At the end of the term, Billy wrote out his current singing marks in a row and put a multiplication sign between some of them. The product of the resulting numbers turned out to be equal to 2007. What is Billy’s term mark for singing? (The marks that he can get are between 2 and 5, where 5 is the highest mark).
Prove that for any natural number \(a_1> 1\) there exists an increasing sequence of natural numbers \(a_1, a_2, a_3, \dots\), for which \(a_1^2+ a_2^2 +\dots+ a_k^2\) is divisible by \(a_1+ a_2+\dots+ a_k\) for all \(k \geq 1\).
Is there a sequence of natural numbers in which every natural number occurs exactly once, and for any \(k = 1, 2, 3, \dots\) the sum of the first \(k\) terms of the sequence is divisible by \(k\)?
Is it possible to arrange the numbers 1, 2, ..., 60 in a circle in such an order that the sum of every two numbers, between which lies one number, is divisible by 2, the sum of every two numbers between which lie two numbers, is divisible by 3, the sum of every two numbers between which lie six numbers, is divisible by 7?
The product of two natural numbers, each of which is not divisible by 10, is equal to 1000. Find the sum of these two numbers.
Does there exist a natural number which, when divided by the sum of its digits, gives a quotient and remainder both equal to the number 2011?
Four people discussed the answer to a task.
Harry said: “This is the number 9”.
Ben: “This is a prime number.”
Katie: “This is an even number.”
And Natasha said that this number is divisible by 15.
One boy and one girl answered correctly, and the other two made a mistake. What is the actual answer to the question?
The sequence of numbers \(a_1, a_2, \dots\) is given by the conditions \(a_1 = 1\), \(a_2 = 143\) and
for all \(n \geq 2\).
Prove that all members of the sequence are integers.
You are given 12 different whole numbers. Prove that it is possible to choose two of these whose difference is divisible by 11.
Prove that amongst numbers written only using the number 1, i.e.: 1, 11, 111, etc, there is a number than is divisible by 1987.