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Find-the-smallest-number-such-that-when-divided-by-18-the-remainder-is-17-When-divided-by-20-the-remainder-is-19-and-when-divided-by-24-the-remainder-is-23-

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Question Number 13154 by tawa tawa last updated on 15/May/17
Find the smallest number such that when divided by 18 the remainder is 17, When divided by 20 the remainder is 19. and when divided by 24 the remainder is 23.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{18}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{17}, \\ $$$$\mathrm{When}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{20}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{19}.\:\mathrm{and}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{24}\:\mathrm{the}\:\mathrm{remainder}\: \\ $$$$\mathrm{is}\:\mathrm{23}.\: \\ $$
Answered by mrW1 last updated on 15/May/17
N=18i+17=20j+19=24k+23 20j+19=18j+18+2j+1=18i+17 ⇒18(j−i)+2j+2=0 ⇒9(j−i)+j+1=0 ⇒j+1=9i′ ⇒j=9i′−1 20(9i′−1)+19=24k+23 20×9i′=24(k+1) 5×3i′=2(k+1) ⇒i′=2n ⇒k=15n−1 ⇒N=24(15n−1)+23=360n−1 (n=1,2,3...) The number is 359 or 719 or...
$${N}=\mathrm{18}{i}+\mathrm{17}=\mathrm{20}{j}+\mathrm{19}=\mathrm{24}{k}+\mathrm{23} \\ $$$$\mathrm{20}{j}+\mathrm{19}=\mathrm{18}{j}+\mathrm{18}+\mathrm{2}{j}+\mathrm{1}=\mathrm{18}{i}+\mathrm{17} \\ $$$$\Rightarrow\mathrm{18}\left({j}−{i}\right)+\mathrm{2}{j}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{9}\left({j}−{i}\right)+{j}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{j}+\mathrm{1}=\mathrm{9}{i}' \\ $$$$\Rightarrow{j}=\mathrm{9}{i}'−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{20}\left(\mathrm{9}{i}'−\mathrm{1}\right)+\mathrm{19}=\mathrm{24}{k}+\mathrm{23} \\ $$$$\mathrm{20}×\mathrm{9}{i}'=\mathrm{24}\left({k}+\mathrm{1}\right) \\ $$$$\mathrm{5}×\mathrm{3}{i}'=\mathrm{2}\left({k}+\mathrm{1}\right) \\ $$$$\Rightarrow{i}'=\mathrm{2}{n} \\ $$$$\Rightarrow{k}=\mathrm{15}{n}−\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow{N}=\mathrm{24}\left(\mathrm{15}{n}−\mathrm{1}\right)+\mathrm{23}=\mathrm{360}{n}−\mathrm{1}\:\:\left({n}=\mathrm{1},\mathrm{2},\mathrm{3}…\right) \\ $$$${The}\:{number}\:{is}\:\mathrm{359}\:{or}\:\mathrm{719}\:{or}… \\ $$
Commented by RasheedSindhi last updated on 15/May/17
The smallest number is required So I think the answer is only 359.
$$\mathrm{The}\:\mathrm{smallest}\:\mathrm{number}\:\mathrm{is}\:\mathrm{required} \\ $$$$\mathrm{So}\:\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{only}\:\mathrm{359}. \\ $$
Commented by tawa tawa last updated on 15/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by RasheedSindhi last updated on 15/May/17
∗Common diffefence of divisor and its remainder =18−17=20−19=24−23=1 ∗lcm of divisors 18,20,24 : 18=2.3^2 20=2^2 .5 24=2^3 .3 lcm is 2^3 .3^2 .5=360 ∗lcm-(common diff. of divisor &remainder)u =360−1=359
$$\ast\mathrm{Common}\:\mathrm{diffefence}\:\mathrm{of}\:\mathrm{divisor} \\ $$$$\mathrm{and}\:\mathrm{its}\:\mathrm{remainder} \\ $$$$\:\:\:\:=\mathrm{18}−\mathrm{17}=\mathrm{20}−\mathrm{19}=\mathrm{24}−\mathrm{23}=\mathrm{1} \\ $$$$\:\ast\mathrm{lcm}\:\mathrm{of}\:\mathrm{divisors}\:\mathrm{18},\mathrm{20},\mathrm{24}\:: \\ $$$$\:\mathrm{18}=\mathrm{2}.\mathrm{3}^{\mathrm{2}} \\ $$$$\:\mathrm{20}=\mathrm{2}^{\mathrm{2}} .\mathrm{5} \\ $$$$\:\mathrm{24}=\mathrm{2}^{\mathrm{3}} .\mathrm{3} \\ $$$$\mathrm{lcm}\:\mathrm{is}\:\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{2}} .\mathrm{5}=\mathrm{360} \\ $$$$ \\ $$$$\ast\mathrm{lcm}-\left(\mathrm{common}\:\mathrm{diff}.\:\mathrm{of}\:\mathrm{divisor}\:\&\mathrm{remainder}\right)\mathrm{u} \\ $$$$=\mathrm{360}−\mathrm{1}=\mathrm{359} \\ $$$$ \\ $$
Commented by tawa tawa last updated on 15/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mrW1 last updated on 16/May/17
That′s smart!
$${That}'{s}\:{smart}! \\ $$

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