4. Let $a, b, c, d, e$ be positive integers. Their sum is 2018. Let $M=\max (a+b, b+c, c+d, d+e)$. Find the smallest possible value of $M$.
Answer: 673.

Solution. Estimation. We have the inequalities

$$
a+b \leqslant M ; \quad b+c \leqslant M ; \quad c+d \leqslant M ; \quad d+e \leqslant M
$$

Multiply the first and last inequalities by 2 and add the other two. We get:

$$
2(a+b+c+d+e)+b+d \leqslant 6 M ; \quad 6 M \geqslant 4036+b+d \geqslant 4038 ; \quad M \geqslant 673 .
$$

Example. The equality $M=673$ is achieved when $a=c=e=672$, $b=d=1$.

Evaluation. 14 points for a correct solution. If there is only a correct example, 7 points.

## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences  Final Stage 2017-2018 academic year  Problems, answers, and evaluation criteria  9th grade Variant 1 physics