3. Determine for which natural number n the quotient

$$
\frac{33000}{(n-4)(n+1)}
$$

a) is the largest, b) is the smallest natural number.
SOLUTION. We have $33000=1 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 11$ and $(n+1)-(n-4)=5$. Since for every natural $n$ the value $n+1$ is positive, the given ratio is positive only when the value $n-4$ is also positive, hence $n \geq 5$.

a) For every natural $n \geq 5$ we have $n-4 \geq 1$ and $n+1 \geq 6$, so the greatest value of the given ratio is $33000:(1 \cdot 6)=5500$ and it is achieved for $n=5$.

b) When looking for the smallest ratio, let $a, b$ be the numbers $n+1, n-4$ in the order to be specified. First, assume that the factorization of the number $ab$ into prime factors contains the primes 11 and 5. Then $a, b$ are consecutive multiples of five, and exactly one of them, say $a$, is a multiple of 55.

First, consider $a=55$. Of the two possible values $b=50$ and $b=60$, we choose the larger one (to get a smaller value of the ratio we are examining). The value $b=60$ from the equation $n+1=60$ (or the equation $n-4=55$) corresponds to $n=59$ and the examined ratio is then equal to the number 10.

For $a=110$ (or $a=165$), the number 33000 is not divisible by any of the neighboring multiples of five, i.e., the numbers 105 and 115 (or 160 and 170).

For further (larger) multiples $a$ of 55, we get $ab \geq 215 \cdot 220 > 33000$.

If the factorization of the number $ab$ into prime factors does not contain the prime 11 or the prime 5, the examined ratio (assuming it is an integer) is divisible by the number 11 or the number 125, so it is a number greater than the value 10 we found earlier.

Conclusion: The greatest value of the given ratio is 5500 for $n=5$ and the smallest is 10 for $n=59$.

## EXERCISES TO PRACTICE: