## SUBJECT II

Calculate the sum of natural numbers, between 400 and 600, which when divided by 12 give a remainder of 10.
## SUBJECT II

$\mathrm{x}: 12=\mathrm{a}$, remainder 10; thus $\mathrm{x}=12 \mathrm{a}+10$.

$1 p$

Since the numbers are between 400 and 600, the following appear:

$$
12 \cdot 33+10=406
$$

$12 \cdot 34+10$

$12 \cdot 35+10$

$12 \cdot 49+10=598$

2p

- justification that there are 17 numbers $1 p$

By adding the relations, we get 12(33+34+35+..... +49) +10 17 ............. 1p

$$
12\left(\mathrm{~S}_{49}-\mathrm{S}_{32}\right)+170=8534 \quad \text {............................... } 2 p
$$

Grading: $7 p$