The number of natural numbers lying strictly between 212 and 999 with digit sum equal to 15 is
Step-by-Step Solution
Key Concept: For a 3-digit number $\overline{abc}$ with $a+b+c=15$ and $a\geq1$, count solutions by enumerating over $a=2,\ldots,9$ with the range restriction (numbers must be $>212$ and $<999$).
Step 1: Count all 3-digit natural numbers $\overline{abc}$ such that $a+b+c=15$, with $1 \leq a \leq 9$ and $0 \leq b, c \leq 9$.
For a given digit $a$, we need to find the number of pairs $(b,c)$ such that $b+c = 15-a$, with $0 \leq b, c \leq 9$.
The number of solutions for $x+y=k$ with $0 \leq x,y \leq 9$ is given by:
* $k+1$ if $0 \leq k \leq 9$
* $19-k$ if $10 \leq k \leq 18$
Applying this rule for $k=15-a$:
* For $a=1$, $b+c=14$: $19-14=5$ numbers.
* For $a=2$, $b+c=13$: $19-13=6$ numbers.
* For $a=3$, $b+c=12$: $19-12=7$ numbers.
* For $a=4$, $b+c=11$: $19-11=8$ numbers.
* For $a=5$, $b+c=10$: $19-10=9$ numbers.
* For $a=6$, $b+c=9$: $9+1=10$ numbers.
* For $a=7$, $b+c=8$: $8+1=9$ numbers.
* For $a=8$, $b+c=7$: $7+1=8$ numbers.
* For $a=9$, $b+c=6$: $6+1=7$ numbers.
The total number of 3-digit numbers with a digit sum of 15 is:
$$5+6+7+8+9+10+9+8+7 = 69$$
Step 2: Exclude numbers not strictly between 212 and 999.
This requires excluding numbers $\leq 212$ and numbers $\geq 999$.
First, consider numbers $\geq 999$. The number 999 has a digit sum of $9+9+9=27 \neq 15$. Any number greater than 999 would be a 4-digit number or higher, which is outside the scope of 3-digit numbers. Therefore, no numbers with a digit sum of 15 are $\geq 999$.
Next, consider numbers $\leq 212$. We examine the numbers counted in Step 1:
* **Numbers with $a=1$**: These are numbers of the form $\overline{1bc}$ where $b+c=14$. The numbers are 159, 168, 177, 186, 195. All these 5 numbers are less than 212 and must be excluded.
* **Numbers with $a=2$**: These are numbers of the form $\overline{2bc}$ where $b+c=13$. The numbers are 249, 258, 267, 276, 285, 294. The smallest of these numbers is 249. Since $249 > 212$, none of these numbers need to be excluded.
Thus, a total of 5 numbers must be excluded.
Step 3: Calculate the final count.
The total number of 3-digit numbers with a digit sum of 15 is 69.
The number of these that are not strictly between 212 and 999 is 5.
The number of natural numbers strictly between 212 and 999 with a digit sum of 15 is:
$$69 - 5 = 64$$
Correct Answer: 64