Let's break down the problem step by step.

Let:

• $N = 100a + 10b + c$, where $a$, $b$, and $c$ are the hundreds, tens, and units digits of $N$, respectively.

Given conditions:

1. $N$ is a three-digit number.

2. $2N$ is equal to the sum of four quantities:

   • The sum of the digits of $N$, which is $a + b + c$.
   • The two-digit number formed by the first two digits of $N$, which is $10a + b$.
   • The two-digit number formed by the last two digits of $N$, which is $10b + c$.
   • The constant 961.

3. The hundreds digit $a$ is 3 more than the tens digit $b$, so we have: $a = b + 3$

Using these conditions, let's form the main equation from condition (2): $2N = (a + b + c) + (10a + b) + (10b + c) + 961$

Expanding $N = 100a + 10b + c$, we substitute $2N$ in terms of $a$, $b$, and $c$: $2(100a + 10b + c) = a + b + c + 10a + b + 10b + c + 961$

Simplifying both sides: $200a + 20b + 2c = 11a + 12b + 2c + 961$

Subtract $2c$ from both sides: $200a + 20b = 11a + 12b + 961$

Now, let's further simplify: $189a + 8b = 961$

Given that $a = b + 3$, we can substitute $a$ as $b + 3$: $189(b + 3) + 8b = 961$

Expanding and solving for $b$: $189b + 567 + 8b = 961$ $197b = 394$ $b = 2$

Thus, the tens digit of $N$ is 2.

Would you like a more detailed explanation or have any questions?

Here are some related questions you might find helpful:

1. How would this approach change if the hundreds digit of $N$ were not specifically defined in relation to the tens digit?
2. How can we verify the answer by plugging the values back into the original equation?
3. What if $N$ had a condition involving its digits adding up to a specific number?
4. How would the solution process change if we were looking for a two-digit number instead?
5. Can you find examples of other numbers that satisfy similar conditions for different values?

Tip: Always double-check initial conditions in word problems—they often simplify your equations greatly when substituted early.