Olympiad Maths Prep

Track / Stage 5 / 195 of 400 #795 of 2000

Problem 795

AIME late
Number theory Difficulty 5.5 Find the answer

2.1. Let A={3+10k,6+26k,5+29k,k=1,2,3,4,}A=\{3+10 k, 6+26 k, 5+29 k, k=1,2,3,4, \cdots\}. Determine the smallest positive integer rr such that there exists an integer bb with the property that the set B={b+rk,k=1,2,3,4,}B=\{b+r k, k=1,2,3,4, \cdots\} is disjoint from AA

Official solution

2.1 The answer is 290 . First observe that the arithmetic sequences {p+mk,kN}\{p+m k, k \in \mathbb{N}\} and {q+nk,kN}\{q+n k, k \in \mathbb{N}\} are disjoint if and only if pqlnkmp-q \neq l n-k m for all integers k,lk, l, which holds if and only if gcd(m,n)\operatorname{gcd}(m, n) does not divide pqp-q. Therefore, the required rr cannot be relatively prime to 10=(2)(5),26=(2)(13)10=(2)(5), 26=(2)(13) and 29 . We start by choosing rr to be the smallest lcm\mathrm{lcm} of d1,d2,d3d_{1}, d_{2}, d_{3} where d1,d2,d3d_{1}, d_{2}, d_{3} are factors (greater than 1 ) of 10,26 and 29 respectively. The smallest such rr is 58=(2)(29)58=(2)(29). In this case, gcd(10,58)=2,gcd(26,58)=2\operatorname{gcd}(10,58)=2, \operatorname{gcd}(26,58)=2 and gcd(29,58)=29\operatorname{gcd}(29,58)=29. We also require b3≢0(mod2),b6≢0(mod2)b-3 \not \equiv 0(\bmod 2), b-6 \not \equiv 0(\bmod 2) and b5≢0b-5 \not \equiv 0 (mod 29). But there is no solution for bb from the first two equations. Therefore we cannot take r=58r=58. The next smallest lcm would be 290=(5)(2)(29)290=(5)(2)(29). In this case, a simple checking using the above criterion shows that {1+290k,kN}\{1+290 k, k \in \mathbb{N}\} is disjoint from AA

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.