A multiple-choice quiz
by VickiSilver.
Estimated time: 23 mins.

Most Recent Scores

Sep 20 2024
:
Guest 222: 1/5Sep 18 2024 : Guest 99: 0/5

Sep 16 2024 : wellenbrecher:

Score Distribution

Quiz Answer Key and Fun Facts

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.

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.

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.

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.

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.

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.

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.

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.

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!

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!

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

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.