Q. 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
Number
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The problem should read as follows:
Any value less than 0.00000001 will have MORE than 7 zeroes to the right of the decimal point.
Since (t/1000)� cannot have more than 7 zeroes to the right of the decimal point, (t/1000)� must be GREATER THAN OR EQUAL TO 0.00000001:
(t/1000)� ≥ 0.00000001
(t/10³)� ≥ 1/10�
t�/10¹² ≥ 1/10�
10�t� ≥ 10¹²
t� ≥ 10�.
None of the three options for t satisfies the constraint that t� ≥ 10�.
The correct answer is A.
The smallest value that has 7 zeroes to the right of the decimal point is 0.00000001.If there are fewer than 8 zeroes between the decimal point and the first
nonzero digit in the decimal expansion of (t/1000)�, 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
Any value less than 0.00000001 will have MORE than 7 zeroes to the right of the decimal point.
Since (t/1000)� cannot have more than 7 zeroes to the right of the decimal point, (t/1000)� must be GREATER THAN OR EQUAL TO 0.00000001:
(t/1000)� ≥ 0.00000001
(t/10³)� ≥ 1/10�
t�/10¹² ≥ 1/10�
10�t� ≥ 10¹²
t� ≥ 10�.
None of the three options for t satisfies the constraint that t� ≥ 10�.
The correct answer is A.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
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I unlock the best way for YOU to solve problems.
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Maybe an easier way:
If t is a single digit integer, then (t/1000) = .00t
If we raise that to the fourth power, we get (.00t)�. That will give us at least eight zeros (.00 four times is .00000000), so no single digit t will suffice.
If t is a single digit integer, then (t/1000) = .00t
If we raise that to the fourth power, we get (.00t)�. That will give us at least eight zeros (.00 four times is .00000000), so no single digit t will suffice.
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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 Roman numeral:Joy Shaha wrote:Q. 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
I. 3
If t = 3, then (t/1000)^4 = (3/1000)^4 = (.003)^4 has twelve decimal places with the digits 81 (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 with the digits 625 (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 with the digits 6561 (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
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