Find the last number which when divided by 6, 8, 15 and 30 leaves remainder 2, 4, 11 and 26 respectively?a)124b)116c)120d)134e)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

To find the last number that satisfies the given conditions, we need to find the least common multiple (LCM) of the four divisors: 6, 8, 15, and 30.

Finding the LCM involves finding the smallest number that is divisible by each of the given divisors. We can start by finding the prime factorizations of each number:

- 6: 2 x 3
- 8: 2 x 2 x 2
- 15: 3 x 5
- 30: 2 x 3 x 5

Now, we identify the highest power of each prime factor that appears in any of the prime factorizations:

- 2: appears in 2 x 2 x 2 (highest power = 2^3 = 8)
- 3: appears in 2 x 3 and 3 x 5 (highest power = 3^1 = 3)
- 5: appears in 3 x 5 (highest power = 5^1 = 5)

To find the LCM, we multiply the highest powers of each prime factor:

LCM = 2^3 x 3^1 x 5^1 = 8 x 3 x 5 = 120

So, the last number that satisfies the given conditions is 120.

Answer: Option B) 116

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