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
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Solutions to Exercise 3.4
Question 1: 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:
Question 2: 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.
Question 3: 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.
Question 5: 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.
Question 6: 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:
Question 9: 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:
- $(1, 2^5 \times 3 \times 11)$
- $(2, 2^4 \times 3 \times 11)$
- $(2^2, 2^3 \times 3 \times 11)$
- $(2^3, 2^2 \times 3 \times 11)$
- $(2^4, 2 \times 3 \times 11)$
- $(2^5, 3 \times 11)$
- $(3, 2^5 \times 11)$
- $(11, 2^5 \times 3)$
- $(2 \times 3, 2^4 \times 11)$
- $(2^2 \times 3, 2^3 \times 11)$
- $(2^3 \times 3, 2^2 \times 11)$
- $(2^4 \times 3, 2 \times 11)$
Question 10: 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.
Question 11: 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.