Top 10 Most Important Real Numbers Questions with Solutions
Step 1: Apply Euclid’s division lemma to 225 and 135
225 = 135 × 1 + 90
Step 2: Now apply to 135 and remainder 90
135 = 90 × 1 + 45
Step 3: Now apply to 90 and remainder 45
90 = 45 × 2 + 0
Since remainder is 0, the HCF is the last non-zero remainder.
∴ HCF(135, 225) = 45
Let a be any positive integer and b = 6. By Euclid’s division lemma:
a = 6q + r, where 0 ≤ r < 6
Possible remainders (r): 0, 1, 2, 3, 4, 5
For odd integers:
– When r = 1: a = 6q + 1 (odd)
– When r = 3: a = 6q + 3 (odd)
– When r = 5: a = 6q + 5 (odd)
r cannot be 0, 2, or 4 as those would make a even (6q, 6q+2, 6q+4).
Hence proved that any odd integer is of form 6q+1, 6q+3 or 6q+5.
Prime factorizations:
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
LCM = Product of greatest power of all prime factors
HCF = 3 (only common prime factor)
LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
∴ HCF = 3, LCM = 420
Assume √3 is rational. Then it can be written as √3 = a/b where a, b are coprime integers.
Squaring both sides: 3 = a²/b² ⇒ 3b² = a²
This shows a² is divisible by 3 ⇒ a is divisible by 3 (by Fundamental Theorem of Arithmetic).
Let a = 3k. Then 3b² = (3k)² ⇒ 3b² = 9k² ⇒ b² = 3k²
Now b² is divisible by 3 ⇒ b is divisible by 3.
But this contradicts our assumption that a and b are coprime (both divisible by 3).
Hence, √3 is irrational.
For a number to end with 0, its prime factorization must contain both 2 and 5.
6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
The prime factors are only 2 and 3. There is no factor of 5.
∴ 6ⁿ cannot end with digit 0 for any natural number n.
Using the relationship: HCF × LCM = Product of two numbers
9 × LCM = 306 × 657
LCM = (306 × 657) / 9
LCM = 34 × 657 = 22,338
∴ LCM(306, 657) = 22,338
7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78 = 13 × 2 × 3 × 13
The expression has prime factors other than 1 and itself.
By definition, a composite number has more than two distinct prime factors.
Hence, it is a composite number.
Assume 3 + 2√5 is rational. Then it can be written as p/q where p, q are integers.
3 + 2√5 = p/q ⇒ 2√5 = (p/q) – 3 ⇒ √5 = (p – 3q)/(2q)
This implies √5 is rational (since right side is rational).
But we know √5 is irrational. This is a contradiction.
Hence, 3 + 2√5 is irrational.
This problem requires finding the LCM of their time taken.
Prime factorizations:
12 = 2² × 3
18 = 2 × 3²
LCM = 2² × 3² = 4 × 9 = 36 minutes
∴ They will meet after 36 minutes.
Using factor tree method:
156 ÷ 2 = 78
78 ÷ 2 = 39
39 ÷ 3 = 13
156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
∴ Prime factorization: 156 = 2² × 3 × 13