## 5. Division with Remainder

Determine the sum of all natural numbers whose quotient when divided by 7 is less than the remainder.

## Result: $\quad 336$
## Solution.

Let $n$ be a natural number that when divided by 7 gives a quotient $k$ and a remainder $o$. When dividing by 7, the remainder $o \in \{0,1,2,3,4,5,6\}$, so $n = 7k + o$, with $k < o < 7$. The possibilities can be represented in a table:

| $k$ | $o$ | $n=7k+o$ |
| :---: | :---: | :---: |
| 0 | 1 | 1 |
| 0 | 2 | 2 |
| 0 | 3 | 3 |
| 0 | 4 | 4 |
| 0 | 5 | 5 |
| 0 | 6 | 6 |
| 1 | 2 | 9 |
| 1 | 3 | 10 |
| 1 | 4 | 11 |
| 1 | 5 | 12 |
| 1 | 6 | 13 |
| $Z b r$ |  |  |


| $k$ | $o$ | $n=7k+o$ |
| :---: | :---: | :---: |
| 2 | 3 | 17 |
| 2 | 4 | 18 |
| 2 | 5 | 19 |
| 2 | 6 | 20 |
| 3 | 4 | 25 |
| 3 | 5 | 26 |
| 3 | 6 | 27 |
| 4 | 5 | 33 |
| 4 | 6 | 34 |
| 5 | 6 | 41 |

The sum of all numbers is $21+55+74+78+67+41=336$.