How many numbers have unit digit $1$? 
Let $f(n)$ be the number of positive integers that have exactly $n$ digits and whose
  digits have a sum of $5$. Determine, with proof, how many of the $2014$ integers
  $f(1), f(2), . . . , f(2014)$ have a units digit of $1.$

HINTS ONLY
EDIT:
for $f(2)$ we have: $x + y = 5$ with $x \ne 0$, which gives: $\binom{5}{1}$
$f(5)$ gives: $x + y + z + t +w = 5$ so: $\binom{8}{4}$
$f(4)$ gives, $\binom{7}{3}$
For $f(n)$ we have: $f(n) = \binom{n + 3}{n-1}$
$f(n) = \frac{(n+3)(n+2)(n+1)(n)(n-1)!}{24(n-1)!} = \frac{(n+3)(n+2)(n+1)(n)}{24}$
PROOF CHECK:

Consider $N = \overline{a_1 a_2 ... a_n}$ with $a_i$ as digits such that $a_1 \ge 1$. We have a string of $n$ characters, $a_1, a_2, a_3, ..., a_n$ with $a_1 \ne 0$, thus this string with $a_n$ gives the $n$th digit numbers. the number of numbers corresponds to the solutions of: $a_1 + a_2 + ... + a_n = 5$ for nonnegative integers with $a_1 > 0$. By the stars-and-bars principle there exist: $\binom{n + 4 - 1}{n-1} = \binom{n+3}{n-1} = \frac{(n+3)(n+2)(n+1)(n)}{24}$ numbers, thus giving an explicit formula for $f(n)$. 

Is this enough for a proof? Or is induction needed for a stronger argument?
So we have: $f(n) \equiv 1\pmod{10}$. 

Since $f(n)$ is a combinatoric coefficient, $f(n) \in \mathbb{Z+}$, which implies that: $\frac{(n+3)(n+2)(n+1)(n)}{24} \equiv (n+3)(n+2)(n+1)(n) \pmod{10}$. Since the numerator was divisible by the denominator for all $n$. 

Now help is needed:

How to solve this modular arithemetic expression?


If you want to solve $$\frac{n(n+1)(n+2)(n+3)}{24}\equiv 1\pmod{10},$$
You can multiple both sides by $3$, since it is relatively prime to $10$, but when you multiply by $8$, you have to apply that to the modulus, to. So, you are trying to solve:
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{80}$$
Solve this in pairs, using Chinese remainder theorem:
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{5}$$
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{16}$$
As it turns out, the second only depends on $n\pmod 8$.
