Find the remainder when the smallest 6-digit number divisible by 12, 15 and 25 is divided by 9.

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SSC GD Constable (2022) Official Paper (Held On : 13 Jan 2023 Shift 3)
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  1. 3
  2. 1
  3. 2
  4. 0

Answer (Detailed Solution Below)

Option 1 : 3
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Given values: A number divisible by 12, 15, and 25

Concept:

The number is the least common multiple (LCM) of 12, 15, and 25. The remainder when a number is divided by 9 is the same as the remainder when the sum of its digits is divided by 9.

Calculation:

⇒ The LCM of 12, 15, and 25 is 300. The smallest 6-digit number divisible by 300 is 100200.

⇒ Sum of digits of 100200 = 1 + 0 + 0 + 2 + 0 + 0 = 3

⇒ Remainder when 3 is divided by 9 = 3

Therefore, the remainder when the smallest 6-digit number divisible by 12, 15, and 25 is divided by 9 is 3.
Alternate Method 
LCM of 12, 15 and 25 is = 300.

and

300 ×  334 = 100200

So, the smallest 6-digit number which is divisible by 12, 15 and 25 is = 100200

When 100200 is divided by 9 then the remainder is:

100200 = 9 × 11133 + 3

remainder is = 3.

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