Question

A natural number, when divided by 3, 4, 6 and 7, leaves a remainder of 2 in each case. What is the smallest of all such numbers?

• A. 213
• B. 843
• C. 86
• D. 63

Answer 1

Answer:

C. 86

⇒ LCM of 3, 4, 6 and 7 = $3\times{2^2}\times7=84$

∴ The required number = 84 + 2 = 86