CBSE Class 10 Maths Advanced Real Numbers Long Answer Questions Set 01

Long Answer Questions

Question 1. Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.
Answer:
Let n be a positive integer. By Euclid's division lemma n = 4q + r with r = 0, 1, 2 or 3.
r = 0: n³ = 64q³ = 4(16q³) = 4m.
r = 1: n³ = 64q³ + 48q² + 12q + 1 = 4(16q³ + 12q² + 3q) + 1 = 4m + 1.
r = 2: n³ = 64q³ + 96q² + 48q + 8 = 4(16q³ + 24q² + 12q + 2) = 4m.
r = 3: n³ = 64q³ + 144q² + 108q + 27 = 4(16q³ + 36q² + 27q + 6) + 3 = 4m + 3.
So every cube is of the form 4m, 4m + 1 or 4m + 3.
In simple words: Cube the four possible forms 4q, 4q + 1, 4q + 2, 4q + 3 and pull out a 4 each time.

Teacher's Note:
1. A cube can never leave remainder 2 on division by 4.
2. Test: 3³ = 27 = 4 × 6 + 3 and 5³ = 125 = 4 × 31 + 1.

Question 2. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.
Answer:
Let n be a positive integer and write n = 5k + r with r = 0, 1, 2, 3 or 4.
r = 0: n² = 25k² = 5(5k²), of the form 5q.
r = 1: n² = 25k² + 10k + 1 = 5(5k² + 2k) + 1, of the form 5q + 1.
r = 2: n² = 25k² + 20k + 4 = 5(5k² + 4k) + 4, of the form 5q + 4.
r = 3: n² = 25k² + 30k + 9 = 5(5k² + 6k + 1) + 4, of the form 5q + 4.
r = 4: n² = 25k² + 40k + 16 = 5(5k² + 8k + 3) + 1, of the form 5q + 1.
A square is always 5q, 5q + 1 or 5q + 4, so it cannot be 5q + 2 or 5q + 3.
In simple words: List all five possible remainders; squares only leave 0, 1 or 4 on division by 5.

Teacher's Note:
1. Last digits of squares (0, 1, 4, 5, 6, 9) tell the same story.
2. Try 7² = 49 = 5 × 9 + 4.

Question 3. If n is an odd integer, then show that n² − 1 is divisible by 8.
Answer:
An odd integer has the form n = 2k + 1 (equivalently 4q + 1 or 4q + 3).
n² − 1 = (2k + 1)² − 1 = 4k² + 4k = 4k(k + 1).
k and k + 1 are consecutive integers, so one of them is even and k(k + 1) = 2t for some integer t.
Hence n² − 1 = 4 × 2t = 8t, which is divisible by 8.
In simple words: For odd n, n² − 1 factorises as 4k(k + 1) and k(k + 1) is always even.

Teacher's Note:
1. Alternative: n² − 1 = (n − 1)(n + 1) is a product of two consecutive even numbers.
2. Check: 7² − 1 = 48 = 8 × 6.

Question 4. Using Euclid's division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.
Answer:
Subtract the remainders: 1251 − 1 = 1250, 9377 − 2 = 9375 and 15628 − 3 = 15625. The required number is H.C.F.(1250, 9375, 15625).
H.C.F.(9375, 1250):
9375 = 1250 × 7 + 625
1250 = 625 × 2 + 0
So H.C.F.(1250, 9375) = 625. Now H.C.F.(15625, 625): 15625 = 625 × 25 + 0
The H.C.F. of all three is 625, which exceeds the remainders 1, 2 and 3.
Answer: 625.
In simple words: Remove the remainders, then find the HCF of the three reduced numbers.

Teacher's Note:
1. Find the HCF of two numbers first and then of that result and the third.
2. The divisor 625 is larger than every remainder, as required.

Question 5. Prove that \( \sqrt{7} \) is irrational.
Answer:
Suppose, to the contrary, that \( \sqrt{7} \) is rational. Then \( \sqrt{7} = \frac{a}{b} \) where a and b are co-prime positive integers.
Squaring: \( a^2 = 7b^2 \), so 7 divides a². Since 7 is prime, it divides a. Write a = 7c.
Then \( 7^2c^2 = 7b^2 \), so \( b^2 = 7c^2 \). Now 7 divides b², hence divides b.
So 7 is a common factor of a and b, contradicting that they are co-prime. Therefore \( \sqrt{7} \) is irrational.
In simple words: If √7 were a fraction in lowest terms, both parts would be multiples of 7, which is impossible.

Teacher's Note:
1. The whole proof rests on: if a prime divides a², it divides a.
2. Contradiction proofs begin with 'suppose it is rational'.

Question 6. Show that 12ⁿ cannot end with the digit 0 or 5 for any natural number n.
Answer:
12ⁿ = (2² × 3)ⁿ = 2²ⁿ × 3ⁿ, so its only prime factors are 2 and 3.
A number ending in 0 or 5 is divisible by 5, so 5 would have to appear in its prime factorisation.
By the uniqueness of prime factorisation, 5 is not a factor of 12ⁿ. Hence 12ⁿ never ends with 0 or 5.
In simple words: Numbers ending in 0 or 5 are multiples of 5, but 12ⁿ has only the prime factors 2 and 3.

Teacher's Note:
1. Both 0 and 5 as last digit mean divisibility by 5.
2. 12ⁿ ends in 2, 4, 8 or 6.

Question 7. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
Answer:
The distance must be a multiple of each step length, so the smallest such distance is the L.C.M. of 40, 42 and 45.
40 = 2³ × 5, 42 = 2 × 3 × 7 and 45 = 3² × 5.
L.C.M. = 2³ × 3² × 5 × 7 = 2520.
The minimum distance is 2520 cm (25.2 m).
In simple words: The smallest common distance in complete steps is the LCM of the three step lengths.

Teacher's Note:
1. Take the highest power of each prime that appears.
2. They take 63, 60 and 56 steps respectively.

Question 8. Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.
Answer:
(6q + r)³ = 216q³ + 108q²r + 18qr² + r³ = 6(36q³ + 18q²r + 3qr²) + r³.
It is enough to show that r³ − r is divisible by 6 for r = 0, 1, ..., 5, because then r³ = 6k + r for an integer k and the cube equals 6(36q³ + 18q²r + 3qr² + k) + r.
Compute r³ − r: for r = 0, 1, 2, 3, 4, 5 it is 0, 0, 6, 24, 60, 120, all multiples of 6.
So (6q + r)³ = 6m + r with m = 36q³ + 18q²r + 3qr² + (r³ − r)/6. The cube has the form 6m + r.
In simple words: Expand the cube and check that r³ − r is a multiple of 6 for each of the six values of r.

Teacher's Note:
1. r³ − r = (r − 1)r(r + 1) is a product of three consecutive numbers.
2. Example: 5³ = 125 = 6 × 20 + 5.

Question 9. Prove that one of any three consecutive positive integers must be divisible by 3.
Answer:
Let the three consecutive integers be n, n + 1, n + 2, and write n = 3q + r with r = 0, 1 or 2.
r = 0: n = 3q is divisible by 3.
r = 1: n + 2 = 3q + 3 = 3(q + 1) is divisible by 3.
r = 2: n + 1 = 3q + 3 = 3(q + 1) is divisible by 3.
So one of the three numbers is always a multiple of 3.
In simple words: Whatever the remainder of n on division by 3, one of n, n + 1, n + 2 makes the remainder 0.

Teacher's Note:
1. Split into the three cases r = 0, 1, 2.
2. Try 7, 8, 9: only 9 works.

Teacher's Note:
1. The remainders r, r + 4, r + 3, r + 2, r + 1 cover 0 to 4 once each (mod 5).
2. Test n = 1: 1, 5, 9, 13, 17; only 5.

Question 11. Prove that \( 3+\sqrt2 \) is an irrational number. OR Prove that \( 5\sqrt2 \) is irrational number. [CBSE 10th Board 2013, 3 Marks]
Answer:
First part: 3 + √2. Suppose it is a rational number r. Then \( \sqrt2 = r-3 \), which is rational. This contradicts the irrationality of \( \sqrt{2} \), so \( 3+\sqrt2 \) is irrational.
Second part: 5√2. Suppose \( 5\sqrt2 = \frac{a}{b} \) with integers a, b (b ≠ 0). Then \( \sqrt2 = \frac{a}{5b} \) is rational, a contradiction. So \( 5\sqrt2 \) is irrational.
In simple words: If either number were rational, √2 would turn out rational, which is false.

Teacher's Note:
1. Use the already proved fact that √2 is irrational.
2. Attempt either option, since the question allows a choice.

Question 12. Show that 5ⁿ can't end with the digit 2 for any natural number n. [CBSE 10th Board 2013, 3 Marks]
Answer:
A number ending in 2 is even, so 2 would be a prime factor of it.
5ⁿ has only one prime factor, namely 5 (this is the only factorisation, by the Fundamental Theorem of Arithmetic).
So 2 is not a factor of 5ⁿ, and 5ⁿ cannot end with the digit 2.
In simple words: Numbers ending in 2 are even, but 5ⁿ is a product of 5s only and so is always odd.

Teacher's Note:
1. In fact 5ⁿ always ends in 5.
2. Use prime factors instead of listing last digits.

Question 13. Check whether 6ⁿ can end with the digit 0 for any natural number n. [CBSE 10th Board 2014, 3 Marks]
Answer:
If 6ⁿ ended with 0 it would be divisible by 5, so 5 would be a prime factor of 6ⁿ.
But 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ, whose only prime factors are 2 and 3.
By the uniqueness of prime factorisation, 5 cannot be a factor. Hence 6ⁿ never ends with the digit 0.
In simple words: A number ending in 0 has 5 as a factor, but 6ⁿ has only the primes 2 and 3.

Teacher's Note:
1. The last digit of every power of 6 is 6.
2. Same method as for 4ⁿ and 12ⁿ.

Question 14. Prove that \( \sqrt3+\sqrt5 \) is an irrational number. OR Prove that \( 5+\sqrt2 \) is an irrational number. [CBSE 10th Board 2014, 3 Marks]
Answer:
First part: √3 + √5. Suppose \( \sqrt3+\sqrt5 = r \) is rational (r ≠ 0). Then \( \sqrt5 = r-\sqrt3 \). Squaring: \( 5 = r^2-2r\sqrt3+3 \), so \( \sqrt3 = \frac{r^2-2}{2r} \), a rational number. This contradicts the irrationality of \( \sqrt{3} \). Hence \( \sqrt3+\sqrt5 \) is irrational.
Second part: 5 + √2. Suppose it equals a rational number r. Then \( \sqrt2 = r-5 \) is rational, which is false. Hence \( 5+\sqrt2 \) is irrational.
In simple words: Assuming the sum is rational lets us write one of the roots as a fraction, which cannot be.

Teacher's Note:
1. For two roots, isolate one root and square.
2. For a number plus a root, isolate the root directly.

Question 15. Use Euclid's division algorithm to find the HCF of 10224 and 9648. [CBSE 10th Board 2015, 3 Marks]
Answer:
10224 = 9648 × 1 + 576
9648 = 576 × 16 + 432
576 = 432 × 1 + 144
432 = 144 × 3 + 0
The remainder becomes 0 at the last step, and the last divisor is 144.
So H.C.F.(10224, 9648) = 144.
In simple words: Divide repeatedly, each time using the previous divisor and remainder, until the remainder is zero.

Teacher's Note:
1. The last non-zero remainder, 144, is the HCF.
2. Check: 10224 = 144 × 71 and 9648 = 144 × 67.

Question 16. Prove that \( \sqrt{7} \) is an irrational number. OR Prove that \( 3+\sqrt5 \) is an irrational number. [CBSE 10th Board 2015, 3 Marks]
Answer:
First part: √7.
Suppose, to the contrary, that \( \sqrt{7} \) is rational. Then \( \sqrt{7} = \frac{a}{b} \) where a and b are co-prime positive integers.
Squaring: \( a^2 = 7b^2 \), so 7 divides a². Since 7 is prime, it divides a. Write a = 7c.
Then \( 7^2c^2 = 7b^2 \), so \( b^2 = 7c^2 \). Now 7 divides b², hence divides b.
So 7 is a common factor of a and b, contradicting that they are co-prime. Therefore \( \sqrt{7} \) is irrational.
Second part: 3 + √5. Suppose it equals a rational number r. Then \( \sqrt5 = r-3 \) would be rational, contradicting the irrationality of \( \sqrt{5} \). Hence \( 3+\sqrt5 \) is irrational.
In simple words: Either proof ends by showing that a number known to be irrational would be rational.

Teacher's Note:
1. For √7 use the divisibility argument with the prime 7.
2. For 3 + √5, isolate √5.

Question 17. Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [CBSE 10th Board 2015, 3 Marks]
Answer:
By the division lemma, any positive integer x can be written as 3q, 3q + 1 or 3q + 2.
x = 3q: x² = 9q² = 3(3q²), which is 3m.
x = 3q + 1: x² = 9q² + 6q + 1 = 3(3q² + 2q) + 1, which is 3m + 1.
x = 3q + 2: x² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1, which is 3m + 1.
So a square is of the form 3m or 3m + 1.
In simple words: On division by 3 a square can leave only remainder 0 or 1.

Teacher's Note:
1. Write the three possible forms of x and square each.
2. No square equals 3m + 2: for example 2² = 4 = 3 + 1.

Question 18. Prove that \( \sqrt{5} \) is an irrational number. OR Prove that \( 5+3\sqrt2 \) is an irrational number. [CBSE 10th Board 2016, 3 Marks]
Answer:
First part: √5.
Suppose, to the contrary, that \( \sqrt{5} \) is rational. Then \( \sqrt{5} = \frac{a}{b} \) where a and b are co-prime positive integers.
Squaring: \( a^2 = 5b^2 \), so 5 divides a². Since 5 is prime, it divides a. Write a = 5c.
Then \( 5^2c^2 = 5b^2 \), so \( b^2 = 5c^2 \). Now 5 divides b², hence divides b.
So 5 is a common factor of a and b, contradicting that they are co-prime. Therefore \( \sqrt{5} \) is irrational.
Second part: 5 + 3√2. Suppose it equals a rational number r. Then \( 3\sqrt2 = r-5 \), so \( \sqrt2 = \frac{r-5}{3} \) is rational, a contradiction. Hence \( 5+3\sqrt2 \) is irrational.
In simple words: Assuming a rational value always leads back to a rational √5 or √2, which is false.

Teacher's Note:
1. Choose whichever alternative you find easier.
2. Rational combinations of an irrational number remain irrational.

Question 19. Prove that for any positive integer n, n³ − n is divisible by 6. [CBSE 10th Board 2016, 3 Marks]
Answer:
Factorise: n³ − n = n(n² − 1) = (n − 1)n(n + 1), a product of three consecutive integers.
Among any two consecutive integers one is even, so the product is divisible by 2.
Among any three consecutive integers one is a multiple of 3 (write n = 3q + r with r = 0, 1, 2 and check each case), so the product is divisible by 3.
2 and 3 are co-prime, so the product is divisible by 2 × 3 = 6. Hence n³ − n is divisible by 6.
In simple words: n³ − n is (n − 1) × n × (n + 1), which always contains a multiple of 2 and a multiple of 3.

Teacher's Note:
1. Test n = 4: 64 − 4 = 60 = 6 × 10.
2. If a number is divisible by two co-prime numbers, it is divisible by their product.

Question 20. Prove that \( \frac73\sqrt5 \) is irrational number. OR Prove that \( 5-2\sqrt3 \) is an irrational number. [CBSE 10th Board 2017, 3 Marks]
Answer:
First part: (7/3)√5. Suppose \( \frac73\sqrt5 = r \) is rational. Then \( \sqrt5 = \frac{3r}{7} \) is rational, which contradicts the irrationality of \( \sqrt{5} \). Hence \( \frac73\sqrt5 \) is irrational.
Second part: 5 − 2√3. Suppose it equals a rational number r. Then \( 2\sqrt3 = 5-r \) and \( \sqrt3 = \frac{5-r}{2} \) would be rational, contradicting the irrationality of \( \sqrt{3} \). Hence \( 5-2\sqrt3 \) is irrational.
In simple words: If either number were rational, a number known to be irrational (√5 or √3) would be rational.

Teacher's Note:
1. Move all rational parts to one side.
2. The coefficient 7/3 is a non-zero rational, so it can be divided away.

Question 21. Prove that n² − n is divisible by 2 for any positive integer n. [CBSE 10th Board 2017, 3 Marks]
Answer:
n² − n = n(n − 1), a product of two consecutive integers.
Case 1: n is even, so n(n − 1) has the even factor n.
Case 2: n is odd, so n − 1 is even and n(n − 1) has an even factor.
In both cases the product is even. Hence n² − n is divisible by 2.
In simple words: Of two consecutive integers n − 1 and n, one is always even.

Teacher's Note:
1. Alternatively write n = 2q or n = 2q + 1 and expand.
2. Try n = 5: 25 − 5 = 20.

Question 22. Prove that \( \sqrt2+\sqrt3 \) is irrational.
Answer:
Suppose \( \sqrt2+\sqrt3 = r \) is rational (r ≠ 0). Then \( \sqrt3 = r-\sqrt2 \).
Squaring both sides: \( 3 = r^2-2r\sqrt2+2 \), so \( \sqrt2 = \frac{r^2-1}{2r} \).
The right side is rational, so \( \sqrt{2} \) would be rational. But \( \sqrt{2} \) is irrational (Example 14). This contradiction shows that \( \sqrt2+\sqrt3 \) is irrational.
In simple words: Isolate one root, square, and the other root comes out as a fraction, which is impossible.

Teacher's Note:
1. Same steps as in the solved example for √2 + √5.
2. Remember r ≠ 0 because √2 + √3 is positive.