Shortcut Trick
For a number to be divisible by 6, it must be even and the sum of its digits must be divisible by 3.
Sum of digits of 3426x = 3 + 4 + 2 + 6 + x = 15 + x. For this to be a multiple of 3, x can be 0, 3, 6, or 9.
Since the number must be even, x must be 0 or 6. Since it is NOT divisible by 5, x cannot be 0.
∴ The correct answer is 6.
Alternate Method
Given: The number is 3426x. It is divisible by 6 but NOT divisible by 5.
Formula Used: Divisibility rules for 2, 3, 5, and 6.
⇒ For divisibility by 2, the last digit x must be even: {0, 2, 4, 6, 8}.
⇒ For divisibility by 3, sum of digits = 3 + 4 + 2 + 6 + x = 15 + x. This sum must be divisible by 3.
⇒ Possible values for x from the sum condition: {0, 3, 6, 9}.
⇒ Combining both (divisibility by 6), x must be in the intersection: {0, 6}.
⇒ For divisibility by 5, the last digit must be 0 or 5. Since 3426x is NOT divisible by 5, x ≠ 0.
⇒ From the remaining possible values {0, 6}, we exclude 0, leaving x = 6.
∴ The correct answer is 6.
Additional Information
Divisibility by 3
A number is divisible by 3 if the sum of all its digits is a multiple of 3.
Divisibility by 6
A number is divisible by 6 if it satisfies the divisibility rules for both 2 (ends in 0, 2, 4, 6, 8) and 3 simultaneously.
Divisibility by 5
A number is divisible by 5 if its unit digit is either 0 or 5.
Divisibility by 4
A number is divisible by 4 if the last two digits of the number are divisible by 4 or are 00.