Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question goes as follows:

Let $K$ be a three digit number such that the ratio of the number to the sum of its digit is least. What is the difference between the hundreds and the tens digits of $K$?

Now I was able to do this question by trial and error, assuming hundredth digit place to be 1 and unit as well as tens digit to be 9

So the number is 199, but I am not able to do it logically, any way to do it?

share|cite|improve this question
is at least...? Also: brute force is perfectly logical, just inelegant. – anon Jun 3 '12 at 16:54
that's what I was looking for, an elegant solution, but brute force wont let me prove that yes this is the right answer. – Kartik Anand Jun 3 '12 at 16:57
@Kartik I feel obliged to point out that this is problem where brute force can definitely prove the right answer. There are only 900 possible $K$s, so it is feasible to check each one to find the minimum. – Drew Christianson Jun 3 '12 at 17:54
@DrewChristianson checking just 900 possible K's would be good on a computer, but I had this problem in my exam, so you see I wanted to know if something different is possible :) – Kartik Anand Jun 3 '12 at 18:43
@Kartik. Ah. I didn't know it was an exam problem. Marvis' answer is definitely the better solution then – Drew Christianson Jun 4 '12 at 0:27
up vote 7 down vote accepted

Let the number be $100a + 10 b +c$, where $1 \leq a \leq 9$ and $0 \leq b,c, \leq9$. Hence, we want to minimize $$L=\dfrac{100a + 10b + c}{a+b+c} = 1 + \dfrac{99a+9b}{a+b+c}$$ This means that you should choose $c$ to be maximum as possible since $c$ appears only in the denominator and the term is positive. Hence, $c = 9$. Hence, we want to minimize $$L=\dfrac{100a + 10b + c}{a+b+c} = 1 + \dfrac{99a+9b}{a+b+9} = 1 + \dfrac{90a - 81 + 9a+ 9b + 81}{a+b+9}$$ $$L = 1 + \dfrac{90a-81}{a+b+9} + 9 = 10 + 9 \left(\dfrac{10a-9}{a+b+9} \right)$$ Now again you should choose $b$ to be maximum as possible since $b$ appears only in the denominator and the term is positive. Hence, set $b=9$. Hence, we want to minimize $$L = 10 +9 \left( \dfrac{10a-9}{a+18} \right) = 10 +9 \left( \dfrac{10a + 180 -189}{a+18} \right) = 10 + 90 - \dfrac{9 \times 189}{a+18}$$ Now you ned to choose $a$ to be minimum as possible since $a$ appears in the denominator and the term is negative. Hence, set $a = 1$. Hence, $$L = 100 - \dfrac{9 \times 189}{19} = \dfrac{1900 - 9 \times 189}{19} = \dfrac{199}{19}$$ The number is $199$.

share|cite|improve this answer
seriously thank you :) – Kartik Anand Jun 3 '12 at 17:28

Doing it in a single function, just for fun:

For $x \in \mathbb{Z}$ we want to minimize the ratio $\frac{100a+10b+c}{a+b+c}$ where $ a,b,c\in\mathbb{Z}$ and $100a+10b+c=x$ We can rewrite the ratio as: $$ f(x)=\frac{100\left\lfloor\frac{x}{100}\right\rfloor+10\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rfloor}{10}\right\rfloor+\left\lfloor x-100\left\lfloor\frac{x}{100}\right\rfloor-10\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rfloor}{10}\right\rfloor\right\rfloor}{\left\lfloor\frac{x}{100}\right\rfloor+\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rfloor}{10}\right\rfloor+\left\lfloor x-100\left\lfloor\frac{x}{100}\right\rfloor-10\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rfloor}{10}\right\rfloor\right\rfloor}$$ Which, when graphed, gives: this which clearly is minimized at 199 (or 1-9 if the $x$ is allowed to have two leading $0$s)

share|cite|improve this answer
really interesting, loved it ! – Theorem Jun 3 '12 at 17:42
how did u manage to write it in terms of floor ? – Theorem Jun 3 '12 at 17:44
With positive division, floor is effectively a truncate function; it returns the integer part of a ratio. Lets take $x=152$ as an example. So let, $d_{100}(152)= \left\lfloor\frac{152}{100}\right\rfloor = \left\lfloor 1.52\right\rfloor = 1$ which is the hundreds digit. Then clearly, $\left\lfloor\frac{152-100*1}{10}\right\rfloor = 5$ is the tens digit, because we've subtracted off the hundreds. Just rephrase the $100*1$ to $100*d_{100}(152)$, and you've got a general expression for the tens digit. Repeat the same process for the 1s digit. – Drew Christianson Jun 3 '12 at 17:51
The pic is not showing up – Kartik Anand Jun 3 '12 at 18:46
Your solution is superb indeed, but I wanted something which I could have figured out during my exam :) – Kartik Anand Jun 3 '12 at 18:52

We want $a>0,b\geq 0,c\geq 0$, $a,b,c\leq 9$ s.t. $r = \frac{(100a+10b+c)}{(a+b+c)}$ is minimum.

$r = 1 + \frac{(99a+9b)}{(a+b+c)}$.

For a given $a$ and $b$, this is least when $c$ is maximum, i.e. $9$.

Now given $c=9, r = 1 + \frac{(99a+9b)}{(a+b+9)} = 1 + 9 + \frac{(90a-9)}{(a+b+9)}$.

Again, we see this is least given $a$ when $b$ is maximum, i.e. $9$.

Now, given $b=c=9$,

$r = 10+\frac{(90a-9)}{(a+18)}$


$= \frac{(10a+180-179)}{(a+18)}$

$= 10 - \frac{179}{(a+18)}$.

Clearly, this is least when a has the least possible value, i.e. 1.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.