Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify as...

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.

(a) Claim. There is a unique 3-digit number whose digits have sum 8 and product 10.

“Proof.” Let x, y, and z be the digits. Then x + y + z = 8 and xyz = 10. The only factors of 10 are 1, 2, 5, and 10, but since 10 is not a digit, the digits must be 1, 2, and 5. The sum of these digits is 8. Therefore, 125 is the only 3-digit number whose digits have sum 8 and product 10.

(b) Claim. There is a unique set of three consecutive odd numbers that are

all prime.

“Proof.” The consecutive odd numbers 3, 5, and 7 are all prime. Suppose x, y, and z are consecutive odd numbers, all prime, and x≠ 3.Then y = x + 2 and z = x + 4 Since x is prime, when x is divided by 3, the remainder is 1 or 2. In case the remainder is 1, then x = 3k + 1 for some integer k ≥ 1. But then y = x + 2 = 3k + 3 = 3(k + 1), so y is not prime. In case the remainder is 2, then x = 3k + 2 for some integer k ≥ 1. But then z = x + 4 = 3k + 2 + 4 = 3(k + 2), so z is not prime. In either case we reach the contradiction that y or z is not prime. Thus and Therefore, the only three consecutive odd primes are x = 3, y = 5, and z = 7.

(c) Claim. If x is any real number, then either π – x is irrational or π + x is irrational.

“Proof.” It is known that π is an irrational number; that is, π cannot be written in the form for integers a and b. Consider x = π. Then π − x = 0, which is rational, but π + x = 2π. If 2π were rational, then for some integers a and b. Then so is rational. This is impossible, so 2π is irrational. Therefore either π − x or π + x is irrational.

(d) Claim. If x is any real number, then either is irrational or

is irrational.

“Proof.” It is known that π is an irrational number; that is, π cannot be written in the form for integers a and b. Let x be any real number. Suppose both π − x = 0 and π + x = 0, are rational. Then since the sum of two rational numbers is always rational, (π − x) + (π + x) = 2π is rational. Then for some integers a and b. Then so π is rational. This is impossible. Therefore, at least one of π − x = 0 or π + x = 0 is irrational.

(e) Claim. For all natural numbers n, gcd(n, n + 1) = 1.

“Proof.” (i) 1 divides n and 1 divides n + 1. (ii) Suppose c divides n and c divides n + 1. Then 1 divides c. Therefore, gcd(n, n + 1) = 1

(f) Claim. For all natural numbers n, gcd(2n − 1, 2n + 1) = 1.

“Proof.” Obviously 1 divides both and Suppose c divides 2n − 1 and 2n + 1. Then c divides their sum, 4n, so c also divides Furthermore, c divides their product, Since c divides 4n² and 4n² − 1, 4n² c divides 4n² − (4n² − 1) = 1. Therefore, c ≤ 1. Thus 1 is the greatest common divisor. _