Question

5. Some airline tickets have a 15-digit identification number a1a2...a15, where a15 is a check digit that equals a1a2...a14 mod 7. Which errors in a single digit of a 15-digit airline ticket identification number can be detected if a digit x in the nth column from the right is incorrectly replaced with a digit y? We can detect all the errors except when the difference between x and y is 7. We cannot detect any error if a single digit x is replaced with another digit y. We can detect the errors only if the difference between x and y is 7. We can detect all the errors if a single digit x is replaced with another digit y. Identify the correct steps involved in showing that if 2^n - 1 is prime, then n is prime. (Check all that apply.) We will prove by contrapositive. Suppose n is not prime. Then, n = ab, for some integers a > 1 and b > 1. We must prove that 2^ab - 1 is not prime. We will prove by contrapositive. Suppose n is not prime. Then, n = ab, for some integers a < 1 and b < 1. We must prove that 2^ab - 1 is not prime. Consider the identity 2^ab - 1 = (2^a - 1) · (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1). The identity is valid, since we can clearly see on the right-hand side that all terms except 2^ab and -1 cancel. Clearly, (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1) is greater than 1. Since a > 1, the factor 2^a - 1 is greater than 1. Clearly, (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1) is less than 1. Since a < 1, the factor 2^a - 1 is less than 1. Since 2^n - 1 is the product of two integers that are greater than 1, 2^n - 1 is not prime. Identify the correct steps involved in proving that a^m + 1 is composite if a and m are integers greater than 1 and m is odd. (Check all that apply.) As m is odd, we can write a^m + 1 = (a + 1)(a^(m-1) - a^(m-2) + a^(m-3) - a^(m-4) + ... + 1). As m is odd, we can write a^m + 1 = (a - 1)(a^(m-1) - a^(m-2) + a^(m-3) - a^(m-4) + ... - 1). As both a and m are greater than 1, we have 1 < a + 1 < a^m + 1. Thus, a + 1 is a proper factor of a^m + 1. As both a and m are greater than 1, we have 1 < a - 1 < a^m + 1. Thus, a - 1 is a proper factor of a^m + 1. Thus, we can express a^m + 1 as a product of two proper factors; so, a^m + 1 is composite.

          5.

Some airline tickets have a 15-digit identification number a1a2...a15, where a15 is a check digit that equals a1a2...a14 mod 7. Which errors in a single digit of a 15-digit airline ticket identification number can be detected if a digit x in the nth column from the right is incorrectly replaced with a digit y?

We can detect all the errors except when the difference between x and y is 7.
We cannot detect any error if a single digit x is replaced with another digit y.
We can detect the errors only if the difference between x and y is 7.
We can detect all the errors if a single digit x is replaced with another digit y.

Identify the correct steps involved in showing that if 2^n - 1 is prime, then n is prime. (Check all that apply.)

We will prove by contrapositive. Suppose n is not prime. Then, n = ab, for some integers a > 1 and b > 1. We must prove that 2^ab - 1 is not prime.
We will prove by contrapositive. Suppose n is not prime. Then, n = ab, for some integers a < 1 and b < 1. We must prove that 2^ab - 1 is not prime.
Consider the identity 2^ab - 1 = (2^a - 1) · (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1). The identity is valid, since we can clearly see on the right-hand side that all terms except 2^ab and -1 cancel.
Clearly, (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1) is greater than 1. Since a > 1, the factor 2^a - 1 is greater than 1.
Clearly, (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1) is less than 1. Since a < 1, the factor 2^a - 1 is less than 1.
Since 2^n - 1 is the product of two integers that are greater than 1, 2^n - 1 is not prime.

Identify the correct steps involved in proving that a^m + 1 is composite if a and m are integers greater than 1 and m is odd. (Check all that apply.)

As m is odd, we can write a^m + 1 = (a + 1)(a^(m-1) - a^(m-2) + a^(m-3) - a^(m-4) + ... + 1).
As m is odd, we can write a^m + 1 = (a - 1)(a^(m-1) - a^(m-2) + a^(m-3) - a^(m-4) + ... - 1).
As both a and m are greater than 1, we have 1 < a + 1 < a^m + 1. Thus, a + 1 is a proper factor of a^m + 1.
As both a and m are greater than 1, we have 1 < a - 1 < a^m + 1. Thus, a - 1 is a proper factor of a^m + 1.
Thus, we can express a^m + 1 as a product of two proper factors; so, a^m + 1 is composite.

5.

Some airline tickets have a 15-digit identification number a1a2...a15, where a15 is a check digit that equals a1a2...a14 mod 7. Which errors in a single digit of a 15-digit airline ticket identification number can be detected if a digit x in the nth column from the right is incorrectly replaced with a digit y?

We can detect all the errors except when the difference between x and y is 7.
We cannot detect any error if a single digit x is replaced with another digit y.
We can detect the errors only if the difference between x and y is 7.
We can detect all the errors if a single digit x is replaced with another digit y.

Identify the correct steps involved in showing that if 2^n - 1 is prime, then n is prime. (Check all that apply.)

We will prove by contrapositive. Suppose n is not prime. Then, n = ab, for some integers a > 1 and b > 1. We must prove that 2^ab - 1 is not prime.
We will prove by contrapositive. Suppose n is not prime. Then, n = ab, for some integers a < 1 and b < 1. We must prove that 2^ab - 1 is not prime.
Consider the identity 2^ab - 1 = (2^a - 1) · (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1). The identity is valid, since we can clearly see on the right-hand side that all terms except 2^ab and -1 cancel.
Clearly, (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1) is greater than 1. Since a > 1, the factor 2^a - 1 is greater than 1.
Clearly, (2^a(b-1) + 2^a(b-2) + ... + 2^a + 1) is less than 1. Since a < 1, the factor 2^a - 1 is less than 1.
Since 2^n - 1 is the product of two integers that are greater than 1, 2^n - 1 is not prime.

Identify the correct steps involved in proving that a^m + 1 is composite if a and m are integers greater than 1 and m is odd. (Check all that apply.)

As m is odd, we can write a^m + 1 = (a + 1)(a^(m-1) - a^(m-2) + a^(m-3) - a^(m-4) + ... + 1).
As m is odd, we can write a^m + 1 = (a - 1)(a^(m-1) - a^(m-2) + a^(m-3) - a^(m-4) + ... - 1).
As both a and m are greater than 1, we have 1 < a + 1 < a^m + 1. Thus, a + 1 is a proper factor of a^m + 1.
As both a and m are greater than 1, we have 1 < a - 1 < a^m + 1. Thus, a - 1 is a proper factor of a^m + 1.
Thus, we can express a^m + 1 as a product of two proper factors; so, a^m + 1 is composite.

Added by Tammy B.

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