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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?

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    @Kartik. Ah. I didn't know it was an exam problem. Marvis' answer is definitely the better solution then2012-06-04

3 Answers 3

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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 $199$.

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    seriously thank you :)2012-06-03
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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)

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    Your solution is superb indeed, but I wanted something which I could have figured out during my exam :)2012-06-03
1

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 $b=c=9$,

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

$\frac{(r-10)}{9}=\frac{(10a-1)}{(a+18)}$

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

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

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