We call a \(10\)-digit number interesting if it is divisible by \(11111\), and all its digits are different. How many interesting numbers does there exist?
Note that a number \(k = a_0 + 10a_1 + \dots +10^9 a_9\) is divisible by \(11111\) if and only if a number \(m = (a_0+a_5) +10(a_1+a_6) + \dots + 10^4 (a_4+a_9)\) is also divisible by \(11111\). This is because \(100000=1+9 \times 11111\) and we subtract \(99999 (a_5 + 10a_6 + 100a_7 + 1000a_8 +10000a_9)\) from the original number.
Is it true that if a natural number is divisible by 4 and by 6, then it must be divisible by \(4\times6=24\)?
And what if a natural number is divisible by 5 and by 7? Should it be divisible by 35?
The number \(A\) is not divisible by 3. Is it possible that the number \(2A\) is divisible by 3?
Lisa knows that \(A\) is an even number. But she is not sure if \(3A\) is divisible by 6. What do you think?
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?
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?