Given k∈Z prove that there exist infinite pairs of distinct natural numbers such that n+s(2n)=m+s(2m)kn+s(n2)=km+s(m2).(s(n) denotes the sum of digits of n.) Proposed by Mohammadamin Sharifi
Problem
Source: Iran TST1 Day 2 P5
Tags: number theory, sum of digits, Iranian TST
24.02.2020 05:14
We will prove a slightly stronger statement. Indeed, we'll show that there are infinitely many pairs (m,n)∈N2 satisfying the conditions above, as well as m \equiv 99 \pmod{1000} and n = m+9. Observe then that "we get the first condition for free," i.e. n + s(2n) = m + s(2m) is implied by the constraints we have just added. So let's just show that for any k \in \mathbb{Z}, there are infinitely many pairs of positive integers (m, n) so that n = m+9, m \equiv 99 \pmod{1000}, and kn + s(n^2) = km + s(m^2). Call k tasty if there is at least one such pair (m, n), and tangy if there are infinitely many such pairs (m, n). We seek to show that all positive integers are tangy. Lemma 1. There are arbitrarily small tasty integers. Proof. Consider m = \frac{10^{2t+1} + 98198}{198} and n = m+9 for sufficiently large t. The corresponding k can be as large as needed. \blacksquare Lemma 2. If k is tasty, then k+1 is tangy. Proof. Suppose that k \in \mathbb{Z} is tasty. Let (m, n) be a pair so that n = m+9, m \equiv 99 \pmod{1000}, and kn + s(n^2) = km + s(m^2). Then consider (10^{N} + m, 10^{N} + n) for sufficiently large N. These pairs imply that k+1 is tangy. \blacksquare By the previous two lemmas, all integers are tangy and so we're done. \square
22.08.2020 12:59
What does this s(n) mean exactly?
22.08.2020 13:20
Sum of digits of n (Iin base 10)