CBSE Class 10 – Mathematics
Chapter 1: Real Numbers
Answers & Solutions (Terminal / copy-paste friendly)

1) Using Euclid’s Division Lemma, find the HCF of 210 and 55, and show that it can be expressed as a linear combination of 210 and 55.
Solution (Euclid’s algorithm):
210 = 55 × 3 + 45 (since 55*3 = 165, 210-165 = 45)
55 = 45 × 1 + 10 (45*1 = 45, 55-45 = 10)
45 = 10 × 4 + 5 (10*4 = 40, 45-40 = 5)
10 = 5 × 2 + 0
HCF = 5

Express as linear combination (back-substitute):
5 = 45 – 10×4
10 = 55 – 45×1 => 5 = 45 – 4*(55 – 45) = 5*45 – 4*55
45 = 210 – 55×3 => 5 = 5*(210 – 55×3) – 4*55 = 5*210 – 15*55 – 4*55 = 5*210 – 19*55

So HCF(210,55) = 5 = 210×5 + 55×(-19).

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2) Prove that 5/√3 is an irrational number.
Proof (contradiction):
Assume 5/√3 is rational. Then √3 = 5 / (rational) which is rational.
But √3 is known to be irrational. Contradiction.
Therefore 5/√3 is irrational.

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3) A sweet seller has 420 kaju barfis and 130 badam barfis. He wants to stack them so each stack has same number and none left. Maximum pieces per stack?
Solution:
Required = HCF(420, 130).
420 = 130 × 3 + 30 (130*3=390, 420-390=30)
130 = 30 × 4 + 10 (30*4=120, 130-120=10)
30 = 10 × 3 + 0
HCF = 10

Maximum number of pieces in each stack = 10.

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4) Show that the square of any positive integer is of the form 3m or 3m+1, where m is an integer.
Proof (mod 3 cases):
Any integer n ≡ 0, 1, or 2 (mod 3).
– If n ≡ 0 (mod 3) then n^2 ≡ 0^2 ≡ 0 (mod 3) ⇒ n^2 = 3m.
– If n ≡ 1 (mod 3) then n^2 ≡ 1^2 ≡ 1 (mod 3) ⇒ n^2 = 3m + 1.
– If n ≡ 2 (mod 3) then n^2 ≡ 2^2 ≡ 4 ≡ 1 (mod 3) ⇒ n^2 = 3m + 1.
Hence every square is of form 3m or 3m+1.

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5) Prove that there are infinitely many prime numbers of the form 6n + 5.
Proof (Euclid-type):
Suppose finitely many primes p1, p2, …, pk are ≡ 5 (mod 6).
Consider N = 6·(p1·p2·…·pk) – 1.
Then N ≡ -1 (mod 6) ⇒ N ≡ 5 (mod 6).
Let q be any prime factor of N. q ≠ 2 and q ≠ 3 (since N ≡ 5 mod 6).
If every prime factor of N were ≡ 1 (mod 6), their product ≡ 1 (mod 6) — but N ≡ 5 (mod 6). Contradiction.
So at least one prime divisor q of N satisfies q ≡ 5 (mod 6).
Also q is not among p1..pk (since N ≡ -1 mod pi for each pi), so q is a new prime ≡ 5 (mod 6).
Hence infinitely many primes of form 6n + 5.

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6) Find the HCF of 96 and 404 by Euclid’s Division Lemma. Hence find integers x, y such that HCF = 96x + 404y.
Euclid’s steps:
404 = 96 × 4 + 20 (96*4=384, 404-384=20)
96 = 20 × 4 + 16 (20*4=80, 96-80=16)
20 = 16 × 1 + 4
16 = 4 × 4 + 0
HCF = 4

Back-substitute:
4 = 20 – 16×1
16 = 96 – 20×4 => 4 = 20 – (96 – 20×4) = 5*20 – 96
20 = 404 – 96×4 => 4 = 5*(404 – 96×4) – 96 = 5*404 – 20*96 – 96 = 5*404 – 21*96

So HCF(96,404) = 4 = 96×(-21) + 404×5.

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7) Show that 7 + 5√2 is an irrational number.
Proof (contradiction):
Suppose 7 + 5√2 is rational. Then 5√2 = (rational) – 7 ⇒ √2 is rational.
But √2 is irrational. Contradiction.
Therefore 7 + 5√2 is irrational.

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8) Prove that √7 is irrational.
Proof (Euclid-style):
Assume √7 = a/b in lowest terms (gcd(a,b)=1).
Then a^2 / b^2 = 7 ⇒ a^2 = 7 b^2.
So a^2 is divisible by 7 ⇒ a is divisible by 7. Let a = 7k.
Substitute: (7k)^2 = 7 b^2 ⇒ 49 k^2 = 7 b^2 ⇒ b^2 = 7 k^2.
Hence b^2 divisible by 7 ⇒ b divisible by 7.
So both a and b divisible by 7, contradicting gcd(a,b)=1.
Therefore √7 is irrational.

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9) Find the HCF of 867 and 255, and express it as a linear combination.
Euclid’s steps:
867 = 255 × 3 + 102 (255*3=765, 867-765=102)
255 = 102 × 2 + 51 (102*2=204, 255-204=51)
102 = 51 × 2 + 0
HCF = 51

Back-substitute to express 51:
51 = 255 – 102×2
102 = 867 – 255×3 => 51 = 255 – 2*(867 – 255×3) = 255 – 2*867 + 6*255 = 7*255 – 2*867

So 51 = 867×(-2) + 255×7.

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10) Show that one and only one of the numbers n, n+2, n+4 is divisible by 3, where n is any positive integer.
Proof (mod 3):
Consider n (mod 3). Three cases:
– If n ≡ 0 (mod 3) then n divisible by 3; n+2 ≡ 2, n+4 ≡ 1 (not divisible).
– If n ≡ 1 (mod 3) then n+2 ≡ 0 (mod 3) so n+2 divisible; n and n+4 not divisible.
– If n ≡ 2 (mod 3) then n+4 ≡ 2+1 ≡ 0 (mod 3) so n+4 divisible; n and n+2 not divisible.
In each case exactly one of {n, n+2, n+4} is divisible by 3. QED.

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End of solutions.