## Task 4 - 340724

a) For five consecutive natural numbers, it is required that their sum is 230.

Show that there is exactly one way to satisfy this requirement with five consecutive natural numbers! What is the first of these five numbers?

b) Now it is required that the sum be divisible by 23 and have the smallest possible value.

What is the first of the five consecutive natural numbers that satisfy these requirements?
a) Five numbers of the form $n, n+1, \ldots, n+4$ satisfy the condition $n+n+1+\ldots+n+4=230$ if and only if $5n+10=230$ holds. This is true if and only if the first of these numbers is $n=44$.

b) For every natural number $n$, the considered sum $5n+10$ is divisible by 5. It is exactly then, as required, also divisible by 23 if it is divisible by $5 \cdot 23=115$; because 5 and 23 are coprime.

The smallest value divisible by 115 that $5n+10$ can achieve is when $5n+10=115$. This is true if and only if the first of the five numbers is $n=21$.

Solutions of the 2nd Round 1994 adopted from [5]

### 4.36.3 3rd Round 1994, Class 7