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Question Number 217244 by ArshadS last updated on 07/Mar/25
 Find all two-digit numbers such that  when the number is divided by   the sum of its digits the quotient   is 4 and the remainder is 3.
$$\:\mathrm{Find}\:\mathrm{all}\:\mathrm{two}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{digits}\:\mathrm{the}\:\mathrm{quotient}\: \\ $$$$\mathrm{is}\:\mathrm{4}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{3}. \\ $$
Answered by Frix last updated on 07/Mar/25
((10a+b)/(a+b))=4+(3/(a+b)) ∧ a+b>3  ⇒  b=2a−1∧a+b>3  a=2∧b=3  a=3∧b=5  a=4∧b=7  a=5∧b=9  S={23, 35, 47, 59}
$$\frac{\mathrm{10}{a}+{b}}{{a}+{b}}=\mathrm{4}+\frac{\mathrm{3}}{{a}+{b}}\:\wedge\:{a}+{b}>\mathrm{3} \\ $$$$\Rightarrow \\ $$$${b}=\mathrm{2}{a}−\mathrm{1}\wedge{a}+{b}>\mathrm{3} \\ $$$${a}=\mathrm{2}\wedge{b}=\mathrm{3} \\ $$$${a}=\mathrm{3}\wedge{b}=\mathrm{5} \\ $$$${a}=\mathrm{4}\wedge{b}=\mathrm{7} \\ $$$${a}=\mathrm{5}\wedge{b}=\mathrm{9} \\ $$$${S}=\left\{\mathrm{23},\:\mathrm{35},\:\mathrm{47},\:\mathrm{59}\right\} \\ $$
Commented by ArshadS last updated on 10/Mar/25
Grateful sir!
$$\mathrm{Grateful}\:\mathrm{sir}! \\ $$

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