Which is the least number which when divided by 4, 5, 7, and 8 leaves remainder 1 in each case but when divided by 11 leaves no remainder?

Which is the least number which when divided by 4, 5, 7, and 8 leaves remainder 1 in each case but when divided by 11 leaves no remainder? Correct Answer 561

Calculation:

According to the question:

The number should b 11x; where x is any natural number

Least number which when divided by 4, 5, 7, and 8 leaves remainder 1 = LCM(4, 5, 7, 8) + 1

LCM(4, 5, 7, 8) + 1 = 280 + 1 = 281

But 281 is not divisible by 11

So, we have to try next smallest multiple

⇒ 560 + 1 = 561; which is a multiple of 11

∴ The required number = 561