find-the-smallest-number-which-when-we-divide-by-12-15-18-and-27-leaves-a-remainder-of-8-11-14-and-23-respectively-

Question Number 136040 by zakirullah last updated on 18/Mar/21

$$\:\:\:\:\:\:{find}\:{the}\:{smallest}\:{number}\:{which}\:{when}\:{we}\:{divide}\:{by}\:\mathrm{12},\mathrm{15},\mathrm{18},\:\mathrm{and}\:\mathrm{27}\:\mathrm{leaves}\:\mathrm{a}\:\mathrm{remainder} \\ $$$$\:\:\:\:\:\:\mathrm{of}\:\mathrm{8},\mathrm{11},\mathrm{14}\:\mathrm{and}\:\mathrm{23}\:\mathrm{respectively}? \\ $$

Answered by mr W last updated on 18/Mar/21

N=27a+23=27a+9+14 27a+9=18b ⇒3a+1=2b ⇒a=2c+1 N=27(2c+1)+23=54c+39+11 54c+39=15d ⇒18c+13=5d ⇒c=5e−1 N=54(5e−1)+50=270e−12+8 270e−12=12f ⇒45e−2=2f ⇒e=2n N=270×2n−4=540n−4 N_(min) =536

$${N}=\mathrm{27}{a}+\mathrm{23}=\mathrm{27}{a}+\mathrm{9}+\mathrm{14} \\ $$$$\mathrm{27}{a}+\mathrm{9}=\mathrm{18}{b}\:\Rightarrow\mathrm{3}{a}+\mathrm{1}=\mathrm{2}{b}\:\Rightarrow{a}=\mathrm{2}{c}+\mathrm{1} \\ $$$${N}=\mathrm{27}\left(\mathrm{2}{c}+\mathrm{1}\right)+\mathrm{23}=\mathrm{54}{c}+\mathrm{39}+\mathrm{11} \\ $$$$\mathrm{54}{c}+\mathrm{39}=\mathrm{15}{d}\:\Rightarrow\mathrm{18}{c}+\mathrm{13}=\mathrm{5}{d}\:\Rightarrow{c}=\mathrm{5}{e}−\mathrm{1} \\ $$$${N}=\mathrm{54}\left(\mathrm{5}{e}−\mathrm{1}\right)+\mathrm{50}=\mathrm{270}{e}−\mathrm{12}+\mathrm{8} \\ $$$$\mathrm{270}{e}−\mathrm{12}=\mathrm{12}{f}\:\Rightarrow\mathrm{45}{e}−\mathrm{2}=\mathrm{2}{f}\:\Rightarrow{e}=\mathrm{2}{n} \\ $$$${N}=\mathrm{270}×\mathrm{2}{n}−\mathrm{4}=\mathrm{540}{n}−\mathrm{4} \\ $$$${N}_{{min}} =\mathrm{536} \\ $$

Answered by Rasheed.Sindhi last updated on 19/Mar/21

This is a special case in which difference between a divisor and its corresponding remainder is same. 12−8=15−11=18−14=27−23=4 In this case: Required number=LCM(divisors)−common difference =LCM(12,15,18,27)−4=540−4=536

$${This}\:{is}\:{a}\:{special}\:{case}\:{in}\:{which}\:{difference} \\ $$$${between}\:{a}\:{divisor}\:{and}\:{its}\:{corresponding} \\ $$$${remainder}\:{is}\:{same}. \\ $$$$\mathrm{12}−\mathrm{8}=\mathrm{15}−\mathrm{11}=\mathrm{18}−\mathrm{14}=\mathrm{27}−\mathrm{23}=\mathrm{4} \\ $$$${In}\:{this}\:{case}: \\ $$$${Required}\:{number}={LCM}\left({divisors}\right)−{common}\:{difference} \\ $$$$={LCM}\left(\mathrm{12},\mathrm{15},\mathrm{18},\mathrm{27}\right)−\mathrm{4}=\mathrm{540}−\mathrm{4}=\mathrm{536} \\ $$

Commented by mr W last updated on 19/Mar/21

$${great}\:{sir}! \\ $$

Commented by Rasheed.Sindhi last updated on 19/Mar/21

TH∀NKSss Sir! Please see my answer to Q#135507 if you haven′t seen it & give critical remark.

$$\mathcal{TH}\forall\mathcal{NKS}{ss}\:\mathcal{S}{ir}! \\ $$$$\mathcal{P}{lease}\:{see}\:{my}\:{answer}\:{to} \\ $$$${Q}#\mathrm{135507}\:{if}\:{you}\:{haven}'{t}\:{seen}\:{it} \\ $$$$\&\:{give}\:{critical}\:{remark}. \\ $$

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