10.2. Let $f(n)$ be equal to the product of the even digits of the natural number $\mathrm{n}$ or be zero if there are no even digits. Find the sum $f(1)+f(2)+\cdots+f(100)$.
Answer: 620

Solution: for a single-digit number $n$, $f(n)$ will be equal to $n$ itself if it is even and 0 if it is odd. For single-digit $\mathrm{n}$, we get the sum $2+4+6+8=20$. If $\mathrm{n}$ is a two-digit number, let's consider the cases:

1) If both digits are even. Then the first digit can be $2,4,6$ or 8, and the second digit can be $0,2,4,6$ or 8. The total sum will be $2 * 0+2 * 2+\ldots+2 * 8+4 * 0+\ldots+8 * 8=2 *(0+2+4+6+8)+\ldots+8 *(0+\ldots+8)=(2+4+6+8) *(0+2+4+6+8)=20$ $* 20=400$.
2) If the first digit is even and the second is odd. Then for each of the even first digits ( $2,4,6$ or 8 ), there will be 5 odd second digits, so the total sum is $2 * 5+4 * 5+6 * 5+8 * 5=(2+4+6+8) * 5=20 * 5=100$.
3) If the first digit is odd and the second is even. Then for each of the even last digits ( $0,2,4,6$ or 8 ), there will be 5 odd first digits, so the total sum is $0 * 5+2 * 5+4 * 5+6 * 5+8 * 5=(0+2+4+6+8) * 5=20 * 5=100$.
4) If both digits are odd, the value of the function is 0.

Summing up the obtained values: $20+400+100+100+0=620$.

Criteria: Only for the correct answer - 1 point.