Let N be a two digit natural number divisible by 3. If you reverse the tens and ones digit of N, you will get a new number that is divisible by 2. What will be the smallest possible value of N?

We are tasked with finding the smallest two-digit natural number \( N \) such that:

1. \( N \) is divisible by 3.
2. Reversing the digits of \( N \) produces a new number divisible by 2.

### Step 1: Represent the two-digit number \( N \)
Let \( N \) be represented as \( 10a + b \), where \( a \) is the tens digit and \( b \) is the ones digit. The reversed number would be \( 10b + a \).

### Step 2: Condition 1: \( N \) is divisible by 3
For \( N \) to be divisible by 3, the sum of its digits, \( a + b \), must be divisible by 3. Therefore, we require:
\[
a + b \equiv 0 \pmod{3}
\]

### Step 3: Condition 2: The reversed number is divisible by 2
For the reversed number \( 10b + a \) to be divisible by 2, the ones digit, which is \( a \), must be even. Therefore, \( a \) must be an even number. This gives us possible values of \( a = 0, 2, 4, 6, 8 \).

### Step 4: Find the smallest \( N \) satisfying both conditions
We now check for the smallest possible value of \( N = 10a + b \) such that:
- \( a + b \equiv 0 \pmod{3} \)
- \( a \) is even.

#### Case 1: \( a = 0 \)
If \( a = 0 \), then \( b \) must satisfy \( 0 + b \equiv 0 \pmod{3} \), which implies \( b \equiv 0 \pmod{3} \). The smallest \( N \) with \( a = 0 \) and \( b \equiv 0 \pmod{3} \) is \( N = 30 \). However, this is not a two-digit number, so we discard it.

#### Case 2: \( a = 2 \)
If \( a = 2 \), then \( 2 + b \equiv 0 \pmod{3} \), which implies \( b \equiv 1 \pmod{3} \). Therefore, \( b = 1, 4, 7 \).

- For \( b = 1 \), \( N = 21 \), and the reversed number is \( 12 \), which is divisible by 2. So, \( N = 21 \) satisfies all conditions.

Since 21 is a two-digit number, divisible by 3, and its reversed number 12 is divisible by 2, the smallest possible value of \( N \) is:
\[
\boxed{21}
\]

Would you like to go over any part of this solution or have further questions?

### 5 Related Questions:
1. Can a number divisible by both 2 and 3 be divisible by 6?
2. What is the general rule for divisibility by 2 and 3?
3. How do the digits of a number relate to its divisibility by 9?
4. How can you quickly find the reverse of any two-digit number?
5. What would change in the problem if \( N \) had three digits?

### Tip:
When checking divisibility rules, breaking down numbers into their digits often reveals useful patterns.

Let N be a two-digit natural number divisible by 3. If you reverse the tens and ones digit of N, you will get a new number that is divisible by 2. What will be the smallest possible value of N?

Find the smallest two-digit number divisible by 3 such that when its digits are reversed, the new number is divisible by 2. This problem teaches the rules of divisibility for 2 and 3, and covers concepts like reversing digits and checking divisibility patterns. Ideal for Grades 6-8.

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