Exercise 545Number theory › PrimesHigh schoolAdded 3 Nov 2026 Primes in GCD and LCM #gcd#lcm#prime factorization StatementLet a=23⋅32⋅5a=2^3\cdot3^2\cdot5a=23⋅32⋅5 and b=22⋅33⋅7b=2^2\cdot3^3\cdot7b=22⋅33⋅7. Compute gcd(a,b)\gcd(a,b)gcd(a,b) and lcm(a,b)\operatorname{lcm}(a,b)lcm(a,b). HintGCD takes the minimum exponent, LCM the maximum exponent of each prime. Show solutionHide solutionSolutionGCD. gcd(a,b)=2min(3,2)⋅3min(2,3)⋅5min(1,0)⋅7min(0,1)=22⋅32=36\gcd(a,b)=2^{\min(3,2)}\cdot3^{\min(2,3)}\cdot5^{\min(1,0)}\cdot7^{\min(0,1)}=2^2\cdot3^2=36gcd(a,b)=2min(3,2)⋅3min(2,3)⋅5min(1,0)⋅7min(0,1)=22⋅32=36.LCM. lcm(a,b)=23⋅33⋅5⋅7=8⋅27⋅35=7560\operatorname{lcm}(a,b)=2^3\cdot3^3\cdot5\cdot7=8\cdot27\cdot35=7560lcm(a,b)=23⋅33⋅5⋅7=8⋅27⋅35=7560.Check. gcd⋅lcm=36⋅7560=272160\gcd\cdot\operatorname{lcm}=36\cdot7560=272160gcd⋅lcm=36⋅7560=272160, and ab=(8⋅9⋅5)(4⋅27⋅7)=360⋅756=272160ab=(8\cdot9\cdot5)(4\cdot27\cdot7)=360\cdot756=272160ab=(8⋅9⋅5)(4⋅27⋅7)=360⋅756=272160. Correct.Conclusion. gcd=36\gcd=36gcd=36, lcm=7560\operatorname{lcm}=7560lcm=7560.