Given a prime number $p$. We need to determine the smallest integer $N$ that is divisible by $p$, but when divided by any integer $n$ smaller than $p$,

$1^{0}$. the remainder is $n-1, \quad 2^{0}$. the remainder is 1.

For example, $p=11$.
The sequence of prime numbers up to $p$

$$
p_{1}, p_{2}, \ldots, p_{r}, p
$$

Let $\alpha_{k}$ be the exponent such that $p_{k}^{\alpha_{k}}<p<p_{k}^{\alpha_{k+1}}$.

With the thus determined exponents $\alpha_{k}$, we form the product

$$
P=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{r}^{\alpha_{r}}
$$

Then $l P$, where $l=1,2,3, \ldots$, is divisible by any $n$ less than $p$, so $l P \pm 1$ gives a remainder of 1 or $n-1$ when divided by $n$.

If $P=p Q+r$, where $0<r<p$, then after a finite number of trials, we will find the smallest positive value of $l$ for which $\frac{l r+1}{p}$ or $\frac{l r-1}{p}$ is an integer; however, the latter can also be zero.

If the smallest value of $l$ is $\lambda_{1}$ or $\lambda_{2}$, then $N_{1}=\lambda_{1} P+1$ is the smallest number that is divisible by $p$, but when divided by any $n$ less than $p$, the remainder is 1, while $N_{2}=\lambda_{2} P-1$ is the smallest number that is divisible by $p$, but when divided by any $n$ less than $p$, the remainder is $n-1$.

For example, for $p=11$, the primes less than $p$ are $2,3,5,7$ and

$$
P=2^{3} \cdot 3^{2} \cdot 5 \cdot 7=2520
$$

Now $2520=229 \cdot 11+1$, that is, $r=1$. Clearly

$$
\lambda_{1}=10 \quad \text { and } \quad \lambda_{2}=1
$$

so that

$$
N_{1}=25201 \quad \text { and } \quad N_{2}=2519
$$

25201 : 11 = 2291 ; when $N_{1}$ is divided by any number less than 11, the remainder is +1.

2519 : 11 = 229 ; when $N_{2}$ is divided by any number less than 11, the remainder is $n-1$.

János Harsányi (Ág. ev. g. VIII. o. Bp.).

I. Note. The number $P$ is nothing other than the least common multiple of the numbers less than $p$.

II. Note. For example, for $p=7$, $P=2^{2} \cdot 3 \cdot 5=60$.

However,

$$
60=8 \cdot 7+4, \quad \text { that is } \quad r=4
$$

Clearly $2 \cdot 4-1=7$, so $\lambda_{2}=2$ and $N_{2}=2 \cdot 60-1=119$ is the smallest number that is divisible by 7, but when divided by any number less than 7, the remainder is $n-1$.

On the other hand, $5 \cdot 4+1=21$ is the smallest number in the sequence of numbers of the form $l r+1$ that is divisible by 7, so $N_{1}=5 \cdot 10+1=301$ is the smallest number that is divisible by 7, but when divided by any number less than 7, the remainder is 1. [^0]

[^0]:    ${ }^{1}$ The number of trials is $<p$; or we need to find the smallest positive integer solutions of the indeterminate equations $r x \pm 1=p y$ (in the case of $r x-1=p y$, $y=0$ is also possible).