Shortcut Trick
For a number to be divisible by 12, it must be divisible by both 3 and 4.
For divisibility by 4, the last two digits 9c must be divisible by 4 ⇒ c = 2 or c = 6.
For divisibility by 3, the sum of all digits (28 + b + c) must be a multiple of 3.
If c = 6, the digit sum is 34 + b. For this to be a multiple of 3, b can be 2, 5, or 8.
To find the lowest positive value of (c − b), we take c = 6 and b = 5 ⇒ 6 − 5 = 1.
∴ The correct answer is 1.
Alternate Method
Given:
The 7-digit number 1738b9c is completely divisible by 12.
Concept Used:
A number is divisible by 12 if it is divisible by both 3 (sum of digits is a multiple of 3) and 4 (last two digits form a multiple of 4).
Calculation:
⇒ Divisibility by 4: The number formed by the last two digits is 9c.
⇒ The only numbers in the 90s divisible by 4 are 92 and 96. Thus, c = 2 or c = 6.
⇒ Divisibility by 3: The sum of digits = 1 + 7 + 3 + 8 + b + 9 + c = 28 + b + c must be divisible by 3.
⇒ Case 1: If c = 2, the sum = 30 + b. For the sum to be completely divisible by 3, b can be 0, 3, 6, or 9.
⇒ We need a positive value for (c − b), so c must be greater than b. Thus, b = 0 ⇒ (c − b) = 2 − 0 = 2.
⇒ Case 2: If c = 6, the sum = 34 + b. For the sum to be completely divisible by 3, b can be 2, 5, or 8.
⇒ For a positive value of (c − b), b can be 2 or 5.
⇒ If b = 2, (c − b) = 6 − 2 = 4.
⇒ If b = 5, (c − b) = 6 − 5 = 1.
⇒ Comparing all valid positive outcomes (2, 4, and 1), the lowest possible positive value is 1.
∴ The correct answer is 1.
Additional Information
Divisibility Rule of 3 and 9
A number is divisible by 3 or 9 if the sum of all its digits is completely divisible by 3 or 9 respectively.
Divisibility Rule of 4 and 8
A number is divisible by 4 if the number formed by its last two digits is divisible by 4, and divisible by 8 if the number formed by its last three digits is divisible by 8.
Divisibility Rule of 11
A number is divisible by 11 if the difference between the sum of digits at odd places and the sum of digits at even places is either 0 or a multiple of 11.