Problems

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 $k!$ ($k!=k\times (k-1)\times \dots \times 1$) 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?

Prove the divisibility rule for $4$: a number is divisible by $4$ if and only if the number made by the last two digits of the original number is divisible by $4$;
Can you come up with a divisibility rule for $8$?

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?

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?

Can a $300$-digit number, with one hundred $0$s, one hundred $1$s, and one hundred $2$s among its digits, be a square number?

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.)