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29 Jul 2024, 01:28
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A library assigns unique identification codes to each of its books. Each code consists of a 3-letter key followed by 6 digits, where the same digit can appear more than once. If the library needs to assign codes to 23,154,000 books, at least how many 3-letter keys are needed?
A. 4
B. 23
C. 24
D. 26
E. 22,154,000
A. 4
B. 23
C. 24
D. 26
E. 22,154,000
29 Jul 2024, 01:28
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Official Solution:
A library assigns unique identification codes to each of its books. Each code consists of a 3-letter key followed by 6 digits, where the same digit can appear more than once. If the library needs to assign codes to 23,154,000 books, at least how many 3-letter keys are needed?
A. 4
B. 23
C. 24
D. 26
E. 22,154,000
Since each position in the 6-digit number can be any digit from 0 to 9, there are \(10^6\) (or 1,000,000) possible combinations for the 6 digits. Hence, each 3-letter key will accommodate \(10^6\) identification codes. For instance, two 3-letter keys will be enough for \(2*10^6\) books, and nine 3-letter keys will be enough for \(9*10^6\) books. Thus, to find out how many 3-letter keys are required, we divide the total number of books by the number of combinations possible with the 6 digits:
\(\frac{23,154,000}{10^6} = 23.154\)
Therefore, at least 24 3-letter keys are needed.
Answer: C
A library assigns unique identification codes to each of its books. Each code consists of a 3-letter key followed by 6 digits, where the same digit can appear more than once. If the library needs to assign codes to 23,154,000 books, at least how many 3-letter keys are needed?
A. 4
B. 23
C. 24
D. 26
E. 22,154,000
Since each position in the 6-digit number can be any digit from 0 to 9, there are \(10^6\) (or 1,000,000) possible combinations for the 6 digits. Hence, each 3-letter key will accommodate \(10^6\) identification codes. For instance, two 3-letter keys will be enough for \(2*10^6\) books, and nine 3-letter keys will be enough for \(9*10^6\) books. Thus, to find out how many 3-letter keys are required, we divide the total number of books by the number of combinations possible with the 6 digits:
\(\frac{23,154,000}{10^6} = 23.154\)
Therefore, at least 24 3-letter keys are needed.
Answer: C
18 Mar 2025, 22:06
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I did not quite understand the solution. I think the question could have clarified whether the letters can be arranged and used again or not. foe ex ABC can be rearranged to BAC,CAB,CBA,BCA and so on
