Problem 2. Consider the natural numbers $a, b, c, d$ such that $a<b<c<d$ and $a \cdot b \cdot c \cdot d=2014$. Find the last digit of the number $n=a^{2014}+b^{2014}+c^{2014}+d^{2014}$.

Nicolae Tomescu, Corabia
## Problem 2.

$2014=1 \cdot 2 \cdot 19 \cdot 53 \quad(1 p)$

Since $a<b<c<d$ and 2, 19, 53 are prime numbers, we have $\mathrm{a}=1 ; \mathrm{b}=2 ; \mathrm{c}=19 ; \mathrm{d}=53$ and $n=1^{2014}+$ $2^{2014}+19^{2014}+53^{2014}(1 p)$

$$
U\left(1^{2014}\right)=1, U\left(2^{2014}\right)=4, U\left(19^{2014}\right)=1 \text { and } U\left(53^{2014}\right)=9 \quad(4 p)
$$

Then $U(n)=U(1+4+1+9)=U(15)=5 \quad$ (1p)