The Lost Address Book Trivia Quiz

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.