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.


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Question Number 217359 by ArshadS last updated on 11/Mar/25

$Find all three - digit numbers such that when the number is$ $divided by the sum of its digits the quotient is 7 and the$ $remainder is 5 .$
	Answered by Rasheed.Sindhi last updated on 12/Mar/25

	$100 h + 10 t + u = 7 (h + t + u) + 5$ $93 h + 3 t - 6 u = 5$ $3 (31 h + t - 2 u) = 5$ $l h s i s m u l t i p l e o f 3 w h i l e r h s i s n o t$ $\Rightarrow d i g i t s h, t, u a r e i m p o s s i b l e .$
	Commented by ArshadS last updated on 13/Mar/25

	$O k s i r!$
	Answered by nikif99 last updated on 11/Mar/25
	$Let numbers a^− b^− c^− , where a∈{1,..,9} and b, c∈{0,..,9} 100a+10b+c=7(a+b+c)+5⇔ 93a+3b−6c=5 (1) if a≥2⇒93×2+3b−6c=5 impossible ⇒a=1 (1): 93+3b−6c=5⇔6c−3b=88⇒ 6c≥88⇒c≥14 impossible No such three digit numbers.$
	$L e t n u m b e r s \bar{a} \bar{b} \bar{c},$ $w h e r e a \in {1, . ., 9} a n d b, c \in {0, . ., 9}$ $100 a + 10 b + c = 7 (a + b + c) + 5 \Leftrightarrow$ $93 a + 3 b - 6 c = 5 (1)$ $i f a ⩾ 2 \Rightarrow 93 \times 2 + 3 b - 6 c = 5 i m p o s s i b l e$ $\Rightarrow a = 1$ $(1) : 93 + 3 b - 6 c = 5 \Leftrightarrow 6 c - 3 b = 88 \Rightarrow$ $6 c ⩾ 88 \Rightarrow c ⩾ 14 i m p o s s i b l e$ $N o s u c h t h r e e d i g i t n u m b e r s .$
	Commented by ArshadS last updated on 13/Mar/25

	$Thanks sir!$
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