A special 3-digit number

We first show that $N/S(N) \ge 199/19 > 10$ . To see this, let $N = 100a + 10b + c$ for integers $1 \le a \le 9$ , $0 \le b, c \le 9$ . Then $\frac{N}{S(N)} = \frac{100a+10b+c}{a+b+c} = 100 - 9\frac{10b+11c}{a+b+c},$ clearly showing that the expression is a strictly increasing function of $a$ , whose minimum is attained when $a = 1$ . Similarly, $\frac{N}{S(N)} = 1 + 9 \frac{11a + b}{a+b+c},$ which shows that the expression is minimized when $c$ is as large as possible. So with the choices $a = 1, c = 9$ , we can now write $\begin{aligned} \frac{N}{S(N)} &\ge \frac{10b + 109}{b+10} = 10 + \frac{9}{b+10} \\ &\ge 10 + \frac{9}{19} = \frac{199}{19}, \end{aligned}$ with equality attained when $N = 199$ . Hence we must have $R \ge 11$ ; but by inspection, we find $\frac{198}{S(198)} = 11,$ so $R = 11$ .

Next, suppose by contradiction that there exists $N' = 100a' + 10b' + c' > 198$ such that $N'/S(N') = 11$ . Since $N = 199$ does not result in an integer value of this ratio, it follows that $a' \ge 2$ . But the above inequalities remain valid, so we find $\begin{aligned} \frac{N'}{S(N')} &= \frac{100a'+10b'+c'}{a'+b'+c'} \ge \frac{10b' + 209}{b'+11} \\ &= 10 + \frac{99}{b'+11} > 10 + \frac{99}{21} \\ &> 14. \end{aligned}$ Therefore no such $N'$ exists, and the largest 3-digit number for which $N/S(N) = R$ is $\boxed{198}$ .

let [xyz] denote a 3 digit number such that [xyz]=100x+10y+z ; S(N)=x+y+z N/S(N)=(100x+10y+z)/(x+y+z)=(99x+9y)/(x+y+z) + 1 as numerator is independent of z for minimum value of R, Z=9 (99x+9y)/(x+y+9)=(90x-81)/(x+y+9) + 9 for minimum value of R ,y=9, x=1 min{N/S(N)}=199/19,hence its least integral value is 11 (100x+10y+z)/(x+y+z)=11 89x=10z+y 89x=[zy], hence only solution is [xyz]=198

Is a number abc where a,b and c are its digits. So R=(100 a+10 b+c)/(a+b+c) Suppose that R<=10 So (a+b+c) R<=10 (a+b+c) if and only if (100 a+10 b+c)<=10 (a+b+c) if and only if 90 a<=9 c if and only if 10 a<=c but a>=1 and c<=9 so 10<=9 wich is absurd so R>=11 For R=11: So 11 divides (100 a+10 b+c) and a+c=b But (100 a+10 b+c)=11 (a+b+c) so (99 a+11 b)=11 (2 b) if and only if 9 a=b but a>=1 and b<=9 so a=1 and b=9 and then c=8 so 11 is the smallest value for N/(S(N) and R=11 and the only number N satisfying N/S(N) being a integer is N=198.

N=100a+10b+c S(N)=a+b+c R=N/S(N) let's consider R as function of a,b,c (i)Partial derivative of R with respect to a will give 90b+99c/(a+b+c)^2 (ii)Partial derivative of R with respect to b will give -90a+9c/(a+b+c)^2 (iii)Partial derivative of R with respect to c will give -99a-9b/(a+b+c)^2 Since a,b,c are integers with the following constraints 1<=a<=9 0<=b<=9 0<=c<=9 (i) is strictly increasing ,(ii) and (iii) are strictly decreasing. So to minimize R,we must decrease a,increase b and increase c. a=1,b=9,c=9 we'll get R=199/19=10.47 as minimum value. But R is an integer,so we'll decrease c until we'll get integer value and at c=8,we'll get integer value of R as 11. Lets cross check if 198 is the only value which gives 11 as integer value of R 100a+10b+c/a+b+c=11 89a=b+10c 10c+b represents two digit number thus a cannot be greater than 1. Thus 198 is only number that gives R as 11. Note:Here,first we have considered R as a continuous function of a,b and c so as to know the behaviour of this function and then apply restrictions.

We will first show that the minimum value is at least 11. This proof is borrowed from Ahaan R. from a problem 2 weeks ago calling for the minimum value (not minimum integral value) of $R$ .

We call $M$ as the minimum real value for the given fraction.

The number $N$ in question is a 3-digit number, so we can call the digits $A, B, C,$ , where $A \neq 0$ . This means: $N=100A+10B+C$ and $S(N)=A+B+C$ . So, by the definition of $M$ , we have: $M=\frac{100A+10B+C}{A+B+C}$ .

Remark that this can be re-written as $M=\frac{99A+9B}{A+B+C}+1$ . Consider this as a function of C and notice that C only appears in the denominator. We want to minimize this, so we want to maximize C; i.e. C=9 . (Note: It doesn't matter what positive values A and B are.)

Substituting this into the equation above, we obtain: $M=\frac{99A+9B}{A+B+9}+1=\frac{(90A-81)+(9A+9B+81)}{A+B+9}+1$ . Now, M can be rewritten as $M=\frac{90A-81}{A+B+9}+9+1=\frac{90A-81}{A+B+9}+10$ . Again, we wish to minimize this so, in the same manner, we note that B only appears in the denominator in this function of B. Clearly, we want to maximize B to minimize this fraction. So, B=9 .

Substituting B=9 into this, we get: $M=\frac{90A-81}{A+18}+10$ . Now, haste makes waste! We cannot just conclude that A=9 as well. In fact, it's quite clear that 999 isn't the best we can do. Note that $M=\frac{90A-81}{A+18}+10=\frac{90A+90×18-90×18-81}{A+18} +10=\frac{90-90×18+81}{A+18}+10$ Aha! We want to minimize this expression, so we want to maximize $\frac{1}{A+18}$ . And, to do this, we need to minimize A, since A is in the denominator. This is reverse logic from what we did with B and C. Since A can't be 0 (if it was, we wouldn't have a 3-digit number), A must be 1.

This shows us that in order to produce the minimum value of $M$ , $N$ must be 199. After some computation, we find this minimum value is $10\frac{9}{19}$ . Thus the smallest integral value for $R$ is $11$ , from which we find $100A+10B+C=11A+11B+11C$ , or $-89A+B+10C=0$ .

Let $n$ be a positive integer, and we substitute $B=9A+10n$ so that $-80A+10n+10C=0$ . This yields $C=8A-n$ . As B and C are both digits, they are both less than 10, and we have the two inequalities $9A+10n<10$ $8A-n<10$ Multiplying the second inequality by 10 and adding both inequalities yields $89A<110$ , and as A is a positive integer, we know that $A=1$ , so $B=9+10n$ and $C=8-n$ . From this we find the only appropriate value of n is 0, so B=9 and C=8, and N=198. We test N and we find that R=11, which was proven to be the minimal value of the fraction. We know that N cannot be any greater because of the above inequalities. Thus $N=\boxed{198}$ .

We can write the number N in this way: 100A + 10B + C Where A is the hundreds digit, b the tens digit of C and the units digit. Shortly S (N) = A + B + C If R is equal to 9, we will have

$\frac {100A + 10B + C} {A + B + C}$ = 9

9A +9B + 9C = C + 100A + 10B

81A + B = 8C

If the value of A is equal to 0 (because it is the figure in the hundreds), and the value of C is at most 9, I have this equality would never be true. For replacing extreme values​​, I would have to: 81 (1) = 0 + 8 (9) 81 = 72 .

Now, if the value of R is 10, we have:

100A + C = 10A + 10B + 10B + 10C.

10C = 90A.

10A = C.

Assigning the minimum value for A, we have that :
C = 10 What is impossible, since C is a digit.

Finally, with R = 11, we have:

100A + 10B + C = 11A +11 B +11 C

10C = 89A + B

This time, replacing the values ​​of A, B, C respectively by 1,9,8, equality becomes true:

89 (1) = 9 + 10 (8)

89 = 89

However, we have the lowest possible R is 11, and most N possible in these conditions would be the number 198.

Experimenting a little, we find that $R=11$ is the smallest possible integer value for $R$ , which occurs when $N=198$ . Now we have the find the largest $3$ -digit $N$ that also works for $R=11$ . To make the searching process easier, we note that the greatest value that $N$ could be is $297$ , because the largest possible digit sum of a $3$ -digit number is $27$ . Then, we just need to check the multiples of $11$ between $198$ and $297$ . Nome work except $198$ , so this is the answer.

Since the minimum value of $\frac{N}{S(N)}$ is $\frac{199}{19}$ (because a should be 1 when $N=100a+10b+c$ since we want to find the minimum value of (\frac{N}{S(N)}) and $b+c$ should be as big as possible, so both $b$ and $c$ are $9$ ) which is in between $10$ and $11$ , so the value of $R$ is $11$ because $R$ is the minimum integer value of $\frac{N}{S(N)}$ .

From the equation $\frac{N}{S(N)}=R$ , we can change it into $\frac{100a+10b+c}{a+b+c}=11$ when $N=100a+10b+c$ .

Then, multiply $a+b+c$ to the both side and we will get $100a+10b+c=11a+11b+11c$ $\rightarrow$ $89a-b=10c$ .

Since $a$ , $b$ ,and $c$ are the digit number, they must be the integer and cannot exceed $9$ .

From that restriction, $a=1$ , $b=9$ and $c=8$ .

So the $198$ is the largest and the only number that we can get $\frac{N}{S(N)}=R.$

Let $N = \overline{abc} = 100a + 10b + c$ where $1 \leq a \leq 9$ , $0 \leq b \leq 9$ and $0 \leq c \leq 9$ .

Lemma. $N>10(a+b+c)$

Proof. If $100a+10b+c\leq 10(a+b+c),$ then $90a\leq 9c,$ so $10a\leq c.$ But $a\geq 1$ and $c\leq 9,$ a contradiction.

So the smallest possible integer value of the ratio $\frac{N}{a+b+c},$ is $11.$ In that case, $100a+10b+c=11a+11b+11c,$ so $89a=b+10c.$ Because $b\leq 9$ and $c\leq 9,$ $b+10c\leq 99,$ thus $a= 1.$ Then, clearly, $b=9$ and $c=8,$ so $N=198.$ Therefore, the answer is $198.$

If you count the numbers, you will find it.