SEBA Class 10 Advanced Maths Chapter 3.4 Solutions: Arithmetic of Integers
In this article, we have solved all the questions from Chapter 3.4 of SEBA Class 10 Advanced Mathematics in a simple way. 100 % reliable solutions for all the questions of SEBA Class 10 Advanced Maths Chapter 3.4
See More
Chapter 3.1 of SEBA Class 10 Advanced Mathematics
Chapter 3.2 of SEBA Class 10 Advanced Mathematics
Chapter 3.3 of SEBA Class 10 Advanced Mathematics
Chapter 3.4 of SEBA Class 10 Advanced Mathematics
Solutions to Exercise 3.4
Q1: Express the following numbers as product of primes
(i) 1234 Solution:
$1234 = 2 \times 617$
(ii) 10140 Solution:
10140 = 2 × 2 × 3 × 5 × 13 × 13
(iii) 72149 Solution:
72149 = 7 × 11 × 937
(iv) 41531 Solution:
41531 = 7 × 17 × 349
(v) 1020305 Solution:
Q2: Examine if the numbers in Ex. 1 are prime or composite.
Solution:
Since each number mentioned in example 1 has more than one prime factor (as shown above), each has divisors other than 1 and itself. So they are composite numbers.
Q3: Mention 5 prime numbers of the form $n^{2}-2$ where $n\in\mathbb{N}$
Solution:
For $n=1$, $n^{2}-2 = 1^{2}-2 = -1$
For $n=2$, $n^{2}-2 = 2^{2}-2 = 2$
For $n=3$, $n^{2}-2 = 3^{2}-2 = 7$
For $n=4$, $n^{2}-2 = 4^{2}-2 = 14$
For $n=5$, $n^{2}-2 = 5^{2}-2 = 23$
For $n=6$, $n^{2}-2 = 6^{2}-2 = 34$
For $n=7$, $n^{2}-2 = 7^{2}-2 = 47$
For $n=8$, $n^{2}-2 = 8^{2}-2 = 62$
For $n=9$, $n^{2}-2 = 9^{2}-2 = 79$
Thus, 5 prime numbers of the form $n^{2}-2$ are 2, 7, 23, 47, and 79.
Q4: If p is a prime number and $p|a^{n}$ then show that $p^{n}|a^{n}$, $n\ge1$
Proof: Let p be a prime number such that $p|a^{n}$, where $a^{n} = a \cdot a \cdot a \cdots a$ ($n$ times).
We know that if a prime $p$ divides the product of integers, then $p$ must divide at least one of the factors. [By Corollary 3 of Theorem 9 ]
Therefore, by definition of divisibility, there exists an integer $k$ such that:
Since $k^{n}$ is an integer, this clearly shows that $p^{n}$ divides $a^{n}$.
i.e., $p^{n}|a^{n}$ for $n \ge 1$.
Hence proved.
Q5: For all integers $n\ge2$, show that $n^{4}+4$ is a composite number.
Proof: We can write
For $n \ge 2$, :
- $n^{2}+2n+2 \ge 2^{2}+2(2)+2 = 10 > 1$
- $n^2 – 2n + 2 \geq 2^2 – 2(2) + 2 = 2 > 1$
Since $n^{4}+4$ can be expressed as the product of two integers, each greater than 1.
Therefore, $n^{4}+4$ is a composite number for all integers $n\ge2$.
Hence proved.
Q6: For all integers $n\ge1$ show that $8^{n}+1$ is a composite number.
Proof:
Given $8^{n}+1$
$ = (2^{3})^{n} + 1$
$= (2^{n})^{3} + 1^{3}$
For all integers $n \ge 1$:
- $2^{n}+1 \ge 2^{1}+1 = 3 > 1$
- $2^{2n} – 2^{n} + 1 = 2^{2(1)} – 2^{1} + 1 = 4 – 2 + 1 = 3 > 1$
Since $8^{n}+1$ is expressed as a product of two integers each greater than 1.
Thus, $8^{n}+1$ is a composite number for all integers $n\ge1$.
Hence proved.
Q7: Find the smallest positive integer multiplication by which makes the following number a perfect square.
(i) 3675 Solution:
$3675 = 3^{1} \times 5^{2} \times 7^{2}$
To make $3675$ a perfect square, we have to multiply it by 3.
(ii) 4374
Solution:
$4374 = 2 \times 3 \times 3^{6} = 2^{1} \times 3^{7}$
To make $4374$ a perfect square, we have to multiply it by $2 \times 3 = 6$
(iii) 18375
Solution:
$18375 = 3^{1} \times 5^{3} \times 7^{2}$
To make $18375$ a perfect square, we have to multiply it by $3 \times 5 = 15$
(iv) 74088
Solution:
$74088 = 2^{3} \times 3^{3} \times 7^{3}$
To make $18375$ a perfect square, we have to multiply it by $2 \times 3 \times 7 = 42$
Q8: What is the smallest positive integral multiplier that makes the following number a perfect cube?
(i) 7623
Solution:
Q9: In how many different ways can the number 7056 be expressed as the product of two factors? Find all of these pairs.
Solution:
The prime factorization of 1056 is:
- $(1, 2^5 \times 3 \times 11)$
- $(2, 2^4 \times 3 \times 11)$
- $(2^2, 2^3 \times 3 \times 11)$
- $(2^3, 2^2 \times 3 \times 11)$
- $(2^4, 2 \times 3 \times 11)$
- $(2^5, 3 \times 11)$
- $(3, 2^5 \times 11)$
- $(11, 2^5 \times 3)$
- $(2 \times 3, 2^4 \times 11)$
- $(2^2 \times 3, 2^3 \times 11)$
- $(2^3 \times 3, 2^2 \times 11)$
- $(2^4 \times 3, 2 \times 11)$
Q10: For $n>1$, show that the integers of form $n^{8}+4$ are composite.
Proof:
Given,
For all integers $n > 1$:
- $n^4 + 2n^2 + 2 \ge 2^4 + 2(2)^2 + 2 = 16 + 8 + 2 = 26 > 1$
- $n^4 – 2n^2 + 2 \ge 2^4 – 2(2)^2 + 2 = 16 – 8 + 2 = 10 > 1$
Since $n^{8}+4$ is expressed as a product of two integers, each greater than 1.
Thus, $n^{8}+4$ is a composite number for all integers $n\ge1$.
Hence proved.
Q11: Examine if the integers 701 and 1009 are prime or not.
(i) 701
Solution: If p be the least of all positive divisors of 701 then $p < \sqrt{701}$.
Now,
Thus $26 < \sqrt{701} < 27$
The prime numbers less than 26 are:
Therefore, if 701 has any prime divisor then it must be one of these numbers. But it can be verified by division that none of these primes is a divisor of 701.
Thus, 701 is a prime number.
(ii) 1009
Solution: If p be the least of all positive divisors of 1009 then $p < \sqrt{1009}$.
Now,
Thus $31 < \sqrt{1009} < 32$
The prime numbers less than 32 are:
Therefore, if 1009 has any prime divisor then it must be one of these numbers. But it can be verified by division that none of these primes is a divisor of 1009.
Thus, 1009 is a prime number.