Answer :
The number of boxes the man has must leave a remainder of 1 when divided by 3, 4, 5, or 6, and no remainder when divided by 7. This means the number is 1 more than a multiple of the least common multiple (LCM) of 3, 4, 5, and 6, and also a multiple of 7.
The LCM of 3, 4, 5, and 6 is 60. So, the number of boxes, \( n \), can be expressed as:
\[ n = 60k + 1 \]
where \( k \) is an integer, and \( n \) must also be divisible by 7.
Looking at the options:
- (A) 106: \( 106 - 1 = 105 \), which is divisible by 7.
- (B) 302: \( 302 - 1 = 301 \), which is not divisible by 7.
- (C) 309: \( 309 - 1 = 308 \), which is divisible by 7.
- (D) 400: \( 400 - 1 = 399 \), which is not divisible by 7.
So, the possible answers are 106 and 309. However, since 106 does not satisfy the condition \( 60k + 1 \) (as \( 105/60 \) is not an integer), the correct answer is:
(C) 309.
The LCM of 3, 4, 5, and 6 is 60. So, the number of boxes, \( n \), can be expressed as:
\[ n = 60k + 1 \]
where \( k \) is an integer, and \( n \) must also be divisible by 7.
Looking at the options:
- (A) 106: \( 106 - 1 = 105 \), which is divisible by 7.
- (B) 302: \( 302 - 1 = 301 \), which is not divisible by 7.
- (C) 309: \( 309 - 1 = 308 \), which is divisible by 7.
- (D) 400: \( 400 - 1 = 399 \), which is not divisible by 7.
So, the possible answers are 106 and 309. However, since 106 does not satisfy the condition \( 60k + 1 \) (as \( 105/60 \) is not an integer), the correct answer is:
(C) 309.
