How many integers $n$ are there for which the value of expression (1) is an integer? How many positive integers $n$ are there for which the value of $K$ is a positive integer?

$$
K=\frac{2610+201 n+110 n^{2}-n^{3}}{10+n}
$$
a) The given expression can be transformed as follows:

$$
\begin{gathered}
K=\frac{\left(n^{3}+10 n^{2}\right)-\left(120 n^{2}+1200 n\right)+(999 n+9990)-12600}{-(n+10)}= \\
=-n^{2}+120 n-999+\frac{12600}{n+10} .
\end{gathered}
$$

Substituting an integer for \( n \), the first three terms on the right side are integers. Therefore, \( K \) is an integer if and only if the fractional expression is an integer, i.e., if \( n+10 \) is equal to one of the divisors of 12600. Here, we consider both positive and negative divisors with the same absolute value separately.

\( 12600=2^{3} \cdot 3^{2} \cdot 5^{2} \cdot 7 \), and according to the 1106th exercise, the number of positive divisors is the product of the exponents increased by 1: \((3+1)(2+1)(2+1)(1+1)=72\). There are as many negative divisors, so there are 144 integers \( n \) for which \( K \) is an integer. Specifically, from the equation

$$
n+10=-12600,-6300, \ldots, 8,9,10,12, \ldots, 12600
$$

we get

$$
n=-12610,-6310, \ldots,-2,-1,0,2, \ldots, 12590
$$

b) To determine the integer pairs \( n>0, K>0 \), we exclude those positive integers \( n \) for which \( K \leq 0 \). Transform (1) as follows: \( K=K_{1}+K_{2} \), where

$$
K_{1}=\frac{12600}{n+10}, \quad K_{2}=(111-n)(n-9)=-(n-60)^{2}+2601
$$

\( K_{1} \) is positive for all \( n>0 \) and monotonically decreases as \( n \) increases. If \( 9 \leq n \leq 111 \), then \( K_{2} \geq 0 \), so here \( K>0 \). For \( n>111 \) and \( 00
\end{gathered}
$$

so here also \( K>0 \).

The second form shows that for \( n>60 \), \( K_{2} \) is also monotonically decreasing, so this holds for \( K \) as well. For \( n=113 \), \( K0 \). Therefore, \( K>0 \) for \( 0<n<113 \) integer values.

For these \( n \), \( 10<n+10<123 \), so we need to find how many divisors of 12600 are between 10 and 123, excluding the boundaries. Two observations simplify the answer. The factorization \( 12600=2^{3} \cdot 3^{2} \cdot 5^{2} \cdot 7 \) shows that all integers from 1 to 10 are divisors, so the number of positive divisors to be excluded due to the lower limit is 10. Furthermore, the square root of 12600 is close to 112.2, and the number of divisors smaller and larger than this is equal, as dividing by a number smaller than the square root results in a quotient larger than the square root and vice versa. For example, 12600/105 = 120, these are so-called paired divisors of 12600. Thus, the number of divisors to be retained up to 112.2 is \( 72 / 2 - 10 = 26 \), and we only need to find the divisors of 12600 between 112.2 and 123. We have already seen that 120 is a suitable divisor, and there are no more, as 113 is prime, and the others are multiples of the prime numbers \( 19, 23, 29, 13, 59, 17, 11, 61 \), which are not present in the factorization of 12600.

Therefore, there are 27 natural numbers \( n \) for which \( K \) is a positive integer.

Based on the work of László Hárson (Budapest, Berzsenyi D. Gymnasium, 3rd grade), with simplifications[^0]


[^0]:    ${ }^{1}$ K. M. L. $35(1967)$ 151. o.