Question

If \(3a25b\) is divisible by 12, find the maximum value of \(|a-b|\).

Hint:

Always break divisibility problems into individual rules and test digit combinations systematically.

Answer 1

Correct Answer: 8

Step 1: Use the divisibility rule of 12.
A number is divisible by 12 if it is divisible by both 3 and 4.
Step 2: Apply divisibility rule of 4.
For divisibility by 4, the last two digits must be divisible by 4.
The last two digits are \(5b\).
Possible values of \(5b\) divisible by 4 are:
\[ 52,\; 56 \]
So, \(b = 2\) or \(b = 6\).
Step 3: Apply divisibility rule of 3.
Sum of digits of \(3a25b\) is:
\[ 3 + a + 2 + 5 + b = 10 + a + b \]
For divisibility by 3:
\[ 10 + a + b \equiv 0 \pmod{3} \]
Step 4: Check valid combinations to maximize \(|a-b|\).
Case 1: \(b = 2\)
\[ 10 + a + 2 = 12 + a \]
For divisibility by 3, \(a = 0, 3, 6, 9\).
Maximum \(|a-b| = |9 - 2| = 7\)
Case 2: \(b = 6\)
\[ 10 + a + 6 = 16 + a \]
For divisibility by 3, \(a = 2, 5, 8\).
Maximum \(|a-b| = |8 - 6| = 2\)
Step 5: Final comparison.
Maximum possible value of \(|a-b|\) is obtained in Case 1.
Final Answer:
\[ \boxed{7} \]