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19 Dec 2023, 07:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
55% (02:17) correct
45%
(02:31)
wrong
based on 246
sessions
History
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Not Attempted Yet
12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun
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
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
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20 Dec 2023, 06:24
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GMAT Club Official Explanation:
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.
General Discussion
19 Dec 2023, 12:55
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Bunuel
total 4 digit numbers = 9999 - 1000 + 1 = 9000
3^0 = 1
3^1 = 3
3^2 = 9
9000 - (numbers that do not contain 1, or 3 or 9)
Numbers that do not contain 1, or 3 or 9 =
_ _ _ _
First place can be filled in 6 ways, the rest places can be filled in 7 ways. Total = 7 * 7 * 7 * 6 = 2058
Total = 9000 - 2058 = 6942
IMO D
