Digit adding problem - 1

First of all, the digit sum of the $3$ -digit number must be between $1$ (i.e. $100$ ) and $27$ (i.e. $999$ ). When we perform the digit sum of the numbers $1$ through $27$ , we find that the only number from $1$ to $27$ which has a digit sum which isn't a single digit is $19$ . This has a digit sum that is $10$ , which, in turn takes us to $1$ . Therefore, any $3$ -digit number whose digit sum does not equal $19$ will have $2$ iterations, and any $3$ -digit number whose digit sum does equal $19$ will have exactly $3$ iterations.

Now we need $3$ digits that add up to $19$ . Note that we can exclude $0$ , and hence the occurrence of the first digit being $0$ , as this would leave $2$ digits which sum to $19$ , and the highest possible sum is $18$ (9,9).

Assume that there is a $9$ . This would leave a sum of $10$ which can be made in the following ways:

• $\color{#3D99F6} 9,9,1$
• $\color{#D61F06} 9,8,2$
• $\color{#D61F06} 9,7,3$
• $\color{#D61F06} 9,6,4$
• $\color{#3D99F6} 9,5,5$

Now assume that the highest digit is an $8$ (no $9$ as this would double count some permutations):

• $\color{#3D99F6} 8,8,3$
• $\color{#D61F06} 8,7,4$
• $\color{#D61F06} 8,6,5$

Now assuming that the highest digit is a $7$ :

• $\color{#3D99F6} 7,7,5$
• $\color{#3D99F6} 7,6,6$

If we now consider the highest digit being a $6$ , the highest digit sum we could make is only $18$ (i.e. $6,6,6$ ), so these are all the permutations. All the permutations highlighted blue have a repeated digit, meaning there are $3$ ways to arrange the digits, and all the permutations highlighted red don't have a repeated digit, meaning there are $6$ ways to arrange the digits. This means that the answer is $(5 \cdot 3) + (5 \cdot 6) = \boxed{45}$ .

In order to get the single digit number in 3 steps,the number formed at the 2nd step should be such that the addition of the digit is more than 9.Now the number possibilities are [19,28,29,37,38,39,46,47...,99], but as we know that the largest sum of digits in any 3 digit number is 27=9+9+9 for 999, so all the possibilities except 19 are ruled out.Now to get the numbers simply start with unit digit as 1, which will give 199.Now unit digit as 2, we will get 289,298.If this is continued, we can see a pattern emerging,(unit digit,total numbers)=(1,1) (2,2) (3,3) (4,4)...(9,9).So total such numbers=9*10/2=45 (as per summation of natural numbers equation with difference=1).