# Task 4.

Masha tightly packed 165 identical balls into the shape of a regular triangular pyramid. How many balls are at the base?
Answer: 45

Solution. We will solve the problem in general for $\frac{1}{6} n(n+1)(n+2)$ balls. At the base of a triangular pyramid of height $n$ rows lies the $n$-th triangular number of balls $1+2+3+\ldots+n=\frac{1}{2} n(n+1)$. Then the total number of balls in the pyramid is $\sum_{k=1}^{n} \frac{k(k+1)}{2}=\frac{1}{6} n(n+1)(n+2)$. The problem can be solved by enumeration.

## B-2

Masha tightly packed 220 identical balls into a regular triangular pyramid. How many balls are at the base?

Answer: 55

## B-3

Masha tightly packed 286 identical balls into a regular triangular pyramid. How many balls are at the base?

Answer: 66

## B-4

Masha tightly packed 364 identical balls into a regular triangular pyramid. How many balls are at the base?

Answer: 78

## Lomonosov Olympiad for Schoolchildren in Mathematics

Preliminary Stage 2020/21 academic year for 9th grade

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