BTGmoderatorDC wrote:If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?
I. 3
II. 5
III. 9
A) None
B) I only
C) II only
D) III only
E) II and III
OA A
Source: Official Guide
If t = 1, then (1/1000)^4 = (1/10^3)^4 = 1/10^12. Therefore, the decimal expansion would have 12 decimal places, with the last (rightmost) digit being a 1. That is, there are 11 zeros between the decimal point and the last digit 1. For any of the given t values, 3, 5 and 9, it will not change the number of decimal places; however, it may change the number of zeros between the decimal point and the first nonzero digit.
If t = 3, then t^4 = 3^4 = 81. So 81 will occupy the last two of the 12 decimal places; that means there are 10 zeros between the decimal point and the first nonzero digit 8.
If t = 5, then t^5 = 5^4 = 625. So 625 will occupy the last three of the 12 decimal places; that means there are 9 zeros between the decimal point and the first nonzero digit 6.
If t = 9, then t^5 = 9^4 = 6561. So 6561 will occupy the last four of the 12 decimal places; that means there are 8 zeros between the decimal point and the first nonzero digit 6.
Since we are looking for fewer than 8 zeros between the decimal point and the first nonzero digit, none of the given t values will make this happen.
Answer: A
