Problems

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Found: 1866

George divided number \(a\) by number \(b\) with the remainder \(d\) and the quotient \(c\). How will the remainder and the quotient change if the dividend and the divisor are increased by a factor of 3?

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?

Can a sum of three different natural numbers be divisible by each of those numbers?

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?

Find such a natural number \(n\) that all the numbers \(n+1\), \(n+2\), ..., \(n+2018\) are composite.

The numbers \(2^{2018}\) and \(5^{2018}\) are expanded and their digits are written out consecutively on one page. How many digits are on the page?

Is it possible for \(n!\) to be written as \(2015000\dots 000\), where the number of 0’s at the end can be arbitrary?

Look again at the divisibility rule for \(3\). Can you come up with a divisibility rule for \(9\)?

A battle of the captains was held at a maths battle. The task was to write the smallest number such that it is divisible by 45 and consists of only 1’s and 0’s as digits. What do you think was the correct answer?

It is known that a natural number is three times bigger than the sum of its digits. Is it divisible by 27?