2. For every natural number $n$, we write the sum $1^{n}+2^{n}+3^{n}+4^{n}$ as a number $\mathrm{v}$ in decimal notation. With how many zeros can these numbers end at most?
IV/2. First, $1+2+3+4=10, 1^{2}+2^{2}+3^{2}+4^{2}=30$ and $1^{3}+2^{3}+3^{3}+4^{3}=100$. Suppose that for some natural number $n>3$ we could express the sum $1^{n}+2^{n}+3^{n}+4^{n}$ as a number that would end with three zeros. This number would be divisible by $1000=8 \cdot 125$. Let's look at the remainder of each term in the mentioned sum when divided by 8. The term $1^{n}$ gives a remainder of 1, and the terms $2^{n}$ and $4^{n}$ give a remainder of 0, since $n>3$. Write $3^{n}=(2+1)^{n}=$ $2^{n}+n \cdot 2^{n-1}+\cdots+\frac{n(n-1)(n-2)}{3 \cdot 2 \cdot 1} \cdot 2^{3}+\frac{n \cdot(n-1)}{2 \cdot 1} \cdot 2^{2}+n \cdot 2+1$. From this, we conclude that the remainder of the term $3^{n}$ when divided by 8 is the same as the remainder of the expression $\frac{n \cdot(n-1)}{2 \cdot 1} \cdot 2^{2}+n \cdot 2+1=2 n^{2}+1$ when divided by 8. We quickly see that the expression $2 n^{2}+1$ has a remainder of 1 if $n$ is an even number, since in this case $2 n^{2}$ is divisible by 8. If $n$ is an odd number, we write $n=2 k+1$ and get $2 \cdot(2 k+1)^{2}+1=8 k^{2}+8 k+3$, from which we see that the remainder is 3.

We have found: if the sum $1^{n}+2^{n}+3^{n}+4^{n}$ for any $n>3$ is divided by 8, we get a remainder of 2 or 4. Since this sum is not divisible by 8, we cannot write it as a number ending with more than two zeros for any natural number $n$. Therefore, a number of the form $1^{n}+2^{n}+3^{n}+4^{n}$ can end with at most two zeros.

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