Chapter 1 - Real Numbers, RD Sharma Solutions - (Part - 11) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 1.61

Q.16. If the least prime factors of two positive integers a and b are 5 and 13 respectively, then the least prime factor of a + b, is _______.
Ans. It is given that the least prime factors of two positive integers a and b are 5 and 13 respectively.
Thus, a = 5p and b = 13q, where p and q are some natural numbers.
Now, a + b = 5p + 13q
Since, 5p and 13q both are odd numbers and sum of two odd numbers is even.
Therefore, a + b is an even number.
Hence, the least prime factor of a + b, is 2.

Q.17. If 2³ × 3^a × b × 7 is the prime factorisation of 2520, then 5a + 2b = ______.
Ans. Prime factorisation of 2520 is
2520 = 2³ × 3² × 5¹ × 7¹ 
Therefore, a = 2 and b = 5
5a + 2b = 5(2) + 2(5)
= 10 + 10
= 20
Hence, 5a + 2b = 20.

Q.18. Given that LCM (91, 26) = 182, then HCF (91, 26) = ______.
Ans. We know that,
Product of two numbers = LCM × HCF
⇒ 91 × 26 = 182 × HCF
⇒ HCF × 182 = 91 × 26
⇒ HCF = 91 × 26/182
⇒ HCF = 26/2
⇒ HCF = 13
Hence, HCF (91, 26) = 13.

Q.19. The decimal expansion of 17 / 8 will terminate after _______ places of decimal.
Ans. 17/8 = 17/2³
= 17 × 5 × 5 × 5 / 2³ × 5³
= 85 × 5 × 5/10³
= 425 × 5/1000
= 2125/1000
= 2.125
Hence, the decimal expansion of 17/8 will terminate after three places of decimal.

Q.20. One of the three consecutive positive integers is always divisible by ________.
Ans. Let the three consecutive numbers be n, n + 1 and n + 2.
Whenever a number is divided by 3, the remainder obtained is 0, 1, or 2. Therefore, n = 3p or n = 3p + 1 or n = 3p + 2, where p is any natural number.
Now,
If n = 3p, then n is divisible by 3
If n = 3p + 1
⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1), then (n + 2) is divisible by 3
If n = 3p + 2 ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1), then (n + 1) is divisible by 3
Thus, we can say that one of the three numbers n, n + 1, n + 2 is always divisible by 3.
Hence, One of the three consecutive positive integers is always divisible by 3.

Q.21. 6ⁿ cannot end with digit 5 for _______ value of n.
Ans. The prime factors of 6 are 2 and 3.
⇒ 6ⁿ = 2ⁿ × 3ⁿ 
Since, 5 is not the factor of 6ⁿ 
Therefore, for any value of n, 6ⁿ will not be divisible by 5.
Hence, 6ⁿ cannot end with digit 5 for any value of n.

Q.22. The value of the remainder, r, when a positive integer 'a' is divided by 3, are ________.
Ans. Let a positive integer 'a' is divided by 3,
a = 3q + r, where q and r both are natural numbers and 0 ≤ r < 3.
Since, r is a natural number and 0 ≤ r < 3
Therefore, r can be 0, 1 or 2 Hence, the value of the remainder, r, when a positive integer 'a' is divided by 3, are 0, 1, 2.

Q.23. 12ⁿ cannot end with the digit 0 to 5 for _______ value of n.
Ans. The prime factors of 12 are 2 and 3.
12 = 2² × 3
⇒ 12ⁿ = 22ⁿ × 3ⁿ 
Since, 5 is not the factor of 12ⁿ 
Therefore, for any value of n, 12ⁿ will not be divisible by 5.
Hence, 12ⁿ cannot end with the digit 0 or 5 for any value of n.
Disclaimer: The question is: 12ⁿ cannot end with the digit 0 or 5 for _______ value of n.

Q.24. Two numbers are in the ratio 3 : 4 and their LCM is 120. The sum of the numbers is ______.
Ans. Let the two numbers be 3x and 4x.
HCF of (3x, 4x) =  x
We know that,
Product of two numbers = LCM × HCF
⇒ 3x × 4x = 120 × x
⇒ 12x² = 120x
⇒ 12x = 120
⇒ x = 120/12
⇒ x = 10
Therefore, first number is 3 × 10 = 30 and second number is 4×10 = 40.
Sum of the numbers = 30 + 40 = 70.
Hence, the sum of the numbers is 70.

Q.25. The LCM of 2x, 5x and 7x is ______, where x is a positive integer.
Ans. LCM of 2x, 5x and 7x = 2 × 5 × 7 × x
= 70x
Hence, the LCM of 2x, 5x and 7x is 70x.

Q.1. State Euclid's division lemma.
Ans. Euclid’s Division Lemma:
Let a and b be any two positive integers.
Then, there exist unique integers q and r such that
a = bq + r, 0 ≤ r < b
If b/a then r = 0.
Otherwise, r satisfies the stronger inequality 0 < r < b.

Q.2. State Fundamental Theorem of Arithmetic.
Ans. FUNDAMENTAL THEOREM OF ARITHMETIC:
Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique except for the order in which the prime factors occur.
While writing a positive integer as the product of primes, if we decide to write the prime factors in ascending order and we combine the same primes, then the integer is expressed as the product of powers of primes and the representation is unique.
So, we can say that every composite number can be expressed as the products of powers distinct primes in ascending or descending order in a unique way.

Q.3. Write 98 as product of its prime factors.
Ans. Using the factor tree for prime factorization, we have:
Therefore,
98 = 2 x 7 x 7
98 = 2 x 7²

Q.4. Write the exponent of 2 in the price factorization of 144.
Ans. Using the factor tree for prime factorization, we have:

Therefore, 144 = 2 x 2 x 2 x 2 x 3 x 3
144 = 24 x 32
Hence the exponent of 2 in 144 is 4.

Q.5. Write the sum of the exponents of prime factors in the prime factorization of 98.
Ans. Using the factor tree for prime factorization, we have:
Therefore,
98 = 2 x 7 x 7
98 = 2 x 7²
The exponents of 2 and 7 are 1 and 2 respectively.
Hence the sum of the exponents is 3.

Page No 1.62

Q.6. If the prime factorization of a natural number n is 2³ ✕ 3² ✕ 5² ✕ 7, write the number of consecutive zeros in n.
Ans. Since, it is given that
n = 2³ x 3² x 5² x 7
n = 8 x 9 x 25 x 7
n = (7 x 9 x 2) x (4 x 25)
n = (7 x 9 x 2) x 100
n = 12600
Hence the number of consecutive zeroes are 2.

Q.7. If the product of two numbers is 1080 and their HCF is 30, find their LCM.
Ans. It is given that the product of two numbers is 1080.
Let the two numbers be a and b.
Therefore,
a x b = 1080
HCF is 30.
We need to find the LCM
We know that the product of two numbers is equal to the product of the HCF and LCM.
Thus,
LCM = a x b/HCF
LCM = 1080/30
LCM = 36
Hence the LCM is 36.

Q.8. Write the condition to be satisfied by q so that a rational number p / q has a terminating decimal expansions.
Ans. We need to find the condition to be satisfied by q so that a rational number p / q has a terminating decimal expression.
For the terminating decimal expression, we should have a multiple of 10 in the denominator.
Hence, the prime factorization of q must be of the form 2^m x 5ⁿ, where m and n are non-negative integers.

Q.9. Write the condition to be satisfied by q so that a rational number p / q has a terminating decimal expansion.
Ans. We need to find the condition to be satisfied by q so that a rational number p / q has a non-terminating decimal expression.
For the terminating decimal expression, we should not have a multiple of 10 in the denominator.
Hence, the prime factorization of q must not be of the form 2^m x 5ⁿ, where m and n are non-negative integers.

Q.10. Complete the missing entries in the following factor tree.
Ans. We need to fill the values for a and b in the following factor tree:
It is clear from the factor tree above that
b = 3 x 7
b = 21
Also,
a = 2 x b
a = 2 x 21
a = 42
Thus, the missing entries are 21 and 42.

Q.11. The decimal expansion of the rational number 43/2⁴ × 5³ will terminate after how many places of decimals?
Ans. We have,
43/2⁴ × 5³
Theorem states:
Let x = p / q be a rational number, such that the prime factorization of q is of the form 2^m x 5ⁿ, where m and n are non-negative integers.
Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.
This is clear that the prime factorization of the denominator is of the form 2^m x 5ⁿ.
Hence, it has terminating decimal expansion which terminates after 4 places of decimal.

Q.12. Has the rational number 441/2² × 5⁷ × 7² a terminating or a non-terminating decimal representation?
Ans. We have,
441/2² × 5⁷ × 7²
Theorem states:
Let x = p / q be a rational number, such that the prime factorization of q is not of the form 2^m x 5ⁿ, where m and n are non-negative integers.
Then, x has a decimal expression which is non-terminating repeating.
This is clear that the prime factorization of the denominator is not of the form
2^m x 5ⁿ.
Hence, it has non-terminating decimal expansion.

Q.13. Write whether 2√45 + 3√20 / 2√5 on simplification gives a rational or an irrational number.
Ans. Let us simplify 2√45 + 3√20 / 2√5
2√45 + 3√20 / 2√5 = 6√5 + 6√5 / 2√5
= 12√5 / 2√5
= 6
6 is rational number.

Q.14. What is an algorithm?
Ans. Algorithm is a step-by-step procedure for calculations.
For example:
Euclid’s Division Algorithm: In order to compute the HCF of two positive integers say a and b, with a > b by using Euclid’s algorithm, we follow the following steps:
STEP I : Apply Euclid’s Division Lemma to a and b and obtain whole numbers q₁  and r₁,such that a = bq₁ + r₁, 0 ≤ r₁ < b
STEP II: If r₁ = 0, b is the HCF of a and b.
STEP III: If r₁ ≠ 0, apply Euclid’s division lemma to b and r₁ and obtain whole numbers q₁ and r₂,such that b = q₁r₁ + r₂
STEP IV: If r₂ = 0, then r₁ is the HCF of a and b.
STEP V: If r₂ ≠ 0, then apply Euclid’s division lemma to r₁ and r₂ and continue the above process till the remainder r_n is zero. The divisor at this stage i.e; r_n-1, or the non-zero remainder at the previous stage is the HCF of a and b.

Q.15. What is a lemma?
Ans. A proven statement used as a stepping-stone toward the proof of another statement is called lemma.
For example:
Euclid’s Division Lemma: Let a and b be any two positive integers.
Then, there exist unique integers q and r such that
a = bq + r, 0 ≤ r < b
If b / a then r = 0 .
Otherwise, r satisfies the stronger inequality 0 < r < b.