1. The problem asks for the number of one-digit nonnegative integers $x$ such that the sum $$S = x! + \sum_{i=0}^{2013} i!$$ is divisible by 4. 2. Note that the sum $$\sum_{i=0}^{2013} i!$$ includes very large factorials. Factorials from 4! onwards are multiples of 4 since $$4! = 24$$ and $24$ is divisible by 4. 3. Therefore, $$i! \equiv 0 \pmod{4}$$ for all $$i \geq 4$$. So the large sum modulo 4 depends only on the first few factorials: $$\sum_{i=0}^{2013} i! \equiv 0! + 1! + 2! + 3! \pmod{4}$$ 4. Calculate these factorials modulo 4: - $$0! = 1$$ - $$1! = 1$$ - $$2! = 2$$ - $$3! = 6 \equiv 2 \pmod{4}$$ 5. Summing them modulo 4: $$1 + 1 + 2 + 2 = 6 \equiv 2 \pmod{4}$$ 6. So, $$\sum_{i=0}^{2013} i! \equiv 2 \pmod{4}$$ 7. The total sum is then: $$S \equiv x! + 2 \pmod{4}$$ 8. We want $S$ divisible by 4, so: $$x! + 2 \equiv 0 \pmod{4} \implies x! \equiv 2 \pmod{4}$$ 9. Find $x$ (0 to 9) such that $$x! \equiv 2 \pmod{4}$$: - $0! = 1 \not\equiv 2$ - $1! = 1 \not\equiv 2$ - $2! = 2 \equiv 2$ - $3! = 6 \equiv 2$ - For $x \geq 4, x!$ is multiple of 4, so: $$x! \equiv 0 \not\equiv 2$$ 10. Only $x=2$ and $x=3$ satisfy the condition. Final answer: 2 values of $x$ make $S$ divisible by 4. Hence, the correct choice is (a) 2.