Question

A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?

• A. 1
• B. 2
• C. 3
• D. None of these

Hint:

Use the Chinese Remainder Theorem or solve the system of congruences systematically to find the values that satisfy all conditions.

Answer 1

The Correct Option is B

The number must satisfy the system of congruences: \[ N \equiv 1 \, (\text{mod} \, 2), \quad N \equiv 2 \, (\text{mod} \, 3), \quad N \equiv 3 \, (\text{mod} \, 4), \quad N \equiv 4 \, (\text{mod} \, 5), \quad N \equiv 5 \, (\text{mod} \, 6). \] This means \( N + 1 \) must be divisible by 2, 3, 4, 5, and 6. The least common multiple of these numbers is 60, so \( N + 1 = 60k \), or \( N = 60k - 1 \). Now, check for values of \( k \) such that \( N \) lies between 0 and 100. - If \( k = 1 \), then \( N = 60 \times 1 - 1 = 59 \). - If \( k = 2 \), then \( N = 60 \times 2 - 1 = 119 \), which is greater than 100. Thus, the only solution is \( N = 59 \), so there is only 1 such integer.