Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Sure, let's go through each of the questions and provide detailed explanations for each.
### Question (i)
For what value of \( n, 4^n \) ends in 0?
Options:
(a) 10
(b) when \( n \) is even
(c) when \( n \) is odd
(d) no value of \( n \)
To answer this question, we need to understand the form of \( 4^n \). The number \( 4^n \) is \( (2^2)^n = 2^{2n} \). Since \( 2^{2n} \) is a power of 2, and powers of 2 end in either 2, 4, 8, or 6, but never in 0. Therefore, there is no value of \( n \) for which \( 4^n \) ends in 0.
Answer: (d) no value of \( n \)
### Question (ii)
If \( a \) is a positive rational number and \( n \) is a positive integer greater than 1, then for what value of \( n, a^n \) is a rational number?
Options:
(a) when \( n \) is any even integer
(b) when \( n \) is any odd integer
(c) for all \( n > 1 \)
(d) only when \( n = 0 \)
A positive rational number raised to any positive integer power is always rational. Therefore, for any \( n > 1 \), \( a^n \) will be rational.
Answer: (c) for all \( n > 1 \)
### Question (iii)
If \( x \) and \( y \) are two odd positive integers, then which of the following is true?
Options:
(a) \( x^2 + y^2 \) is even
(b) \( x^2 + y^2 \) is not divisible by 4
(c) \( x^2 + y^2 \) is odd
(d) both (a) and (b)
For two odd integers \( x \) and \( y \), their squares \( x^2 \) and \( y^2 \) will both be odd because the square of an odd number is always odd. The sum of two odd numbers is even, so \( x^2 + y^2 \) is even.
Moreover, if you look at the remainders when squaring odd numbers (they are always 1 modulo 4), the sum \( x^2 + y^2 \) will be \( 1 + 1 = 2 \) modulo 4, which means \( x^2 + y^2 \) is not divisible by 4.
Answer: (d) both (a) and (b)
### Question (iv)
The statement 'One of every three consecutive positive integers is divisible by 3' is
Options:
(a) always true
(b) always false
(c) sometimes true
(d) None of these
For any set of three consecutive integers \( n, n+1, n+2 \), one of these numbers must be divisible by 3 due to the properties of consecutive integers and the nature of division by 3.
Answer: (a) always true
### Question (v)
If \( n \) is any odd integer, then \( n^2 - 1 \) is divisible by
Options:
(a) 22
(b) 55
(c) 88
(d) 8
For any odd integer \( n \), \( n^2 - 1 \) can be factored as \( (n-1)(n+1) \). This expression consists of two consecutive even integers because \( n \) is odd, and one of those integers will be a multiple of 4 and the other will be an even number. Thus, their product will always be divisible by 8.
Answer: (d) 8
So, the detailed step-by-step answers are:
(i) (d) no value of \( n \)
(ii) (c) for all \( n > 1 \)
(iii) (d) both (a) and (b)
(iv) (a) always true
(v) (d) 8
### Question (i)
For what value of \( n, 4^n \) ends in 0?
Options:
(a) 10
(b) when \( n \) is even
(c) when \( n \) is odd
(d) no value of \( n \)
To answer this question, we need to understand the form of \( 4^n \). The number \( 4^n \) is \( (2^2)^n = 2^{2n} \). Since \( 2^{2n} \) is a power of 2, and powers of 2 end in either 2, 4, 8, or 6, but never in 0. Therefore, there is no value of \( n \) for which \( 4^n \) ends in 0.
Answer: (d) no value of \( n \)
### Question (ii)
If \( a \) is a positive rational number and \( n \) is a positive integer greater than 1, then for what value of \( n, a^n \) is a rational number?
Options:
(a) when \( n \) is any even integer
(b) when \( n \) is any odd integer
(c) for all \( n > 1 \)
(d) only when \( n = 0 \)
A positive rational number raised to any positive integer power is always rational. Therefore, for any \( n > 1 \), \( a^n \) will be rational.
Answer: (c) for all \( n > 1 \)
### Question (iii)
If \( x \) and \( y \) are two odd positive integers, then which of the following is true?
Options:
(a) \( x^2 + y^2 \) is even
(b) \( x^2 + y^2 \) is not divisible by 4
(c) \( x^2 + y^2 \) is odd
(d) both (a) and (b)
For two odd integers \( x \) and \( y \), their squares \( x^2 \) and \( y^2 \) will both be odd because the square of an odd number is always odd. The sum of two odd numbers is even, so \( x^2 + y^2 \) is even.
Moreover, if you look at the remainders when squaring odd numbers (they are always 1 modulo 4), the sum \( x^2 + y^2 \) will be \( 1 + 1 = 2 \) modulo 4, which means \( x^2 + y^2 \) is not divisible by 4.
Answer: (d) both (a) and (b)
### Question (iv)
The statement 'One of every three consecutive positive integers is divisible by 3' is
Options:
(a) always true
(b) always false
(c) sometimes true
(d) None of these
For any set of three consecutive integers \( n, n+1, n+2 \), one of these numbers must be divisible by 3 due to the properties of consecutive integers and the nature of division by 3.
Answer: (a) always true
### Question (v)
If \( n \) is any odd integer, then \( n^2 - 1 \) is divisible by
Options:
(a) 22
(b) 55
(c) 88
(d) 8
For any odd integer \( n \), \( n^2 - 1 \) can be factored as \( (n-1)(n+1) \). This expression consists of two consecutive even integers because \( n \) is odd, and one of those integers will be a multiple of 4 and the other will be an even number. Thus, their product will always be divisible by 8.
Answer: (d) 8
So, the detailed step-by-step answers are:
(i) (d) no value of \( n \)
(ii) (c) for all \( n > 1 \)
(iii) (d) both (a) and (b)
(iv) (a) always true
(v) (d) 8
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
