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Found: 53

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?

Robinson found a chest with books and instruments after the ship wreck. Not all the books were in readable condition, but some of the books he managed to read. One sentence read “72 chickens cost *619* p”. (The starred digits were not readable). He has not tasted a chicken for quite some time, and it was pleasant to imagine a properly cooked chicken in front of him. He also was able to decipher the cost of one chicken. Can you?

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.

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.

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?