- Athe first statement forces a remainder of 1 on division by 3, while the second forces a remainder of 0
- Ba remainder must always be less than half of its divisor
- C6 and 9 are not co-prime, so two such conditions can never hold together
- Dthe two remainders would have to add up to a multiple of 3
Step-by-Step Solution
Both 6 and 9 are multiples of 3, so each statement also fixes the remainder on division by 3. From N = 6k + 4 = 3(2k + 1) + 1 the remainder is 1; from N = 9m + 3 = 3(3m + 1) the remainder is 0. A number cannot leave two different remainders on division by 3, so no such number exists. Note that non-co-prime divisors are not the problem by themselves — 10 leaves 4 on division by 6 and 1 on division by 9 quite happily.
Answer: A) the first statement forces a remainder of 1 on division by 3, while the second forces a remainder of 0