What-is-the-minimum-value-obtained-when-an-arbitrary-number-of-three-different-non-zero-digits-is-divided-by-the-sum-of-its-digits-

Question Number 111537 by Aina Samuel Temidayo last updated on 04/Sep/20

What is the minimum value obtained

when an arbitrary number of three

different non - zero digits is divided

by the sum of its digits ?

Answered by 1549442205PVT last updated on 04/Sep/20

Suppose \overset{―}{abc} = 100 a + 20 b + c . We need

find the smallest value of

P = \frac{100 a + 10 b + c}{a + b + c} where

a, b, c \in {1, 2, 3, \dots, 9}; a \neq b; a \neq c; b; c

P = 1 + \frac{99 a + 9 b}{a + b + c} . We will prove that

\frac{99 a + 9 b}{a + b + c} ⩾ \frac{171}{18} \Leftrightarrow \frac{11 a + b}{a + b + c} ⩾ \frac{19}{18} (*)

\Leftrightarrow 198 a + 18 b ⩾ 19 a + 19 b + 19 c

\Leftrightarrow 179 a ⩾ b + 19 c . Since b \neq c ⩽ 9

This last inequality is always true since

b, c ⩽ 9 \Rightarrow b + 20 c ⩽ 8 + 19.9 = 179 ⩽ 179 a

due to a ⩾ 1 . Hence the inequality is

proved . Consequently, P ⩾ 1 + \frac{171}{18} = \frac{21}{2}

Which means \frac{\overset{―}{abc}}{a + b + c} has least value

equal to \frac{21}{2} when \overset{―}{abc} = 189

Commented by Her_Majesty last updated on 04/Sep/20

“ t h r e e d i f f e r e n t {d i g i t s}^{″}

Commented by 1549442205PVT last updated on 04/Sep/20

Thank you, Sir . I missed that hypothesis

and corrected

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

Thanks .

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

How did you know you are to prove

that \frac{99 a + 9 b}{a + b + c} ⩾ \frac{171}{18} ?

Commented by 1549442205PVT last updated on 05/Sep/20

Had in the above solution since it is

equivalent \frac{11 a + b}{a + b + c} ⩾ \frac{19}{18}

\Leftrightarrow 198 a + 18 b ⩾ 19 a + 19 b + 19 c

\Leftrightarrow 179 a ⩾ 19 c + b . The inequality is

always true \forall a, b, c \in {1, 2, \dots, 9}, a \neq b,

b \neq c, c \neq a . Indeed, 19 c + b ⩽ 19.9 + 8

= 179.1 ⩽ 179 a due to a ⩾ 11

Commented by Aina Samuel Temidayo last updated on 05/Sep/20

Thanks but you {didn}^{'} t really answer

my question .

Answered by Her_Majesty last updated on 04/Sep/20

\frac{100 a + 10 b + c}{a + b + c}

(1) l o w e s t n u m b e r w i t h a < b < c

(2) a a s l o w a s p o s s i b l e \Rightarrow a = 1

(3) c a s h i g h a s p o s s i b l e \Rightarrow c = 9

\frac{10 b + 109}{b + 10} = 10 w i t h b \to \infty \Rightarrow b a s h i g h a s

p o s s i b l e \Rightarrow b = 8 (a \neq b \neq c)

\Rightarrow \frac{189}{18} = 10.5

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

How did you know \frac{10 b + 109}{b + 10} is

equal to 10 ?