## Subject III (7 p)

Let the set $A=\{\overline{a b c d}, \overline{a b}=\overline{c d}+4\}$

a) Determine whether 2016 and 2024 belong to the set A.

b) Find the remainder of the division of any number from A by 101.

c) If $\mathrm{S}$ is the sum of all numbers in $\mathrm{A}$, show that $\mathrm{S}$ is not a perfect square.
## Subject III

Let the set $A=\{\overline{a b c d}, \overline{a b}=\overline{c d}+4\}$

a). Determine whether 2016 and 2024 belong to the set A.

b). Find the remainder of the division of any number from $\mathrm{A}$ by 101.

c). If $\mathrm{S}$ is the sum of all numbers in $\mathrm{A}$, show that $\mathrm{S}$ is not a perfect square.

## Solution

a). 2016 has $20=16+4$, while 2024 does not satisfy $20=24+4$. Therefore, 2016 belongs to the set, and 2024 does not belong to the set. $1 p$
b). $\overline{a b c d}=\overline{a b} \cdot 100+\overline{c d}=(\overline{c d}+4) \cdot 100+\overline{c d}=\overline{c d} \cdot 101+400$ $.2 p$
Since the first term is divisible by 101, the remainder of the division of this number by 101 is the remainder of the division of 400 by 101, which is 97. $1 p$

It is easier to write $\overline{a b c d}=\overline{a b} \cdot 100+\overline{a b}-4=\overline{a b} \cdot 101-4$. Since $a$ cannot be 0, there are a total of 90 numbers, and thus the sum is:
$S=10 \cdot 101-4+11 \cdot 101-4+\ldots+99 \cdot 101-4=101 \cdot(10+11+\ldots+99)-4 \cdot 90=$ $101 \cdot(99 \cdot 100: 2-9 \cdot 10: 2)-360=101 \cdot(45 \cdot 110-45)-360=45 \cdot(101 \cdot 110-101-$ 8). $.2 p$

We observe that the number in the last parenthesis has the last digit 1, so it is not divisible by 5. Since S is divisible by 5, but not by 25, we conclude that $\mathrm{S}$ is not a perfect square.