Among all natural numbers we can distinguish $\textit{prime}$ and $\textit{composite}$ numbers.

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A number is $\textit{composite}$ if it is a product of two smaller natural numbers. For example, $6 = 2\times3$. Otherwise, and if the number is not equal to 1, it is called $\textit{prime}$. The number 1 is neither prime nor composite.

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The well-known fact, called the $\textbf{Fundamental Theorem of Arithmetic}$, says that any natural number greater than 1 can be uniquely expressed as a product of prime numbers in non-decreasing order. For example, $$630=2\times3\times3\times5\times7=2\times3^2\times5\times7$$

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$\textit{Modulo operation}$: Given any two natural numbers $a$ and $b$, called the dividend and the divisor respectively, we can divide $a$ by $b$ with the remainder. That is to find such non-negative integer numbers $c$ and $d$ ($d<b$), called the quotient and the remainder respectively, that $a=c\times b+d$. For example, $41=2\times15+11$ is the division of 41 by 15 with the remainder 11 and $5=0\times7+5$ is the division of 5 by 7 with the remainder 5.

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If $a$ is divided by $b$ with zero remainder (without a remainder) we say that “$a$ is divisible by $b$”$\;$or “$b$ divides $a$”. From the definition of \textit{modulo operation} for $a$ the property to be divisible by $b$ is equivalent to the existence of non-negative integer $c$ such that $a=c\times b$. We denote it by $a÷b$ for “$a$ is divisible by $b$”$\;$and $b|a$ for “$b$ divides $a$”. For example, $105÷7$ and $9|111111111$ because $105=15\times7$ and $111111111=12345679\times9$.

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We immediately deduce from the Fundamental Theorem of Arithmetic that if a product of two natural numbers is divisible by a prime number, then one of these numbers is divisible by this prime number.

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Problems:

Let us introduce the notation – we denote the product of all natural numbers from 1 to $n$ by $n!$. For example, $5!=1\times2\times3\times4\times5=120$.

a) Prove that the product of any three consecutive natural numbers is divisible by 3!=6. b) What about the product of any four consecutive natural numbers? Is it always divisible by 4!=24?

A young mathematician felt very sad and lonely during New Year’s Eve. The main reason for his sadness ( have you guessed already?) was the lack of mathematical problems. So he decided to create a new one on his own. He wrote the following words on a small piece of paper: “Find the smallest natural number n such that n! is divisible by 2018″$\;$, but unfortunately he immediately forgot the answer. What is the correct answer to this question?