Question

Ashish has a certain number of boxes to pack into a parcel. If he packs 3, 4, 5 or 6 in a parcel, he is left with 1 and if he packs 7 in a parcel, none is left over. What is the number of boxes he has?

• A. \(108\)
• B. \(301\)
• C. \(309\)
• D. \(400\)

Hint:

To solve multiple modular conditions, combine congruences using LCMs and solve using the Chinese Remainder Theorem or substitution method.

Answer 1

The Correct Option is B

Step 1: Define the congruences. We are told: \[ x \equiv 1 \pmod{3}, \quad x \equiv 1 \pmod{4}, \quad x \equiv 1 \pmod{5}, \quad x \equiv 1 \pmod{6}, \quad x \equiv 0 \pmod{7} \] Step 2: Combine the first four congruences. Since \(x\) leaves a remainder of 1 when divided by 3, 4, 5, or 6, we find the LCM of these numbers: \[ \text{LCM}(3, 4, 5, 6) = 60 \] So, \(x \equiv 1 \pmod{60} \Rightarrow x = 60k + 1\) Step 3: Apply the condition \(x \equiv 0 \pmod{7}\) Substitute: \[ 60k + 1 \equiv 0 \pmod{7} \Rightarrow 60k \equiv -1 \pmod{7} \Rightarrow 4k \equiv 6 \pmod{7} \] Multiply both sides by the inverse of 4 modulo 7, which is 2: \[ k \equiv 12 \equiv 5 \pmod{7} \Rightarrow k = 7m + 5 \Rightarrow x = 60(7m + 5) + 1 = 420m + 301 \] Step 4: Smallest value for \(x\) For \(m = 0\), \(x = 301\), which satisfies all the conditions.