If there are fewer than \(8\) zeroes between the decimal point and the first nonzero digit in the decimal expansion of

Gmat_mission wrote: ↑

Fri Jan 08, 2021 4:28 am

If there are fewer than \(8\) zeroes between the decimal point and the first nonzero digit in the decimal expansion of \(\left(\dfrac{t}{1000}\right)^4,\) 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

Answer: A

Source: Official Guide

First: (t/1000)^4 = (t^4)/(1000^4)
Now recognize that 1000^4 = (10^3)^4 = 10^12
So, (t/1000)^4 = (t^4)/(1000^4) = (t^4)/(10^12)

IMPORTANT: When we divide a number by 10^12, we must move the decimal point 12 spaces to the left
So, for example, 1234567/10^12 = 0.000001234567
Likewise, 8888/10^12 = 0.000000008888
And, 66666666666666/10^12 = 66.666666666666

Now let's check each option

I. 3
It t = 3, then (t^4)/(10^12) = (3^4)/(10^12)
= 81/(10^12)
= 0.000000000081
There are 10 zeroes between the decimal point and the first nonzero digit
Since the question tells us that there are fewer than 8 zeroes between the decimal point and the first nonzero digit, we can ELIMINATE statement I


II. 5
It t = 5, then (t^4)/(10^12) = (5^4)/(10^12)
= 625/(10^12)
= 0.000000000625
There are 9 zeroes between the decimal point and the first nonzero digit
Since the question tells us that there are fewer than 8 zeroes between the decimal point and the first nonzero digit, we can ELIMINATE statement II


III. 9
It t = 9, then (t^4)/(10^12) = (9^4)/(10^12)
= 6561/(10^12)
= 0.000000006561
There are 8 zeroes between the decimal point and the first nonzero digit
Since the question tells us that there are fewer than 8 zeroes between the decimal point and the first nonzero digit, we can ELIMINATE statement III

Answer: A

Cheers,
Brent

Gmat_mission wrote: ↑

Fri Jan 08, 2021 4:28 am

If there are fewer than \(8\) zeroes between the decimal point and the first nonzero digit in the decimal expansion of \(\left(\dfrac{t}{1000}\right)^4,\) 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

Answer: A

Source: Official Guide

Solution:

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