Question

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

Hint:

First use the last two digits $5b$ to restrict $b$ through divisibility by $4$. For each possible $b$, use the digit-sum rule for $3$ and compare all possible values of $|a-b|$.

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:
\[ \boxed{7} \]

Answer 2

Concept:

• Divisibility by $12$ requires divisibility by both $4$ and $3$.
• The last two digits determine divisibility by $4$, while the digit sum determines divisibility by $3$.

Step 1: Restrict the last digit.
The ending $5b$ must be divisible by $4$. Among $50$ to $59$, only $52$ and $56$ qualify, so $b=2$ or $b=6$.

Step 2: Test $b=2$.
The digit sum is $3+a+2+5+2=12+a$. Hence $a\in\{0,3,6,9\}$, and the largest difference is $|9-2|=7$.

Step 3: Test $b=6$.
The digit sum is $3+a+2+5+6=16+a$. Hence $a\in\{2,5,8\}$. The differences are $4,1,2$, so this case has maximum $4$.

Step 4: Compare both cases.
The overall maximum is $\max(7,4)=7$.

Final Answer: $7$