Prove the divisibility rule for \(3\): the number is divisible by \(3\) if and only if the sum of its digits is divisible by \(3\).
Find the smallest \(k\) such that \(k!\) (\(k!= k\times(k-1)\times \ldots \times 1\)) is divisible by \(2024\).
While studying numbers and its properites, Robinson came across a 3-digit prime number with the last digit being equal to the sum of the first two digits. What was the last digit of that number if among the number did not have any zeros among it’s digits?
When Robinson Crusoe’s friend and assistant named Friday learned about divisibility rules, he was so impressed that he proposed his own rule:
a number is divisible by 27 if the sum of it’s digits is divisible by 27.
Was he right?
One day Friday multiplied all the numbers from 1 to 100. The product appeared to be a pretty large number, and he added all the digits of that number to receive a new smaller number. Even then he did not think the number was small enough, and added all the digits again to receive a new number. He continued this process of adding all the digits of the newly obtained number again and again, until finally he received a one-digit number. Can you tell what number was it?
Robinson Crusoe’s friend Friday was looking at \(3\)-digit numbers with the same first and third digits. He soon noticed that such number is divisible by \(7\) if the sum of the second and the third digits is divisible by \(7\). Prove that he was right.
Prove the following divisibility rule by 37: divide the number starting from the right end of the number into blocks of three digits. Now, the original number is divisible by 37 if the sum of all three digit numbers obtained in this way is divisible by 37.
(It might be the case that the number of digits is not divisible by 3, and you cannot divide the original number into blocks of three digits. To overcome this problem we allow a block of three digits to start from 0, for example number 2345678 should be divided into blocks of three digits as 002, 345, and 678.)
2016 digits are written in a circle. It is known, that if you make a number reading the digits clockwise, starting from some particular place, then the resulting 2016-digit number is divisible by 27. Show that if you start from some other place, and moving clockwise make up another 2016-digit number, then this new number is also divisible by 27.
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.
Can you pay 20 p in seven coins?