Shortcut Trick
To check divisibility by 99, the number must be divisible by its co-prime factors, 9 and 11.
Sum of digits of 699138 = 6 + 9 + 9 + 1 + 3 + 8 = 36 (Divisible by 9)
Difference of sum of alternating digits = (8 + 1 + 9) − (3 + 9 + 6) = 18 − 18 = 0 (Divisible by 11)
Since the number satisfies both rules, it is exactly divisible by 99.
∴ The correct answer is 99.
Alternate Method
Given:
The number to test is 699138.
The options are 117, 99, 102, and 108.
Method Used:
We test divisibility rules for the factors of the given options. We analyze Option 2 (99), which requires checking divisibility by 9 and 11.
Calculations:
⇒ Divisibility by 9: Sum of digits = 6 + 9 + 9 + 1 + 3 + 8 = 36.
⇒ Since 36 ÷ 9 = 4, the number is divisible by 9.
⇒ Divisibility by 11: Sum of digits at odd positions (from right to left) = 8 + 1 + 9 = 18.
⇒ Sum of digits at even positions (from right to left) = 3 + 9 + 6 = 18.
⇒ Difference = 18 − 18 = 0.
⇒ Since the difference is 0, the number is perfectly divisible by 11.
⇒ As 699138 is divisible by both 9 and 11, it is exactly divisible by their product, 99.
Checking other options for confirmation:
⇒ 108 is a multiple of 4, but the last two digits (38) are not divisible by 4. So, 108 is eliminated.
⇒ 102 is a multiple of 17. 699138 ÷ 102 leaves a non-zero remainder. So, 102 is eliminated.
∴ The correct answer is 99.
Additional Information
Divisibility by 9
A number is divisible by 9 if the sum of its digits is exactly a multiple of 9.
Divisibility by 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 0 or a multiple of 11.
Divisibility by Composite Numbers
To check divisibility for a composite number (like 99), break it down into co-prime factors (like 9 and 11). If the given number is divisible by both factors, it is divisible by the composite number.
Divisibility by 4
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.