Alice has a very simple starting strategy that guarantees a win at least one-third of the time regardless of Bob's strategy.

Randomly choose one of 3, 4, and 5. If she chooses 4 or 5, choose 3 next.

Why this works:

All integers are either $3x$, $3x+1$, or $3x+2$ for some integer $x$. If Bob chooses a multiple of 3 there is a 1 in 3 chance Alice will choose 3 and win right away. If Bob chooses a number that's one more than a multiple of 3, Alice has a 1 in 3 chance of choosing 4 first, so $3x+1-4=3x-3=3(x-1)$ which is a multiple of 3, which is what Alice will choose second. If Bob chooses $3x+2$, then if Alice chooses 5 first $3x+2-5=3(x-1)$ and Alice will get it when she chooses 3 second. So because Bob (nor Alice) knows what Alice will choose first, Bob has no way to defend against Alice's strategy.

I believe that there is a strategy under which Alice can always win by choosing the a certain series of numbers in order, however I have not yet found the right series. My reasoning is as follows:

My strategy for trying to determine the series is similar to how the sliding bolt puzzle works. If you look at this in terms of modular arithmetic, we are simply trying to eliminate possible states and force the solution to move to a single state regardless of where it started from.

The way to analyze a particular series of numbers is to look at in terms of the modulo of the least common multiple of all the numbers. For example, let's look at if 2 were allowed, but the only numbers we could use were 2, 3, 4, 6, 8, and 9. The least common multiple of these numbers is 72, so we are interested in X mod 72. Initially, X mod 72 could be any number between 0 and 71. If we choose 2 first and it's not a multiple of 2, then X-2 mod 72 could be any odd number between 1 and 71. If we then choose 3, then X-2 mod 72 could not be 3, 9, etc. so X-5 mod 72 could not be an odd number or 0, 6, 12, etc. There are many more states than in the sliding bolt puzzle, but I feel like there should be a way to whittle away the possibilities until it has been forced into a single state.

Alice has a very simple starting strategy that guarantees a win at least one-third of the time regardless of Bob's strategy.

Randomly choose one of 3, 4, and 5. If she chooses 4 or 5, choose 3 next.

Why this works:

All integers are either $3x$, $3x+1$, or $3x+2$ for some integer $x$. If Bob chooses a multiple of 3 there is a 1 in 3 chance Alice will choose 3 and win right away. If Bob chooses a number that's one more than a multiple of 3, Alice has a 1 in 3 chance of choosing 4 first, so $3x+1-4=3x-3=3(x-1)$ which is a multiple of 3, which is what Alice will choose second. If Bob chooses $3x+2$, then if Alice chooses 5 first $3x+2-5=3(x-1)$ and Alice will get it when she chooses 3 second. So because Bob (nor Alice) knows what Alice will choose first, Bob has no way to defend against Alice's strategy.

I believe that there is a strategy under which Alice can always win by choosing the a certain series of numbers in order, however I have not yet found the right series. My reasoning is as follows:

My strategy for trying to determine the series is similar to how the sliding bolt puzzle works. If you look at this in terms of modular arithmetic, we are simply trying to eliminate possible states and force the solution to move to a single state regardless of where it started from.

The way to analyze a particular series of numbers is to look at in terms of the modulo of the least common multiple of all the numbers. For example, let's look at if 2 were allowed, but the only numbers we could use were 2, 3, 4, 6, 8, and 9. The least common multiple of these numbers is 72, so we are interested in X mod 72. Initially, X mod 72 could be any number between 0 and 71. If we choose 2 first and it's not a multiple of 2, then X-2 mod 72 could be any odd number between 1 and 71. If we then choose 3, then X-2 mod 72 could not be 3, 9, etc. so X-5 mod 72 could not be an odd number or 0, 6, 12, etc. There are many more states than in the sliding bolt puzzle, but I feel like there should be a way to whittle away the possibilities until it has been forced into a single state.

Alice can always win by choosing the following numbers in order:

3, 7, 6, 16, 12, 5, 4, 9, 11, 10, 8

Examples - 100, 101, and 102:

100 - 3 = 97, 97 - 7 = 90, 90/6 = 15

101 - 3 = 98, 98/7 = 14

102/3 = 34

Explanation to follow

Alice has a very simple starting strategy that guarantees a win at least one-third of the time regardless of Bob's strategy.

Randomly choose one of 3, 4, and 5. If she chooses 4 or 5, choose 3 next.

Why this works:

All integers are either $3x$, $3x+1$, or $3x+2$ for some integer $x$. If Bob chooses a multiple of 3 there is a 1 in 3 chance Alice will choose 3 and win right away. If Bob chooses a number that's one more than a multiple of 3, Alice has a 1 in 3 chance of choosing 4 first, so $3x+1-4=3x-3=3(x-1)$ which is a multiple of 3, which is what Alice will choose second. If Bob chooses $3x+2$, then if Alice chooses 5 first $3x+2-5=3(x-1)$ and Alice will get it when she chooses 3 second. So because Bob (nor Alice) knows what Alice will choose first, Bob has no way to defend against Alice's strategy.

I believe that there is a strategy under which Alice can always win by choosing the a certain series of numbers in order, however I have not yet found the right series. My reasoning is as follows:

My strategy for trying to determine the series is similar to how the sliding bolt puzzle works. If you look at this in terms of modular arithmetic, we are simply trying to eliminate possible states and force the solution to move to a single state regardless of where it started from.

The way to analyze a particular series of numbers is to look at in terms of the modulo of the least common multiple of all the numbers. For example, let's look at if 2 were allowed, but the only numbers we could use were 2, 3, 4, 6, 8, and 9. The least common multiple of these numbers is 72, so we are interested in X mod 72. Initially, X mod 72 could be any number between 0 and 71. If we choose 2 first and it's not a multiple of 2, then X-2 mod 72 could be any odd number between 1 and 71. If we then choose 3, then X-2 mod 72 could not be 3, 9, etc. so X-5 mod 72 could not be an odd number or 0, 6, 12, etc. There are many more states than in the sliding bolt puzzle, but I feel like there should be a way to whittle away the possibilities until it has been forced into a single state.