Let $$x$$​ be the least number of 5 digits, which when divided by 28, 40, 42 and 48, leaves remainder 6 in each case and $$x$$​ is divisible by 246. What is the sum of the digits of $$x$$​ ?

Correct option is A

Solution:

Step 1: Let the required number be

Let the number be:

x = LCM(28, 40, 42, 48) × k + 6

This is because the number leaves a remainder of 6 when divided by each of those numbers. So if we subtract 6, the result must be divisible by the LCM of the divisors.

So:

x - 6 is divisible by LCM(28, 40, 42, 48)

Step 2: Find LCM of 28, 40, 42, and 48

Prime factorization:

• 28 = 2² × 7
• 40 = 2³ × 5
• 42 = 2 × 3 × 7
• 48 = 2⁴ × 3

The LCM takes the highest powers of all primes:

LCM = 2⁴ × 3 × 5 × 7 = 16 × 105 = 1680

So:

x - 6 = 1680 × k
=> x = 1680k + 6

Step 3: Find the least 5-digit number

We want x to be at least 10000:

1680k + 6 ≥ 10000
=> 1680k ≥ 9994
=> k ≥ 9994 ÷ 1680 ≈ 5.95
=> k = 6 (next whole number)

So,

x = 1680 × 6 + 6 = 10080 + 6 = 10086

Step 4: Check if x is divisible by 246

10086 ÷ 246 = 41 → It is divisible.

Step 5: Find the sum of digits of x = 10086

1 + 0 + 0 + 8 + 6 = 15

Final Answer: (a) 15