FREE! Click here to Join FunTrivia. Thousands of games, quizzes, and lots more!
Quiz about Pizza Party On Algebra Avenue
Quiz about Pizza Party On Algebra Avenue

Pizza Party On Algebra Avenue Trivia Quiz


Would you like to come to our pizza party on Algebra Avenue? Great! Just help me with my terrible memory for numbers. All you will need will be simple algebra and some logic.

A multiple-choice quiz by VickiSilver. Estimated time: 23 mins.
  1. Home
  2. »
  3. Quizzes
  4. »
  5. Brain Teasers Trivia
  6. »
  7. Logic Puzzles
  8. »
  9. Difficult Logic Puzzles

Author
VickiSilver
Time
23 mins
Type
Multiple Choice
Quiz #
138,282
Updated
Dec 03 21
# Qns
5
Difficulty
Impossible
Avg Score
1 / 5
Plays
11255
Question 1 of 5
1. First of all we have to find my friends Alice and Bill, who live in two different houses on Algebra Avenue. Each of the houses on Algebra Avenue is numbered with a two-digit number from 10 to 99. I can't remember where Alice and Bill live, but I remember that their house numbers are the reverse of each other (in the sense that "21" is the reverse of "12".) I also remember that the sum of their two house numbers is a perfect square. The difference between their two house numbers is also a perfect square. Alice lives in the house with the smaller house number. What is the number of Bill's house?

Answer: (Just type in the two-digit number.)
Question 2 of 5
2. Now we have to find my friends Charlie and Debra, who live in two different houses on Algebra Avenue. I can't remember their house numbers either, but I remember that they are the reverse of each other. The sum of their house numbers is a perfect square, and the difference between their house numbers is a perfect cube. Charlie lives in the house with the smaller house number. What is the number of Debra's house?

Answer: (Just type in the two-digit number.)
Question 3 of 5
3. Now that we've found everybody, we need to go back to my house to get some money. I keep it in a rectangular box. Unfortunately, I own a lot of rectangular boxes, and I can't remember which one I put my money in. I remember that the three dimensions of the box (height, width, and length) are all whole numbers of inches, that they are all different, and that none of the dimensions is a prime number. I also remember that the volume of the box in cubic inches is the same as the total surface area of all six sides of the box in square inches. What is this volume or surface area?

Answer: (Just type in the number, no units.)
Question 4 of 5
4. Now we've found the right box, but I forgot it has a lock on it. It's one of those locks where you have to punch in a certain number to open it. I can't remember what the number is, but I remember that it's a four-digit number. I also remember that the sum of all four digits is equal to the product of all four digits, and that the four-digit number is evenly divisible by this sum or product. By the way, I also remember that the four-digit number is not "0000". What is it?

Answer: (Just type in the four-digit number.)
Question 5 of 5
5. Now I have my money. I'll let you choose whatever toppings you want on your pizza if you can figure out how much I have. I have a certain number of one-dollar bills (more than zero) and a certain number of ten-dollar bills (more than zero.) The number of one-dollar bills multiplied by the number of ten-dollar bills is equal to the total amount of money I have in dollars. The number of ten-dollar bills is not a prime. How much money do I have?

Answer: (Just type in the number, no dollar sign.)

(Optional) Create a Free FunTrivia ID to save the points you are about to earn:

arrow Select a User ID:
arrow Choose a Password:
arrow Your Email:




Quiz Answer Key and Fun Facts
1. First of all we have to find my friends Alice and Bill, who live in two different houses on Algebra Avenue. Each of the houses on Algebra Avenue is numbered with a two-digit number from 10 to 99. I can't remember where Alice and Bill live, but I remember that their house numbers are the reverse of each other (in the sense that "21" is the reverse of "12".) I also remember that the sum of their two house numbers is a perfect square. The difference between their two house numbers is also a perfect square. Alice lives in the house with the smaller house number. What is the number of Bill's house?

Answer: 65

Let's call Bill's house number "XY" and Alice's house number "YX". The value of Bill's house number will be 10X + Y and the value of Alice's house number will be 10Y + X. The sum of their house numbers will be 10X + Y + 10Y + X. This yields a total sum of 11X + 11Y. We can factor out 11 to express this same value as (11)(X + Y). We want this value to be a perfect square. Since 11 is a prime number, this can only be true if (X + Y) is 11 or 11 times a perfect square; 44, 99, and so on. Since X and Y are single-digit numbers, (X + Y) cannot be greater than 18.

Therefore (X + Y) must be 11. The difference between the two house numbers is (10X + Y) - (10Y + X). This is equal to 9X - 9Y. Factor out 9 to get (9)(X - Y). We also want this value to be a perfect square. Since 9 is a perfect square, this will be true whenever (X - Y) is a perfect square; 1, 4, 9, 16, and so on. Since X and Y are single-digit numbers, (X - Y) cannot be greater than 9. Possible values for (X - Y) are 1, 4, and 9. Now let's take the equation (X + Y) = 11 and add it to each of these three possible equations: (X - Y) = 1, (X - Y) = 4, and (X - Y) = 9.

This yields three new equations: 2X = 12, 2X = 15, and 2X = 20. Dividing both sides of these equations by 2 yields possible values for X of 6, 7.5, and 10. We know that X is a single-digit number, so X must be 6. We know that (X + Y) = 11, so Y must be 5. Therefore Alice lives at house number 56, and Bill lives at house number 65. 65 + 56 = 121 = 11 squared; 65 - 56 = 9 = 3 squared.
2. Now we have to find my friends Charlie and Debra, who live in two different houses on Algebra Avenue. I can't remember their house numbers either, but I remember that they are the reverse of each other. The sum of their house numbers is a perfect square, and the difference between their house numbers is a perfect cube. Charlie lives in the house with the smaller house number. What is the number of Debra's house?

Answer: 74

Let's call Debra's house number "XY" and Charlie's house number "YX". Using the same procedure as in Question 1, we find that (X + Y) = 11. This time, however, we want (9)(X - Y) to be a perfect cube. Since 9 is equal to 3 times 3, this will be true only when (X - Y) = 3 or 3 times a perfect cube; 24, 81, and so on. Since X and Y are single-digit numbers, (X - Y) cannot be greater than 9.

Therefore (X - Y) = 3. Adding the equation (X + Y) = 11 to the equation (X - Y) = 3 yields the new equation 2X = 14. Dividing both sides of this equation by 2 yields X = 7. Since (X + Y) = 11, Y must be 4.

Therefore Charlie lives at house number 47, and Debra lives at house number 74. 74 + 47 = 121 = 11 squared; 74 - 47 = 27 = 3 cubed.
3. Now that we've found everybody, we need to go back to my house to get some money. I keep it in a rectangular box. Unfortunately, I own a lot of rectangular boxes, and I can't remember which one I put my money in. I remember that the three dimensions of the box (height, width, and length) are all whole numbers of inches, that they are all different, and that none of the dimensions is a prime number. I also remember that the volume of the box in cubic inches is the same as the total surface area of all six sides of the box in square inches. What is this volume or surface area?

Answer: 288

Let's call the shortest dimension A, the medium dimension B, and the longest dimension C. The volume of the box in cubic inches will be ABC. The two smallest sides of the box will have a surface area of AB each, the two medium sides will have a surface area of AC each, and the two largest sides will have a surface area of BC each.

The total surface area of all six sides will be 2AB + 2AC + 2BC. We want to find whole, non-prime values of A, B, and C for this equation: ABC = 2AB + 2AC + 2BC. Let's consider possible values of A. If A were 2 or less, the left side of the equation would be 2BC or less.

However, since 2BC is only one of the three values added together on the right side of the equation, the right side of the equation must be greater than 2BC.

Therefore A is greater than 2. If A were 6 or greater, the left side of the equation would be 6BC or greater. However, since 2AB and 2AC are both less than 2BC, the right side of the equation must be less than (2BC + 2BC + 2BC); that is, it must be less than 6BC.

Therefore A must be less than 6. Therefore A can be 3, 4, or 5. We are told that none of the dimensions is a prime number, so A must be 4. Let's enter this value of A into the equation to get: 4BC = 8B + 8C + 2BC. Subtract 2BC from both sides to get: 2BC = 8B + 8C. Subtract 8C from both sides to get: 2BC - 8C = 8B. Divide both sides by 2 to get: BC - 4C = 4B. Factor out C on the left side of the equation to get: C(B - 4)= 4B. Divide both sides by (B - 4) to get: C = 4B/(B - 4). Let's consider possible values of B. If B were equal to or greater than 8, the right side of the equation would be less than or equal to (4)(8)/(8 - 4); that is, less than or equal to 8. [Because we are dividing by (B - 4) to get the value of C, the value of C goes down as the value of B goes up.] However, we know that C is greater than B. Therefore B is less than 8. We know that B is greater than A, so B must be greater than 4. Possible values for B are therefore 5, 6, and 7. We know that none of the dimensions is a prime number, so B must be 6. Let's enter this value of B into our most recent equation to get: C = (4)(6)/(6 - 4). Therefore C equals 12. (4)(6)(12) = 288; (2)(4)(6) + (2)(4)(12) + (2)(6)(12) = 288.
4. Now we've found the right box, but I forgot it has a lock on it. It's one of those locks where you have to punch in a certain number to open it. I can't remember what the number is, but I remember that it's a four-digit number. I also remember that the sum of all four digits is equal to the product of all four digits, and that the four-digit number is evenly divisible by this sum or product. By the way, I also remember that the four-digit number is not "0000". What is it?

Answer: 4112

Let's call the four digits A, B, C, and D, arranged so that A is less than or equal to B, which is less than or equal to C, which is less than or equal to D. (At this point, we don't care what order they come in to form the four-digit number.) We know that ABCD = A + B + C + D. If any digit were 0, the left side of the equation would be 0.

The right side of the equation can only be 0 if all the digits are 0, but we are told that this is not true. Therefore, none of the digits is 0. If all of the digits were 2, the left side of the equation would be 16, and the right side of the equation would be 8.

As the value of the digits increase, the left side of the equation increases more quickly than the right side of the equation, so that they will never be equal.

Therefore, not all the digits can be 2 or greater. Therefore, A must be 1. Let's enter this value of A into the equation to get: BCD = 1 + B + C + D. If the three remaining unknown digits were all 2, the left side of the equation would be 8 and the right side of the equation would be 7. Again, as the values of the digits increase, the left side of the equation increases more quickly than the right side of the equation, so that they will never be equal.

Therefore at least one more digit must be 1. Therefore B must be 1. Let's enter this value of B into the equation to get: CD = 2 + C + D. If C were also 1, the resulting equation would be: D = 3 + D. This is impossible, so C must be greater than 1. If both unknown digits were 3, the left side of the equation would be 9 and the right side of the equation would be 8. As before, the left side of the equation would increase more quickly than the right side of the equation as the values of C and D increase, so they would never be equal. Therefore at least one digit must be less than 3. Therefore C must be greater than 1 but less than 3, so C is equal to 2. Let's enter this value of C into the equation to get: 2D = 4 + D. Subtracting D from both sides of the equation, we get D = 4. Therefore the four digits are 1, 1, 2, and 4. (1)(1)(2)(4) = 8; 1 + 1 + 2 + 4 = 8. Now we have to arrange these four digits to make a four-digit number that is evenly divisible by 8. This must be an even number, so the last digit must be 2 or 4. Therefore the only possible arrangements of these four digits which we need to consider are 1124, 1142, 1214, 1412, 2114, and 4112. Only 4112 is divisible by 8 (4112/8 = 514), so this is the four-digit number we need.
5. Now I have my money. I'll let you choose whatever toppings you want on your pizza if you can figure out how much I have. I have a certain number of one-dollar bills (more than zero) and a certain number of ten-dollar bills (more than zero.) The number of one-dollar bills multiplied by the number of ten-dollar bills is equal to the total amount of money I have in dollars. The number of ten-dollar bills is not a prime. How much money do I have?

Answer: 72

Let's call the number of ten-dollar bills X and the number of one-dollar bills Y. The amount of money I have is therefore 10X + Y. We want to find X and Y such that 10X + Y = XY. Subtract Y from both sides of this equation to get: 10X = XY - Y. Factor out Y on the right side of the equation to get: 10X = Y(X - 1). Divide both sides of the equation by (X - 1) to get: 10X/(X-1) = Y. We want Y to be a whole number.

This is only possible if (X-1) divides evenly into 10X. Since the only factor that X and (X - 1) can have in common is 1, (X - 1) must be 1 or a factor of 10.

Therefore possible values for (X - 1) are 1, 2, 5, and 10. This means that possible values for X are 2, 3, 6, and 11. We are told that X is not a prime, so X must be 6. Let's enter this value of X into the most recent equation: (10)(6)/(6 - 1) = Y.

This yields a value for Y of 12. Therefore I have 6 ten-dollar bills and 12 one-dollar bills, or 72 dollars total. (10)(6) + 12 = 72; (6)(12) = 72. The answers is not $40, because that requires 2 10 dollar bills, and 2 is a prime number, which the rules disallow. I hope you enjoy your pizza!
Source: Author VickiSilver

This quiz was reviewed by FunTrivia editor crisw before going online.
Any errors found in FunTrivia content are routinely corrected through our feedback system.
Related Quizzes
1. 90 Percent of the Game is Half-Mental Tough
2. ASA Quiz About This Quiz 1 Very Difficult
3. ASA Quiz About This Quiz 2 Very Difficult
4. ASA Quiz About This Quiz 3 Difficult
5. ASA Quiz About This Quiz 4 Tough
6. Brain Bogglers Difficult
7. Brain Bogglers 2 Tough
8. Brain Bogglers 6 Difficult
9. Brain Bogglers 7 Tough
10. Brain Bogglers 8 Difficult
11. Chaos, Challenge, Logic, Order Very Difficult
12. Do You Know The Alphabet? Very Difficult

11/21/2024, Copyright 2024 FunTrivia, Inc. - Report an Error / Contact Us