Question

MINIMUM DIGIT SUM FRACTION

Answer 1

Let S(N) denote the digit sum of the integer N. As N ranges over all 3-digit positive numbers, what value of N would give the minimum of
\[\huge \ M=\frac{N}{S(N)}\]

Answer 2

M = (100a + 10b + c)/(a+b+c) = 1 + (99a + 9b)/(a+b+c) = 1+ 9(11a +b)/(a+b+c)
So its min when
(11a +b) is min and a+b+c is max
lets set a= 1 and b=0, then c=9

10.9 ? hmm

Answer 3

I am just guessing

Answer 4

119/11 is lesser. :|

Answer 5

a=1 b=1 c=9

Answer 6

One thing I see a<=b<=c

Answer 7

infact, lest till now I get is 199/19

Answer 8

least*

Answer 9

this should be the least cos b<=9 c<9 a=1

Answer 10

hmm, maybe
But I have only used hit and trial, an explanatory solution will be nice,
@mukushla

Answer 11

199 is correct but as you said trial and error

Answer 12

by the way wat happened to 1+
why dont we say
there \[N=\frac{a+b+c+9(11a+b)}{a+b+c}\]

Answer 13

We have
1+ 9(11a +b)/(a+b+c)
this will be minimum when (11a+b)/(a+b+c) is min.
=> (a + b + c + 10a -c)/(a+b+c) = 1 + (10a-c)/(a+b+c)

10a-c should be min and a+b+c should be max
easily, 1a-c is min for a=1 and c=9

then b=9 to make a+b+c max, and hence the ans

same way, guess we can maximize M
for max, (10a-c) should be max and (a+b+c) should be min

10a -c is max for a=9 and c=0, and a+b+c then is min for b=9
hence max(N) = 900.
hmm.

Answer 14

@mukushla