Question

A number when divided by 4, 5 and 6 leaves remainders 2, 3 and 4 respectively. What is the smallest such number?

• A. 58
• B. 62
• C. 86
• D. 74

Hint:

Convert all congruences to the form \( x \equiv -r \mod m \), then compute LCM of moduli and subtract to get original number.

Answer 1

The Correct Option is A

Let the required number be \( N \) such that:

\[ N \equiv 2 \mod 4, \quad N \equiv 3 \mod 5, \quad N \equiv 4 \mod 6 \]

Rewriting all as:

\[ N \equiv -2 \mod 4 \Rightarrow N \equiv 2 \mod 4 \\ N \equiv -2 \mod 5 \Rightarrow N \equiv 3 \mod 5 \\ N \equiv -2 \mod 6 \Rightarrow N \equiv 4 \mod 6 \]

So in all cases:

\[ N + 2 \equiv 0 \mod 4,5,6 \\ \Rightarrow N + 2 = \text{LCM}(4,5,6) = 60 \\ \Rightarrow N = 60 - 2 = \boxed{58} \]