## Subject III

a) Find the remainder of the division of the number $\mathrm{a}=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 2015+4051$ by the number $\mathrm{b}=4 \cdot\left(5^{3}+3 \cdot 5^{3}\right)+3^{2}+5$.

b) If we divide the natural number n by 8, we get a remainder of 6, and if we divide it by 10, we get a remainder of 4. Calculate the remainder of the division of n by 40.

## Solution

a) $b=2014$
$1 \text{p}$
$\mathrm{a}=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 2014 \cdot 2015+2014 \cdot 2+23$ $.2 \text{p}$
$\mathrm{a}=2014(1 \cdot 2 \cdot 3 \cdot \ldots \cdot 2013 \cdot 2015+2)+23 \Rightarrow$ remainder $=23$
$1 \text{p}$
b) $n=8 \cdot \mathrm{c}_{1}+6$ and $\mathrm{n}=10 \cdot \mathrm{c}_{2}+4 \ldots \ldots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 \text{p}$
$5 n=40 \cdot c_{1}+30,4 n=40 \cdot c_{2}+16 \Rightarrow 5 n-4 n=40 \cdot\left(c_{1}-c_{2}\right)+14 \ldots \ldots \ldots \ldots 1 \text{p}$

![](https://cdn.mathpix.com/cropped/2024_06_07_08f6aaa1ef4f953a6e10g-2.jpg?height=52&width=1205&top_left_y=2110&top_left_x=323)
## Subject III

a) Find the remainder of the division of the number $\mathrm{a}=1 \cdot 2 \cdot 3 \cdot \ldots .2015+4051$ by the number $\mathrm{b}=4 \cdot\left(5^{3}+3 \cdot 5^{3}\right)+3^{2}+5$.

b) If we divide the natural number $n$ by 8, we get a remainder of 6, and if we divide it by 10, we get a remainder of 4. Calculate the remainder of the division of $n$ by 40.