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31 Dec 2023, 08:20
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How many positive four-digit integers contain at least one digit of the form \(3^n\), where \(n\) is a non-negative integer?
A. 2,058
B. 3,584
C. 5,416
D. 6,942
E. 8,000
A. 2,058
B. 3,584
C. 5,416
D. 6,942
E. 8,000
31 Dec 2023, 08:20
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Bookmarks
Official Solution:
How many positive four-digit integers contain at least one digit of the form \(3^n\), where \(n\) is a non-negative integer?
A. 2,058
B. 3,584
C. 5,416
D. 6,942
E. 8,000
There are three digits of the form \(3^n\), where \(n\) is a non-negative integer: \(3^0 = 1\), \(3^1 = 3\), and \(3^2 = 9\).
We need to determine the number of positive four-digit integers that contain at least one of the digits 1, 3, or 9.
Typically, in at least counting problems, it's more effective to subtract the number of opposite cases from the total cases.
The total count of positive four-digit integers is \(9*10*10*10 = 9*10^3\) (the first digit can't be 0, offering 9 options, with 10 options for the remaining digits).
The count of positive four-digit integers without any of the digits 1, 3, or 9 is \(6*7*7*7 = 6*7^3\) (the first digit can be any number except 0 and the three restricted digits, and the remaining digits can be any number except the three restricted digits).
Thus, the count of positive four-digit integers containing at least one digit of the form \(3^n\) is:
(Total count of positive four-digit integers) - (Count of four-digit integers without the digits 1, 3, or 9) =
= \(9*10^3 - 6*7^3\) =
= (a number with the units digit of 0) - (a number with the units digit of 8) =
= (a number with the units digit of 2).
Answer: D
How many positive four-digit integers contain at least one digit of the form \(3^n\), where \(n\) is a non-negative integer?
A. 2,058
B. 3,584
C. 5,416
D. 6,942
E. 8,000
There are three digits of the form \(3^n\), where \(n\) is a non-negative integer: \(3^0 = 1\), \(3^1 = 3\), and \(3^2 = 9\).
We need to determine the number of positive four-digit integers that contain at least one of the digits 1, 3, or 9.
Typically, in at least counting problems, it's more effective to subtract the number of opposite cases from the total cases.
The total count of positive four-digit integers is \(9*10*10*10 = 9*10^3\) (the first digit can't be 0, offering 9 options, with 10 options for the remaining digits).
The count of positive four-digit integers without any of the digits 1, 3, or 9 is \(6*7*7*7 = 6*7^3\) (the first digit can be any number except 0 and the three restricted digits, and the remaining digits can be any number except the three restricted digits).
Thus, the count of positive four-digit integers containing at least one digit of the form \(3^n\) is:
(Total count of positive four-digit integers) - (Count of four-digit integers without the digits 1, 3, or 9) =
= \(9*10^3 - 6*7^3\) =
= (a number with the units digit of 0) - (a number with the units digit of 8) =
= (a number with the units digit of 2).
Answer: D
01 Jun 2024, 07:43
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I'm a bit confused by the initial setup of the question. How do we know that there are only 3 digits of the form 3^n?
