Math Problem Statement

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?

Solution

We are tasked with finding the smallest two-digit natural number NN such that:

  1. NN is divisible by 3.
  2. Reversing the digits of NN produces a new number divisible by 2.

Step 1: Represent the two-digit number NN

Let NN be represented as 10a+b10a + b, where aa is the tens digit and bb is the ones digit. The reversed number would be 10b+a10b + a.

Step 2: Condition 1: NN is divisible by 3

For NN to be divisible by 3, the sum of its digits, a+ba + b, must be divisible by 3. Therefore, we require: a+b0(mod3)a + b \equiv 0 \pmod{3}

Step 3: Condition 2: The reversed number is divisible by 2

For the reversed number 10b+a10b + a to be divisible by 2, the ones digit, which is aa, must be even. Therefore, aa must be an even number. This gives us possible values of a=0,2,4,6,8a = 0, 2, 4, 6, 8.

Step 4: Find the smallest NN satisfying both conditions

We now check for the smallest possible value of N=10a+bN = 10a + b such that:

  • a+b0(mod3)a + b \equiv 0 \pmod{3}
  • aa is even.

Case 1: a=0a = 0

If a=0a = 0, then bb must satisfy 0+b0(mod3)0 + b \equiv 0 \pmod{3}, which implies b0(mod3)b \equiv 0 \pmod{3}. The smallest NN with a=0a = 0 and b0(mod3)b \equiv 0 \pmod{3} is N=30N = 30. However, this is not a two-digit number, so we discard it.

Case 2: a=2a = 2

If a=2a = 2, then 2+b0(mod3)2 + b \equiv 0 \pmod{3}, which implies b1(mod3)b \equiv 1 \pmod{3}. Therefore, b=1,4,7b = 1, 4, 7.

  • For b=1b = 1, N=21N = 21, and the reversed number is 1212, which is divisible by 2. So, N=21N = 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 NN is: 21\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 NN had three digits?

Tip:

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

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Math Problem Analysis

Mathematical Concepts

Divisibility Rules
Number Reversal
Basic Algebra

Formulas

Divisibility by 3: sum of digits divisible by 3
Divisibility by 2: last digit must be even

Theorems

Divisibility Rules

Suitable Grade Level

Grades 6-8