Shortcut Trick
The required number leaves a remainder of 4 when divided by 9, so its sum of digits must also leave a remainder of 4 when divided by 9.
Checking the sum of digits for the given options:
364 → 3 + 6 + 4 = 13 → 13 ÷ 9 leaves remainder 4.
357 → 3 + 5 + 7 = 15 → 15 ÷ 9 leaves remainder 6.
Only 364 satisfies this condition and is also perfectly divisible by 7 (364 ÷ 7 = 52).
∴ The correct answer is 364.
Alternate Method
Given:
Divisors = 6, 9, 15, and 18
Remainder in each case = 4
The number is perfectly divisible by 7.
Formula Used:
Required Number = LCM(Divisors) × k + Remainder
Calculation:
⇒ LCM of 6, 9, 15, and 18 = 90
⇒ The required number must be of the form: N = 90k + 4
⇒ We are given that N is perfectly divisible by 7.
⇒ Let us break 90k into multiples of 7: N = 84k + 6k + 4
⇒ Since 84k is a multiple of 7, the remaining term (6k + 4) must also be completely divisible by 7.
⇒ Let us check for the minimum integer values of k:
⇒ For k = 1, 6(1) + 4 = 10 (Not divisible by 7)
⇒ For k = 2, 6(2) + 4 = 16 (Not divisible by 7)
⇒ For k = 3, 6(3) + 4 = 22 (Not divisible by 7)
⇒ For k = 4, 6(4) + 4 = 28 (Divisible by 7)
⇒ Substitute k = 4 into the general form equation:
⇒ N = 90 × 4 + 4 = 360 + 4 = 364
∴ The correct answer is 364.
Additional Information
LCM Remainder Rule
If a number leaves a common remainder R when divided by x, y, and z, the number is always of the form N = LCM(x, y, z) × k + R.
Divisibility by 9
A number is divisible by 9 if the sum of its digits is a multiple of 9. The remainder when a number is divided by 9 is exactly the same as the remainder when its digital sum is divided by 9.
Divisibility by 7
To check divisibility by 7, double the last digit and subtract it from the rest of the number. If the final result is a multiple of 7, the original number is also divisible by 7.