IMO 1970 SL 7

Origin: USS

Problem

For which digits a do exist integers n \geq4 such that each digit of n(n+1) equals a?

Solution

For a = 5 one can take n = 10, while for a = 6 one takes n = 11. Now assume a ̸\in{5, 6}. If there exists an integer n such that each digit of n(n + 1)/2 is equal to a, then there is an integer k such that n(n + 1)/2 = (10k −1)a/9. After multiplying both sides of the equation by 72, one obtains 36n2 + 36n = 8a \cdot 10k −8a, which is equivalent to 9(2n + 1)2 = 8a \cdot 10k −8a + 9. (1) So 8a \cdot 10k −8a + 9 is the square of some odd integer. This means that its last digit is 1, 5, or 9. Therefore a \in{1, 3, 5, 6, 8}. If a = 3 or a = 8, the number on the RHS of (1) is divisible by 5, but not by 25 (for k \geq2), and thus cannot be a square. It remains to check the case a = 1. In that case, (1) becomes 9(2n + 1)2 = 8 \cdot 10k + 1, or equivalently [3(2n+1)−1][3(2n+1)+1] = 8\cdot10k ⇒(3n+1)(3n+2) = 2\cdot10k. Since the factors 3n + 1, 3n + 2 are relatively prime, this implies that one of them is 2k+1 and the other one is 5k. It is directly checked that their difference really equals 1 only for k = 1 and n = 1, which is excluded. Hence, the desired n exists only for a \in{5, 6}.