Animated Solution for Mathematics - Permutations and Combinations: The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is ______.

Welcome, future engineer! Today, we are not just solving a math problem; we are embarking on a detective mission. We are hunting for three-digit odd numbers that satisfy a very specific, elegant condition: the sum of their digits must be a multiple of $7$.

This is a classic combinatorics problem that tests your ability to handle constraints with precision and patience. Let's break it down.

Setting the Stage

Imagine a three-digit number as a sequence of three slots: $x$, $y$, and $z$. Here, $x$ is the hundreds digit, $y$ is the tens digit, and $z$ is the units digit.

To form a valid three-digit number, we must respect the fundamental rules of our number system: The hundreds digit $x$ cannot be zero, so $x\in \{1,2,\dots ,9\}$. The tens digit $y$ is flexible, so $y\in \{0,1,\dots ,9\}$. * The problem demands that our number be odd, which locks the units digit $z$ into the set $\{1,3,5,7,9\}$.

These are our boundaries. Never rush past these; they are the walls of our arena.

The Search for Multiples

Now, consider the sum of the digits, $S=x+y+z$. We are told $S$ must be a multiple of $7$.

Let's find the range of $S$: The minimum sum occurs when $x=1,y=0,z=1$, giving $S_{min}=1+0+1=2$. The maximum sum occurs when $x=9,y=9,z=9$, giving $S_{max}=9+9+9=27$.

The multiples of $7$ within the range $[2,27]$ are $7,14,$ and $21$. These are our targets.

The Case-by-Case Odyssey

To solve this, we will iterate through each possible value of $z$. This is the most robust way to ensure we don't miss any combinations.

If $z=1$, then $x+y+1\in \{7,14,21\}$, which implies $x+y\in \{6,13,20\}$. Since the maximum value of $x+y$ is $18$, $20$ is impossible.

For $x+y=6$: Pairs are $(1,5),(2,4),(3,3),(4,2),(5,1),(6,0)$ ($6$ ways). For $x+y=13$: Pairs are $(4,9),(5,8),(6,7),(7,6),(8,5),(9,4)$ ($6$ ways).

Total for $z=1$ is $6+6=12$ ways.

If $z=3$, then $x+y+3\in \{7,14,21\}$, which implies $x+y\in \{4,11,18\}$.

For $x+y=4$: Pairs are $(1,3),(2,2),(3,1),(4,0)$ ($4$ ways). For $x+y=11$: Pairs are $(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2)$ ($8$ ways). * For $x+y=18$: Pair is $(9,9)$ ($1$ way).

Total for $z=3$ is $4+8+1=13$ ways.

If $z=5$, then $x+y+5\in \{7,14,21\}$, which implies $x+y\in \{2,9,16\}$.

For $x+y=2$: Pairs are $(1,1),(2,0)$ ($2$ ways). For $x+y=9$: Pairs are $(1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1),(9,0)$ ($9$ ways). * For $x+y=16$: Pairs are $(7,9),(8,8),(9,7)$ ($3$ ways).

Total for $z=5$ is $2+9+3=14$ ways.

If $z=7$, then $x+y+7\in \{7,14,21\}$, which implies $x+y\in \{0,7,14\}$. Since $x\ge 1$, $x+y=0$ is impossible.

For $x+y=7$: Pairs are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),(7,0)$ ($7$ ways). For $x+y=14$: Pairs are $(5,9),(6,8),(7,7),(8,6),(9,5)$ ($5$ ways).

Total for $z=7$ is $7+5=12$ ways.

If $z=9$, then $x+y+9\in \{7,14,21\}$, which implies $x+y\in \{5,12,19\}$. Since $19$ is impossible:

For $x+y=5$: Pairs are $(1,4),(2,3),(3,2),(4,1),(5,0)$ ($5$ ways). For $x+y=12$: Pairs are $(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3)$ ($7$ ways).

Total for $z=9$ is $5+7=12$ ways.

We have meticulously counted every possibility. Now, we simply sum them up:

There are exactly $63$ such numbers. This problem teaches us that even when faced with complex constraints, a systematic, disciplined approach will always lead you to the truth. Keep practicing, and that JEE Advanced seat will be yours!