If there are fewer than 8 zeroes between the decimal point…

Reading Time: < 1 minute

Last Updated on May 11, 2023

GMAT OFFICIAL GUIDE PS

Solution:

We are given that the decimal expansion of (t/1000)^4 has fewer than 8 zeroes between the decimal point and the first nonzero digit. We are also given that 3, 5, and 9 are possible values of t. Let’s test each of these numbers:
I. 3
If t = 3, then (t/1000)^4 = (3/1000)^4 = (.003)^4 has twelve decimal places to the right of the decimal point with the digits 81 as the 2 rightmost digits (notice that 3^4 = 81). So there must be 10 zeros between the decimal point and the first nonzero digit 8 in the decimal expansion. This is not a possible value of t.
II. 5
If t = 5, then (t/1000)^4 = (5/1000)^4 = (.005)^4 has twelve decimal places to the right of the decimal point with the digits 625 as the 3 rightmost digits (notice that 5^4 = 625). So there must be 9 zeros between the decimal point and the first nonzero digit 6 in the decimal expansion. This is not a possible value of t.
III. 9
If t = 9, then (9/1000)^4 = (9/1000)^4 = (.009)^4 has twelve decimal places to the right of the decimal point with the digits 6561 as the 4 rightmost digits (notice that 9^4 = 6561). So there must be 8 zeros between the decimal point and the first nonzero digit 6 in the decimal expansion. This is not a possible value of t.
Recall that we are looking for fewer than 8 zeros between the decimal point and the first nonzero digit in the decimal expansion. So none of the given numbers are possible values of t.

Answer: A

Share
Tweet
WhatsApp
Share