A stoneboard was found on the territory of the ancient Greek Academia as a result of archaeological excavations.
The archeologists decided that this stoneboard belonged to a mathematician who lived in the 7th century BC. The list of unsolved problems was written on the stoneboard. The archaeologists became thrilled to solve the problems but got stuck on the fifth. They were looking for a 10-digit number. The number should consist of only different digits. Moreover, if you cross any 6 digits, the remaining number should be composite. Can you help the archeologists to figure out the answer?
The number \(b^2\) is divisible by \(8\). Show that it must be divisible by \(16\).
Find a number which:
a) It is divisible by \(4\) and by \(6\), is has a total of 3 prime factors, which may be repeated.
b) It is divisible by \(6, 9\) and \(4\), but not divisible by \(27\). It has \(4\) prime factors in total, which may be repeated.
c) It is divisible by \(5\) and has exactly \(3\) positive divisors.
a) The number \(a\) is even. Should \(3a\) definitely also be even?
b) The number \(5c\) is divisible by \(3\). Is it true that \(c\) is definitely divisible by \(3\)?
c) The product \(a \times b\) is divisible by \(7\). Is it true that one of these numbers is divisible by \(7\)?
d) The product \(c \times d\) is divisible by \(26\). Is it true that one of these numbers is divisible by \(26\)?
a) The number \(a^2\) is divisible by \(11\). Is \(a^2\) necessarily also divisible by \(121\)?
b) The number \(b^2\) is divisible by \(12\). Is \(b^2\) necessarily also divisible by \(144\)?
a) Prove that a number is divisible by \(8\) if and only if the number formed by its laast three digits is divisible by \(8\).
b) Can you find an analogous rule for \(16\)? What about \(32\)?
Look at this formula found by Euler: \(n^2 +n +41\). It has a remarkable property: for every integer number from \(1\) to \(21\) it always produces prime numbers. For example, for \(n=3\) it is \(53\), a prime. For \(n=20\) it is \(461\), also a prime, and for \(n=21\) it is \(503\), prime as well. Could it be that this formula produces a prime number for any natural \(n\)?
Denote by \(n!\) (called \(n\)-factorial) the following product \(n!=1\cdot 2\cdot 3\cdot 4\cdot...\cdot n\). Show that if \(n!+1\) is divisible by \(n+1\), then \(n+1\) must be prime. (It is also true that if \(n+1\) is prime, then \(n!+1\) is divisible by \(n+1\), but you don’t need to show that!)
Can the equality \(K \times O \times T = U \times W \times E \times N \times H \times Y\) be true if the numbers from 1 to 9 are substituted for the letters? Different letters correspond to different numbers.
Which five-digit numbers are there more of: ones that are not divisible by 5 or those with neither the first nor the second digit on the left being a five?