarly,

\(\begin{array} {} n_7 = 10k + 7 = 2+ 5 + 5(2k-1) & \Rightarrow \overline{S}_{7} = \{ \space \}\\ n_6 = 10k + 6 = 6 + 5(2k) & \Rightarrow \overline{S}_{6} = \{ \space \}\\ n_1 = 10k + 1 = 6+ 5 + 5(2k-1) & \Rightarrow \overline{S}_{1} = \{ 1 \}\\ n_8 = 10k + 8 = 18+ 5(2k) & \Rightarrow \overline{S}_{8} = \{ 8 \}\\ n_3 = 10k + 3 = 18+ 5 + 5(2k-1) & \Rightarrow \overline{S}_{3} = \{ 3,13 \} \\ n_4 = 10k + 4 = 54+ 5(2k) & \Rightarrow \overline{S}_{4} = \{ 4,14,24,34,44 \}\\ n_9 = 10k + 9 = 54+ 5 + 5(2k-1) & \Rightarrow \overline{S}_{9} = \{ 9,19,29,39,49 \} \end{array} \)

$\overline{S} = \overline{S}_0 \cup \overline{S}_1 \cup \overline{S}_2 \cup \overline{S}_3 \cup \overline{S}_4 \cup \overline{S}_5 \cup \overline{S}_6 \cup \overline{S}_7 \cup \overline{S}_8 \cup \overline{S}_9 \\ \quad = \{ 1,3,4,5,8,9,10,13,14,15,19,20,24,25,29,30,34,35,39, \\ \quad \quad \quad 40,44,45,49,50,55,60,...,1995, 2000, 2005 \}$

Therefore the largest integer that does not $\in S$ is $\boxed{2005}$ .

Unit digits of first few numbers. For (n+5):- $\color{#D61F06}{ 2, 7, 2....repeats}$ .
For 3n:- 2, 6, 18, 54, 162 >>>- $\color{#D61F06}{ 2, 6, 8, 4, 2....repeats}$
Applying (n+5) on these, :-7, 11, 13, 9>>>> $\color{#D61F06}{7, 1, 3, 9}$ .
So all 1, 2, 3, 4, 6, 8, 9 unit digits greater than 54 are covered by (n+5) or 3n.
In fact 3n only supply seed of 6, 8, and 4 for numbers greater than 54.
It is (n+5) that covers all digits except 0 and 5 for numbers greater than 54.
So the highest number is 2005. Second highest is 2000.

We note that each element is of the form 2 (3^a)+5y. Proof; We definitely know that each element is of the form (2+5k) (3^a) +5z= 2*(3^a) + multiple of 5 as required. There is one thing an element will never be... a multiple of 5. Hence 2005 is the largest answer.

2,7,12,17,22,.…,2002,2007_ 6,11,16,21,…2001,2006_ 18,23,28,…,2003,2008_ For 2005 : 1_ have 2005/3 Or 2_ have 2000 . so 1 it is not integer ,2 in S it is not any 0,or 5 in the last digit!

Is there even any reason why 5 cannot be in S? The rule never mentions about what isn't a member of S. Even if it's not derived from a known member of S, why can't I include it just because I want it?

Case work!

Based on the clues, we know that the 2 is going to be important. The other two clues are all the other clues we need. Basically we need an exclusionary system that will take out all of the numbers that we don't want: the numbers in the set S.

Exclude numbers that are: 1) Numbers in form 5n+2 2) Numbers in form 3^n * 2 3) Numbers in form 3^n(5n+2) 4) Numbers in form 3^n*2 + 5n

1) It's simple. No numbers that end with 2 or 7 (since multiples of 5 repeat, so 5+2 or 10+2 = 7 or 12 so the units digit shouldn't be 7 or 2.)

2) No doubles of a power of 3. Seems easy enough to classify.

3) No multiples of three that end with 1 or 6, since (5n+2) ends with 2 or 7, and multiplying that by 3 we get 1 or 6.

4) No numbers that end in 6 8 4 or 2. (It's pretty obvi why)

So we test our luck a little here, it can't be numbers that end in 1,2,3,4,6,7,8. Testing the first number that doesn't have those digits, 2005, we get 2005.

Sorry for the bashy solution, I'm bad at math.