1. The task is a sequence of 5 natural numbers. The arithmetic mean of the first three numbers is 19, and the arithmetic mean of all five numbers is 20. Determine the last two numbers in the sequence if their product is 456.
1. Let $a, b, c, d, e$ be given numbers.

From $\frac{a+b+c}{3}=19$ it follows that $a+b+c=57$.

1 POINT

From $\frac{a+b+c+d+e}{5}=20$ it follows that $a+b+c+d+e=100$.

1 POINT

By comparing, we get $d+e=43$, and it is given that $d \cdot e=456$.

1 POINT

We factorize 456 into prime factors: $456=2 \cdot 2 \cdot 2 \cdot 3 \cdot 19$.

1 POINT

By combining the prime factors, we get the numbers 24 and 19, which satisfy the condition $24+19=43$.

The last two numbers in the sequence are 24 and 19.

2 POINTS

TOTAL 6 POINTS