Question

The number of natural numbers, between 212 and 999 , such that the sum of their digits is 15 , is

Answer 1

64

Answer 2

Let the number be represented as $N = abc$ with $a,b,c$ digits. We need:

$$a+b+c = 15,$$

with $a$ from 2 to 9 (since $N\ge 212$), and $b,c$ from 0 to 9. For $a=2$, the equation becomes:

$$b+c = 13.$$

The allowed pairs $(b,c)$ (with $0\le b,c\le9$) are:

$$(4,9),\,(5,8),\,(6,7),\,(7,6),\,(8,5),\,(9,4).$$

So, there are 6 solutions when $a=2$. (All corresponding numbers are $\geq 212$; e.g. $249$ is the smallest.)

For each $a$ from 3 to 9, set $S = 15 - a$. The number of solutions for $b+c = S$ is given by:

$$f(S) = \begin{cases} S+1, & \text{if } S\le9, \\ 19-S, & \text{if } S>9. \end{cases}$$

Compute for each:

• $a=3:\ S=12 \implies f(12)=19-12=7.$
• $a=4:\ S=11 \implies f(11)=19-11=8.$
• $a=5:\ S=10 \implies f(10)=19-10=9.$
• $a=6:\ S=9 \implies f(9)=9+1=10.$
• $a=7:\ S=8 \implies f(8)=8+1=9.$
• $a=8:\ S=7 \implies f(7)=7+1=8.$
• $a=9:\ S=6 \implies f(6)=6+1=7.$

Adding these together:

$$\text{Total for } a=3\text{ to }9 = 7+8+9+10+9+8+7 = 58.$$

Now, total number of numbers is:

$$6\, (\text{for } a=2) + 58 = 64.$$