BuilderBoi

The smallest number that can be expressed is 1(0,1).

The largest number that can be expressed is 136(0,1,2,3,4,...,16)

Note that because we have a 1 and 2, every number will be reachable between 1 and 136. 

Can you take it from here?

Disregard my answer....

Note that \(3^{2004} \times 5^{2004} = 15^{2004}\)

Because the base ends in a 5, the ones digit will always be \(\color{brown}\boxed{5}\)

So we have this equation: \(3{ n \choose 4} = {n + 1 \choose 4}\)

Simplify to: \(3 \times { n! \over {n -4}! \times 24} = { n+1! \over {n -3}! \times 24} \)

Now, further simplify: \({n! \over (n-4)! \times 8} = {(n+1)! \over (n-3)! \times 24}\)

Now, note that \(n! = (n-4)! \times (n-3) \times (n-2) \times (n-1) \times n\).

Applying this logic to both sides gives us \({(n -3)(n-2)(n-1)n\over 8} = {n(n +1)(n-1)(n-2)\over 24}\)

Cross multiplying gives us \({(n -3)(n-2)(n-1)24n} = {8n(n +1)(n-1)(n-2)}\)

Canceling out like terms gives us \(8(n+1) = 24(n-3)\), meaning \(n = \color{brown}\boxed5\)

Note that there are 5 choices for the final digit. 

Of these, 2 are odd, so the probability is \(\color{brown}\boxed{2 \over 5}\)

Substitute the point into the second equation: \(-4k + 7 = -7\). 

Now, just solve for k. (note that equation m isn't needed)

Note that \(-a^y = a^y\) when y is even. 

This means that all the terms with an even-powered term will cancel out. 

Thus, the resulting expansion will have \(\color{brown}\boxed{6}\) terms: \(a^{11}, a^9, a^7, a^5, a^3, a\)