Finding the Least Digit A for Divisibility by 6

The question asks for the least natural number 'A' such that the six-digit number 87937A is divisible by 6.

A number is divisible by 6 if and only if it is divisible by both 2 and 3.

Divisibility by 2 Rule

• For a number to be divisible by 2, its last digit must be even (0, 2, 4, 6, or 8).
• In the number 87937A, the last digit is A. Therefore, A must be one of {0, 2, 4, 6, 8}.

Divisibility by 3 Rule

• For a number to be divisible by 3, the sum of its digits must be divisible by 3.
• The sum of the digits of 87937A is: $8 + 7 + 9 + 3 + 7 + A = 34 + A$.
• So, $34 + A$ must be divisible by 3.

Determining the Value of A

We need to find the smallest natural number A from the set {0, 2, 4, 6, 8} such that $34 + A$ is divisible by 3.

• Test A = 0: Sum = $34 + 0 = 34$. 34 is not divisible by 3.
• Test A = 2: Sum = $34 + 2 = 36$. 36 is divisible by 3 ($36 \div 3 = 12$). A=2 satisfies both divisibility rules.
• Test A = 4: Sum = $34 + 4 = 38$. 38 is not divisible by 3.
• Test A = 6: Sum = $34 + 6 = 40$. 40 is not divisible by 3.
• Test A = 8: Sum = $34 + 8 = 42$. 42 is divisible by 3 ($42 \div 3 = 14$). A=8 satisfies both rules.

The possible values for A are 2 and 8. Since the question asks for the *least* natural number, the value of A is 2.