A man has a certain number of small boxes to pack into parcels. If he packs 3, 4, 5, or 6 boxes in each parcel, he is always left with one box over; but if he packs 7 boxes in each parcel, none of the boxes is left over. What is the possible total number of boxes he may have to pack?

Introduction / Context:
This arithmetic reasoning problem is based on remainders and divisibility, a typical application of the least common multiple and modular arithmetic. The man packs his boxes in parcels of different sizes. For 3, 4, 5, and 6 boxes per parcel he is left with one extra box, but for 7 boxes per parcel he has no remainder. The task is to determine a total number of boxes that fits all of these conditions using simple number theory ideas.

Given Data / Assumptions:

• When boxes are packed 3 per parcel, the remainder is 1.
• When packed 4, 5, or 6 per parcel, the remainder is also 1 in each case.
• When packed 7 per parcel, there is no remainder, so the total is divisible by 7.
• The answer must be one of the given options and must satisfy all remainder conditions simultaneously.

Concept / Approach:
The statement that the number leaves remainder 1 when divided by 3, 4, 5, and 6 means the total is 1 more than a common multiple of all these divisors. The least common multiple (LCM) of 3, 4, 5, and 6 determines the basic repeating pattern for such numbers. Any valid total must be of the form 60k + 1, where 60 is the LCM. In addition, the number must be divisible by 7. Thus, among numbers of the form 60k + 1 we search for a value that is a multiple of 7 and also appears in the options.

Step-by-Step Solution:
Step 1: Compute the LCM of 3, 4, 5, and 6. The LCM is 60. Step 2: Numbers that leave remainder 1 when divided by 3, 4, 5, and 6 are of the form 60k + 1 for integer k. Step 3: List a few such numbers: 61, 121, 181, 241, 301, and so on. Step 4: Impose the extra condition that the total must be divisible by 7. Check 301: 301 / 7 = 43, so 301 is divisible by 7. Step 5: Check if 301 appears in the options. It is given as option B, so 301 is a valid candidate. Step 6: Other options such as 106, 309, and 400 either do not equal 60k + 1 or are not divisible by 7, so they fail the combined conditions.

Verification / Alternative check:
Verify all conditions for 301: 301 ÷ 3 gives remainder 1, 301 ÷ 4 gives remainder 1, 301 ÷ 5 gives remainder 1, and 301 ÷ 6 also gives remainder 1. Finally, 301 ÷ 7 gives 43 exactly with no remainder. This confirms that 301 satisfies every requirement in the question and is therefore a correct and complete answer.

Why Other Options Are Wrong:
Option A 106 is not 1 more than a multiple of 60, and 106 ÷ 5 does not leave remainder 1. Option C 309 also fails some remainder conditions and does not equal 60k + 1. Option D 400 is neither of the form 60k + 1 nor divisible by 7. Each alternative violates at least one of the divisibility constraints.

Common Pitfalls:
A frequent mistake is to test only one or two divisors, such as checking condition with 3 and 4 but not verifying 5, 6, and 7. Another pitfall is ignoring the power of the LCM and trying trial and error randomly. Using the LCM focuses attention on a small family of possible values and makes it much easier to find the number that meets all conditions.

Final Answer:
The total number of boxes the man may have is 301.