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Question Number 217359 by ArshadS last updated on 11/Mar/25
Find all three-digit numbers such that when the number is divided by the sum of its digits the quotient is 7 and the remainder is 5.
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{three}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mathrm{when}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is} \\ $$$$\mathrm{divided}\:\mathrm{by}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{digits}\:\mathrm{the}\:\mathrm{quotient}\:\mathrm{is}\:\mathrm{7}\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{remainder}\:\mathrm{is}\:\mathrm{5}. \\ $$
Answered by Rasheed.Sindhi last updated on 12/Mar/25
100h+10t+u=7(h+t+u)+5 93h+3t−6u=5 3(31h+t−2u)=5 lhs is multiple of 3 while rhs is not ⇒digits h,t,u are impossible.
$$\mathrm{100}{h}+\mathrm{10}{t}+{u}=\mathrm{7}\left({h}+{t}+{u}\right)+\mathrm{5} \\ $$$$\mathrm{93}{h}+\mathrm{3}{t}−\mathrm{6}{u}=\mathrm{5} \\ $$$$\mathrm{3}\left(\mathrm{31}{h}+{t}−\mathrm{2}{u}\right)=\mathrm{5} \\ $$$${lhs}\:{is}\:{multiple}\:{of}\:\mathrm{3}\:{while}\:{rhs}\:{is}\:{not} \\ $$$$\Rightarrow{digits}\:{h},{t},{u}\:{are}\:{impossible}. \\ $$
Commented by ArshadS last updated on 13/Mar/25
Ok sir!
$${Ok}\:{sir}! \\ $$
Answered by nikif99 last updated on 11/Mar/25
Let numbers a^− b^− c^− , where a∈{1,..,9} and b, c∈{0,..,9} 100a+10b+c=7(a+b+c)+5⇔ 93a+3b−6c=5 (1) if a≥2⇒93×2+3b−6c=5 impossible ⇒a=1 (1): 93+3b−6c=5⇔6c−3b=88⇒ 6c≥88⇒c≥14 impossible No such three digit numbers.
$${Let}\:{numbers}\:\overset{−} {{a}}\overset{−} {{b}}\overset{−} {{c}},\: \\ $$$${where}\:{a}\in\left\{\mathrm{1},..,\mathrm{9}\right\}\:{and}\:{b},\:{c}\in\left\{\mathrm{0},..,\mathrm{9}\right\} \\ $$$$\mathrm{100}{a}+\mathrm{10}{b}+{c}=\mathrm{7}\left({a}+{b}+{c}\right)+\mathrm{5}\Leftrightarrow \\ $$$$\mathrm{93}{a}+\mathrm{3}{b}−\mathrm{6}{c}=\mathrm{5}\:\left(\mathrm{1}\right) \\ $$$${if}\:{a}\geqslant\mathrm{2}\Rightarrow\mathrm{93}×\mathrm{2}+\mathrm{3}{b}−\mathrm{6}{c}=\mathrm{5}\:{impossible} \\ $$$$\Rightarrow{a}=\mathrm{1} \\ $$$$\left(\mathrm{1}\right):\:\mathrm{93}+\mathrm{3}{b}−\mathrm{6}{c}=\mathrm{5}\Leftrightarrow\mathrm{6}{c}−\mathrm{3}{b}=\mathrm{88}\Rightarrow \\ $$$$\mathrm{6}{c}\geqslant\mathrm{88}\Rightarrow{c}\geqslant\mathrm{14}\:{impossible} \\ $$$${No}\:{such}\:{three}\:{digit}\:{numbers}. \\ $$
Commented by ArshadS last updated on 13/Mar/25
Thanks sir!
$$\mathrm{Thanks}\:\mathrm{sir}! \\ $$

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