GMAT Club Olympics 2024 (Day 1): A library assigns unique

Question

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

Bunuel · Accepted Answer

GMAT Club Official Explanation:

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:

23,154,000/10^6 = 23.154

Therefore, at least 24 3-letter keys are needed.

Answer: C.

captain0612 · Answer

Option C

Given: Code consists of two parts: 3 letter key and 6 numerical.

Numbers in the code can repeat.
Total number of books to be encoded with unique identification codes:  23,154,000 books

Minimum number of 3-letter keys needed is?

Unique identification code patterm: ( - - - ) ( - - - - - - )

Since repetition is allowed, total number of numerical patters is  10*10*10*10*10*10 = 10^6

(number of 3 letter keys)(10^6) = 23,154,000

Number of 3 letter keys = 23.154
 Least number of 3 letter keys are 24
Choice C

jksiingh · Answer

The library needs to assign codes to 23154000 books, and the code is a combination of 3 letters and 6 digits.
As the digits can be repeated, we will have a total combination of 10^6 digits

Dividing 23154000 with 10^6, we get 23.154, rounding off to 24
We need atleast 24 3-letter keys to assign codes to all the books

salilbagai · Answer

6 digits means 10^6 combinations can be done using digits with one letter key. Now total books are 2,31,54,000. I need atleast 24 different keys to mark all the books.

temulenbatbayr90 · Answer

23,154,000/1,000,000 = 23,154 and thus 24.

rsrobin864 · Answer

The number of ways of arranging 6 digits with repetition is (10)^6. 
So if we divide 23154000 with (10)^6 then we get 23.154 so this should be the number of 3 letter keys to be formed 
So min should be 24 So ans is C

CatNinja7 · Answer

My approach:

6 digits (repeatable) => 10^6 unique codes possible
To get 23.154*10^6 codes, atleast 24 keys needed

IMO C

Leo_1511 · Answer

Question- 
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?

Solution-

Let K1--K3 represent the letters and D1--D6 represent the digits. So the code will look like below-

K1   K2   K3   D1   D2   D3   D4   D5   D6

K1,K2,K3 can have 26 letters each (A to Z and both inclusive). 
D1,D2,D3,D4,D5,D6 can have 10 digits each (0 to 9 and both inclusive).
Thus the total number of unique codes generated will be- 26*26*26*1,000,000 which is less than 23,154,000.
So, let p be the number of 3 letter keys that are required-
p * 1,00,000 >= 23,154,000 
p>= 23.154

Thus p=24 (At least), Answer is C.

tulee · Answer

Since every digit in the 6 digits can appear more than once, so the digit can go from 000000 to 999999. So for 23,000,000 items, we would need that range of digits 23 times. And since the total items are 23,154,000. We would need to use that range of digit 1 more time (so 23+1=24 ranges of digit in total). To accompany a total of 24 ranges, we need 24 3-letter keys.

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GMAT Club Olympics 2024 (Day 1): A library assigns unique

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