AcDec 24-25 Mathematics Section 1

Jamie’s digital lock keypad is missing the buttons to type 6, 5, and 3. How many different keycodes can Jamie still type if the lock takes a 6-digit code? (Assume the keypad originally had keys for 0-9.)

• 5,040

• 117,649

• 729

• 279,936

• 720

Answer explanation

Jamie can use the digits 0, 1, 2, 4, 7, 8, 9, totaling 7 options. For a 6-digit code, the total combinations are 7^6 = 117,649. Thus, the correct answer is 117,649.

In computer science, a “byte” is a sequence of 8 bits, where each bit is either a 1 or 0. How many unique values could be represented by a byte?

• 8

• 256

• 40,320

• 16

• 255

Answer explanation

A byte consists of 8 bits, and each bit can be either 0 or 1. Therefore, the total number of unique combinations is 2^8 = 256. However, when counting from 0 to 255, there are 256 values, but the highest value is 255.

An IPv4 address is an ordered list of four numbers used to identify devices on a computer network (for example, 127.168.0.255). Each number is at least 0 and at most 255. What is the maximum number of devices that can be uniquely identified by IPv4 addresses?

• 4,294,967,296

• 256!

• 172,061,505

• 4,228,250,625

• 174,792,640

Answer explanation

An IPv4 address consists of four numbers, each ranging from 0 to 255. This gives 256 options per number. Therefore, the total number of unique addresses is 256^4 = 4,294,967,296, making this the maximum number of devices identifiable.

An ice cream shop has 20 available flavors of ice cream. Customers can order ice cream either in a bowl or a cone. They can top it with any one of 5 different flavored syrups, or they can have it plain (i.e. no syrup). How many different possible orders are there?

• 120

• 600

• 38,760

• 30

• 240

Answer explanation

To find the total orders, multiply the number of flavors (20) by the number of serving options (2: bowl or cone) and the syrup options (6: 5 syrups + plain). Thus, 20 * 2 * 6 = 240 different possible orders.

A card dealer gives you a hand of 5 cards, 4 of which are kings. Ignoring the order they were dealt in, how many different hands could they have just dealt you? Assume a standard 52-card deck with no jokers or wildcards.

• 311,875,200

• 2,598,960

• 120

• 48

• 1

Answer explanation

To form a hand with 4 kings, you choose 1 from the 4 kings (4 ways), and then select 1 card from the remaining 48 cards (48 ways). Thus, the total combinations are 4 * 48 = 192, but since the kings are identical, we have 48 unique hands.

A large tech company has a database of clients, where every client is identified by a unique string of English letters A-Z. All strings in the database have the same fixed length, and no letters are repeated. How many clients can the tech company keep track of if each string is 5 letters long?

• 1,490,116,119,384,765,625 clients

• 130 clients

• 7,893,600 clients

• 11,881,376 clients

• 65,780 clients

Answer explanation

To find the number of unique 5-letter strings from 26 letters (A-Z) with no repetitions, calculate 26P5 = 26! / (26-5)! = 26 x 25 x 24 x 23 x 22 = 7,893,600. Thus, the company can track 7,893,600 clients.

Alex likes to hang up the shirts he plans to wear next week. He has 7 hangers, one for each day of the week, but he has 10 unique shirts in his dresser that he could choose for each day. Find the number of ways he could fill his closet with shirts if he plans to wear them from left to right.

• 282,475,249

• 70

• 10,000,000

• 608,400

• 5,040

Answer explanation

To fill 7 hangers with shirts from 10 unique options, we calculate permutations: 10P7 = 10! / (10-7)! = 10! / 3! = 10 x 9 x 8 x 7 x 6 x 5 x 4 = 608,400. Thus, the correct answer is 608,400.

A university club is holding elections for the next year’s executive board. There are 4 open positions (President, Vice President, Treasurer, and Secretary), and there are 6 different members willing to run. None of the positions can be vacant after the election. How many possible administrations are there?

• 360

• 15

• 1296

• 4096

• 24

Answer explanation

To find the number of possible administrations, we calculate the permutations of 6 members taken 4 at a time: 6P4 = 6!/(6-4)! = 6*5*4*3 = 360. Thus, the correct answer is 360.

Trish wrote a computer program that sorts a list of unique numbers, but her program has a bug! Right after sorting the list, it randomizes the order of the numbers again so that it never returns a correct answer. If she gives the program a list of 6 numbers, how many different lists could it possibly return?

• 720

• 60,466,176

• 46,656

• 6!

• 719

Answer explanation

Trish's program sorts the list of 6 unique numbers, which can be arranged in 6! (720) ways. However, one arrangement is the sorted list, so the program can return 719 different lists.

A railroad company uses 3 digits from 0 to 9 to number their locomotives. Next year they’re going to switch to using 3 digits followed by two capital English letters. All existing locomotives will be updated with the new numbering system. How many more locomotives could they potentially index with this system than with just the 3 digit number?

• 1,000

• 676,000

• 675,000

• 17,576

• 18,576

Answer explanation

The 3-digit system allows for 10^3 = 1,000 combinations. The new system allows for 10^3 * 26^2 = 676,000 combinations. The difference is 676,000 - 1,000 = 675,000. Thus, they can index 675,000 more locomotives.