LCM-a-3-5-a-3a-a-

Question Number 93982 by Rasheed.Sindhi last updated on 16/May/20

$$\mathrm{LCM}\left({a},\frac{\mathrm{3}}{\mathrm{5}}{a}\right)=\mathrm{3}{a}\: \\ $$$${a}=? \\ $$

Commented by mr W last updated on 16/May/20

a=5b LCM(5b,3b)=3×5b this is true for any b≥1. therefore a=5, 10, 15, ...

$${a}=\mathrm{5}{b} \\ $$$${LCM}\left(\mathrm{5}{b},\mathrm{3}{b}\right)=\mathrm{3}×\mathrm{5}{b} \\ $$$${this}\:{is}\:{true}\:{for}\:{any}\:{b}\geqslant\mathrm{1}.\:{therefore} \\ $$$${a}=\mathrm{5},\:\mathrm{10},\:\mathrm{15},\:… \\ $$

Commented by PRITHWISH SEN 2 last updated on 16/May/20

the qiestion cannot have any unique solution it can be true for almost every value of a∈R as if a=2 then LCM (2,(6/5))=((LCM of 2,6)/(HCF of 1,5)) = (6/1) = 6=3.2

$$\mathrm{the}\:\mathrm{qiestion}\:\mathrm{cannot}\:\mathrm{have}\:\mathrm{any}\:\mathrm{unique}\:\mathrm{solution} \\ $$$$\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{true}\:\mathrm{for}\:\mathrm{almost}\:\mathrm{every}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\in\mathbb{R} \\ $$$$\mathrm{as}\:\mathrm{if}\:\boldsymbol{\mathrm{a}}=\mathrm{2} \\ $$$$\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{LCM}}\:\left(\mathrm{2},\frac{\mathrm{6}}{\mathrm{5}}\right)=\frac{\mathrm{LCM}\:\mathrm{of}\:\mathrm{2},\mathrm{6}}{\mathrm{HCF}\:\mathrm{of}\:\mathrm{1},\mathrm{5}}\:=\:\frac{\mathrm{6}}{\mathrm{1}}\:=\:\mathrm{6}=\mathrm{3}.\mathrm{2} \\ $$

Commented by mr W last updated on 16/May/20

what i know is that the function LCM applies only to two or more integers. therefore (3/5)a should be integer.

$${what}\:{i}\:{know}\:{is}\:{that}\:{the}\:{function} \\ $$$${LCM}\:{applies}\:{only}\:{to}\:{two}\:{or}\:{more} \\ $$$${integers}.\:{therefore}\:\frac{\mathrm{3}}{\mathrm{5}}{a}\:{should}\:{be} \\ $$$${integer}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 16/May/20

LCM for two or more factors = ((LCM of numerator)/(HCF of denominator)) HCF for two or more factors = ((HCF of numerator)/(LCM of denominator))

$$\mathrm{LCM}\:\mathrm{for}\:\mathrm{two}\:\mathrm{or}\:\mathrm{more}\:\mathrm{factors}\:=\:\frac{\mathrm{LCM}\:\mathrm{of}\:\mathrm{numerator}}{\mathrm{HCF}\:\mathrm{of}\:\mathrm{denominator}} \\ $$$$\mathrm{HCF}\:\mathrm{for}\:\mathrm{two}\:\mathrm{or}\:\mathrm{more}\:\mathrm{factors}\:=\:\frac{\mathrm{HCF}\:\mathrm{of}\:\mathrm{numerator}}{\mathrm{LCM}\:\mathrm{of}\:\mathrm{denominator}} \\ $$

Commented by mr W last updated on 16/May/20