Find the smallest number of five digits which when divided by 15, 30, 38, 40 respectively always leaves 2 as remainder.

Question

Question

This question was previously asked in

Bihar शिक्षक TRE 3.0 (1-5) (Language and General Studies) Official Paper (Held On: 20 Jul, 2024)

Answer 1

Concept Used:

We need to find a number that satisfies the condition N = k (LCM(15, 30, 38, 40)) + 2, where k is an integer.

Formula Used:

LCM(a, b, c, d) = LCM(LCM(LCM(a, b), c), d)

Calculation:

Step 1: Calculate LCM of 15, 30, 38, 40

⇒ LCM(15, 30)

⇒ 15 = 3 × 5

⇒ 30 = 2 × 3 × 5

⇒ LCM(15, 30) = 2 × 3 × 5 = 30

⇒ LCM(30, 38)

⇒ 30 = 2 × 3 × 5

⇒ 38 = 2 × 19

⇒ LCM(30, 38) = 2 × 3 × 5 × 19 = 570

⇒ LCM(570, 40)

⇒ 570 = 2 × 3 × 5 × 19

⇒ 40 = 2³ × 5

⇒ LCM(570, 40) = 2³ × 3 × 5 × 19 = 2280

Step 2: Find the smallest number of five digits

⇒ N = k × 2280 + 2

⇒ The smallest five-digit number is 10000

⇒ k = (10000 - 2) / 2280

⇒ k = 9998 / 2280 ≈ 4.385

⇒ k = 5 (smallest integer greater than 4.385)

⇒ N = 5 × 2280 + 2

⇒ N = 11400 + 2

⇒ N = 11402

∴ The correct answer is option 2.

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