Final answer:
The smallest positive integer n for 4n/105 to be a terminating decimal is 21, as 21 is the smallest integer that ensures the prime factors of the denominator are only 2 or 5.
Step-by-step explanation:
The question is asking what the smallest positive integer n is such that 4n/105 is a terminating decimal. A terminating decimal occurs when the denominator after simplification has prime factors only of 2 and/or 5, since the decimal system is based on powers of 10. The prime factorization of 105 is 3×5×7. To make the fraction terminate, we need to cancel out all prime factors that are not 2 or 5. This can be achieved by multiplying by a factor that includes 3 and 7 in its prime factorization.
To cancel out the 3 and the 7, we need to multiply 105 by 3 and by 7, or that is, by 21. Therefore, n must be a multiple of 21. The smallest positive integer multiple of 21 is 21 itself. Therefore, the smallest positive integer n such that 4n/105 is a terminating decimal is 21.