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Quiz about The Lost Address Book
Quiz about The Lost Address Book

The Lost Address Book Trivia Quiz


So, I'm having a party and want to mail out invitations, but I can't recall anyone's address, and I can't find my address book. I used number theory to try to remember them, but I still can't figure them out. Please help!

A multiple-choice quiz by bfguitarhero. Estimated time: 9 mins.
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Author
bfguitarhero
Time
9 mins
Type
Multiple Choice
Quiz #
313,380
Updated
Feb 16 23
# Qns
10
Difficulty
Difficult
Avg Score
5 / 10
Plays
219
- -
Question 1 of 10
1. The first invitation goes to Adam. The addresses of the street he lives on start at 300 and go up. Also, the number of factors of his address is equal to the total number of possible arrangements of the letters in his name ('A', 'D', 'A', and 'M'). If his house is the first one on his street to have exactly that many factors, then what is his address?

Answer: (One Number, Three Digits)
Question 2 of 10
2. My friend Brian lives in a house which has an address that is a perfect square. No house on his street has an address with more than two digits. When you multiply his address by the number that is one less than his address, the total number of factors of that number is equal to his address. So, what is his address?

Answer: (One Number, One or Two Digits)
Question 3 of 10
3. Chris's address has a unique characteristic. If you square his address, the number of factors of the square will be equal to his address. Also, the square of his address isn't equal the address itself. What is his address?

Answer: (One Number)
Question 4 of 10
4. Devin's address is the first number whose factorial ends in exactly 200 zeros. What is his address?

Answer: (One Number, Three Digits)
Question 5 of 10
5. I always make fun of Evan for being evil because his address is the first multiple of 23 whose digits add up to 23 (and everyone knows that no number is more evil than 23). What's his address?

Answer: (One Number, Four Digits)
Question 6 of 10
6. Frank's address is the first 4-digit number whose number of factors is equal to the number of proper factors (factors other than itself) of the number that is one less than it. What is Frank's address?

Answer: (One Number, Four Digits)
Question 7 of 10
7. George's address can be represented as XYZ, where X represents the 100's digit (greater than 0), Y represents the 10's digit, and Z represents the 1's digit. George's address is one such that X + Y - Z = X^2 + Y^2 - Z^2. If X, Y, and Z are unique integers, and XYZ is not prime, then what is George's address?

Answer: (One Number, Three Digits)
Question 8 of 10
8. Henry's address is the first number that can be represented by the sum of 3, 4, 5, or 7 consecutive positive integers. What is his address?

Answer: (One Number, Three Digits)
Question 9 of 10
9. Ian's address is the 12th multiple of 12 to have 12 factors. What is his address?

Answer: (One Number, Three Digits)
Question 10 of 10
10. Joe's address is the lowest perfect square for which the sum of its digits is a perfect number (A number whose proper factors add up to itself). What is his address?

Answer: (One Number, Five Digits)

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Quiz Answer Key and Fun Facts
1. The first invitation goes to Adam. The addresses of the street he lives on start at 300 and go up. Also, the number of factors of his address is equal to the total number of possible arrangements of the letters in his name ('A', 'D', 'A', and 'M'). If his house is the first one on his street to have exactly that many factors, then what is his address?

Answer: 306

For all solutions:
- x! represents the factorial of x, or x * (x - 1) * (x - 2) * ... * 2 * 1.
- x^y represents x to the power of y, or x times itself y times.
- p represents any prime number. In any prime factorization, all p's must be unique, or different from each other.
- int(x) represents the greatest integer function of x, in which x is truncated, or rounded down to the greatest integer.
- dSum(x) represents the digital sum of x. To find the digital sum of a number, you add all of its digits together. You keep adding the digits of each number together until the number is less than 10. This essentially represents how much greater the number is than a multiple of 9.

The total number of arrangements of "ADAM" is 4!/2!, or 12, meaning that Adam's address must have 12 possible factors. So, the prime factorization must either be in the form p * p * p^2, p^2 * p^3, p * p^5, or p^11. For numbers 300 - 306:

300 = 2^2 * 3 * 5^2
301 = 7 * 43
302 = 2 * 151
303 = 3 * 101
304 = 2^4 * 19
305 = 5 * 61
306 = 2 * 3^2 * 17

Therefore, 306 is the first address with 12 factors, and must also be Adam's address.

For those who don't know, in order to find the number of factors of a number, you must first put it in its prime factorization. For any factor of p^n, the term p can be used from 0 to n times in any given factor. Therefore, the number of factors of p^a * p^b * p^c * ... is (a + 1)(b + 1)(c + 1)... For example, the prime factorization of 24 is 2^3 * 3^1, which means it would have (3 + 1)(1 + 1), or 8, factors.
2. My friend Brian lives in a house which has an address that is a perfect square. No house on his street has an address with more than two digits. When you multiply his address by the number that is one less than his address, the total number of factors of that number is equal to his address. So, what is his address?

Answer: 36

Let Brian's address be represented by n^2. If n is composed of a single p, then n would have to be a multiple of 3 (p^2 has 3 factors). If so, then n would have to equal 3, which obviously doesn't work (9 * 8 has is 72, which has 12 factors, not 9). The next possibility is that n is composed of p * p, which means that it would have to be a multiple of 9 (p^2 * p^2 has 3 * 3, or 9, factors). In that case, n would have to be 3 * p, since 9 is the square of 3. For any p other then 2 or 3, n^2 has 3 digits. If p is 3, than n becomes p^2, which we later find out doesn't work. So, we check for p = 2:

36 * 35 = 2^2 * 3^2 * 5 * 7, which has 36 factors.

Let's examine some other cases:
- If n = (p * p * p), the smallest value for n is 30, which, when squared, ends up being much greater than 100.
- If n = (p^2), then n must be a multiple of 5, since (p^2)^2 has 5 factors, making the smallest value of n 25, which when squared, is also much greater than 100.

Therefore, the only possible value of Brian's address is 36.
3. Chris's address has a unique characteristic. If you square his address, the number of factors of the square will be equal to his address. Also, the square of his address isn't equal the address itself. What is his address?

Answer: 3

Let Chris's address be represented by n. If n is composed of a single p, then n^2, or p^2, is going to have 3 factors. Since 3 is a prime number, it can be n.
Also, n cannot possibly be any other combination of p because:
- if n is composed of (p * p), then n^2 will have 9 factors, and since each p must be a unique prime and 9 is a perfect square, n cannot be composed of (p * p). This applies to (p * p * p), and any other extension of this because of the same reason.
- if n is composed of (p^2), then n^2 will have 5 factors, which is not a perfect square. Futhermore, for any p^n in which n > 2, 2^n, which is the smallest possible value for p^n, will be greater than 2n + 1. Therefore, no exponents can exist in this prime factorization.

So, Chris's address is 3. To show that this works, 3^2 = 9, which has factors of 9, 3, and 1.
4. Devin's address is the first number whose factorial ends in exactly 200 zeros. What is his address?

Answer: 805

To find the number of zeros that a number's factorial must end in, you must find how many powers of 10 the number contains. Since multiples of 2 are far more common than multiples of 5, we look for how many powers of 5 are contained within a factorial, and to do that we use the greatest integer function. To get an idea of how large it needs to be, we'll first look at 100!. int(100/5) + int(100/25) = 20 + 4 = 24, so 100! will end in 24 zeros. Since 24 is about one-eighth of 200, we will look at 800!. int(800/5) + int(800/25) + int(800/125) + int(800/625) = 160 + 32 + 6 + 1 = 199, so 800! will end in 199 zeros. Simply go to the next factorial that is a multiple of 5, and you get that 805! is the first factorial to end in 200 zeros, making Devin's address 805.
5. I always make fun of Evan for being evil because his address is the first multiple of 23 whose digits add up to 23 (and everyone knows that no number is more evil than 23). What's his address?

Answer: 1679

The digital sum of 23 is 5, and Evan's address must have digits that add up to 23, which means that Evan's address also has a digital sum of 5. That means that if 23 * n represents Evan's address, n must have a digital sum of 1. By plugging in numbers, we find that n is 73, and that 1679 is the first multiple of 23 whose digits add up to 23, making it Evan's address.
6. Frank's address is the first 4-digit number whose number of factors is equal to the number of proper factors (factors other than itself) of the number that is one less than it. What is Frank's address?

Answer: 1090

First, the question is pretty much stating, "What is the four-digit number for which the number that is one less than itself has one more factor?" Therefore, one of the numbers must have an even amount of factors and the other must have an odd amount of factors, meaning that one of the numbers must be a perfect square. So, we are looking for the first integer n for which either n^2 + 1 has one less factor than n^2 or n^2 - 1 has one more factor than n^2. Since 32^2 is the first perfect square with 4 digits, let us look at the first two perfect squares, the numbers that are one more and one less than them, their prime factorizations, and their number of factors:

1023 = 3 * 11 * 31 = 8 factors
1024 = 32^2 = 2^10 = 11 factors
1025 = 5^2 * 41 = 6 factors

1088 = 2^6 * 17 = 14 factors
1089 = 33^2 = 3^2 * 11^2 = 9 factors
1090 = 2 * 5 * 109 = 8 factors

Therefore, because 1089 has one more factor than 1090, Frank's address is 1090.
7. George's address can be represented as XYZ, where X represents the 100's digit (greater than 0), Y represents the 10's digit, and Z represents the 1's digit. George's address is one such that X + Y - Z = X^2 + Y^2 - Z^2. If X, Y, and Z are unique integers, and XYZ is not prime, then what is George's address?

Answer: 679

First, by rearranging the equation given, we get that X(X - 1) + Y(Y - 1) = Z(Z - 1). Next, we will let Y = n (you could use X), since Y is a value of interest in the problem. Also, this lets us put a relationship between between Y and Z. For example, if the difference between Y and Z is one, then the difference between Y(Y - 1) and Z(Z - 1) is (n^2 + n) - (n^2 - n), or 2n.

Now, since Y = n, n cannot have a value of 0 or 1, as that would cause X and Z to be the same number. So, we test out some numbers:

Z - Y = 1 (Z must be higher or the equation given is impossible)
(n^2 + n) - (n^2 - n) = 2n

In order to get an integer value for X, 2n needs to be convertable to an X(X - 1) form. For 2n, the lowest such value for n is 3.

n = 3, X = 3 (3 * 2), Y = 3, Z = 4
n = 6, X = 4 (4 * 3), Y = 6, Z = 7
No more possible values for n if Z - Y = 1 (values of n become too large)

Z - Y = 2
(n^2 + 3n + 2) - (n^2 - n) = 4n + 2
n = 7, X = 6 (6 * 5), Y = 7, Z + 9
No more possible values for n if Z - Y = 2

Z - Y = 3
(n^2 + 5n + 6) - (n^2 - n) = 6n + 6
n = 4, X = 6 (6 * 5), Y = 4, Z = 7
n = 6, X = 7 (7 * 6), Y = 6, Z = 9
No more possible values for n if Z - Y = 3

Z - Y = 4
(n^2 + 7n + 12) - (n^2 - n) = 8n + 12
No possible values for n if Z - Y = 4

For all values of Z - Y greater than or equal to 4, there are no possible values of n such that X, Y, and Z are all less than 10. Therefore, we have 5 possibilities:

334
467
679
647
769

Since 334 has a repeating digit of 3, the only composite number left is 679.
8. Henry's address is the first number that can be represented by the sum of 3, 4, 5, or 7 consecutive positive integers. What is his address?

Answer: 210

Let x represent the first of the consecutive numbers. For 3 consecutive numbers, the number can be represented by x + x + 1 + x + 2, which is 3x + 3, or 3(x + 1), meaning the number must be a multiple of 3. For 4 consecutive numbers, the number can be represented by x + x + 1 + x + 2 + x + 3, which is 4x + 6, or 2(2x + 3). Since 2x + 3 must be odd, the number must be a multiple of 2 but not a multiple of 4. For 5 consecutive numbers, the number can be represented by x + x + 1 + x + 2 + x + 3 + x + 4, which is 5x + 10, or 5(x + 2), meaning the number must be a multiple of 5. For 7 consecutive numbers, the number can be represented by 5x + 10 + x + 5 + x + 6, which is 7x + 21, or 7(x + 3), meaning it must be a multiple of 7. So, the number must be the lowest possible multiple of 3, 5, and 7, which is 105, and the lowest multiple of 2 that isn't a multiple of 4, meaning the number must be 210.
9. Ian's address is the 12th multiple of 12 to have 12 factors. What is his address?

Answer: 372

The possible ways to have 12 factors are p^11, p^5 * p, p^3 * p^2, or p^2 * p * p. Since the prime factorization must include 3 * 2^2, p^11 is immediately eliminated. For p^5 * p, the only possibility is 2^5 * 3, which is 96. For p^3 * p^2, the only possibilities are 3^3 * 2^2, which is 108, and 3^2 * 2^3, which is 72. For p^2 * p * p, it must be 2^2 * 3 * p, or 12 * p. Since there are 3 lower possibilities that don't have the prime factorization p^2 * p * p, Ian's address must be 12 times the 9th prime number after 3, which is 31. So, Ian's address must be 372.
10. Joe's address is the lowest perfect square for which the sum of its digits is a perfect number (A number whose proper factors add up to itself). What is his address?

Answer: 17956

First, let's look at some perfect numbers. The first one is 6. However, the sum of the digits can't be 6 because all perfects squares must have a digital sum of 1, 4, 7, or 9. The next perfect number is 28, which has a digital sum of 1. The number we're looking for will have digits that add up to 28, since 496, the next perfect number, is far too large to even consider. Next, we need to know some rules about perfect squares.

As a rule, if dSum(n) = 1 or dSum(n) = 8, then dSum(n^2) = 1, which is what we're looking for. So, Joe's address is going to be the square of a number that is one more or one less than a multiple of 9. And after plugging in a lot of numbers, we find out the lowest perfect square whose digits add up to 28 is 134^2, or 17956.
Source: Author bfguitarhero

This quiz was reviewed by FunTrivia editor crisw before going online.
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