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Given:

(17²⁵ + 19²⁵) ÷ 18

Formula used:

a mod m → reduce base first

Calculation:

⇒ 17 mod 18 = 17

⇒ 19 mod 18 = 1

⇒ 17²⁵ mod 18 = 17²⁵ (17 ≡ −1 mod 18)

⇒ (−1)²⁵ = −1 ⇒ remainder = 17

⇒ 19²⁵ mod 18 = 1²⁵ = 1

⇒ Total remainder = 17 + 1 = 18

⇒ 18 mod 18 = 0

∴ Remainder = 0.

Given:

HCF of x and y = H

p = (x³ + y³) ÷ (x² − xy + y²)

q = (x³ − y³) ÷ (x² + xy + y²)

Formula used:

x³ + y³ = (x + y)(x² − xy + y²)

x³ − y³ = (x − y)(x² + xy + y²)

Calculation:

p = (x + y)(x² − xy + y²) ÷ (x² − xy + y²)

⇒ p = x + y

q = (x − y)(x² + xy + y²) ÷ (x² + xy + y²)

⇒ q = x − y

HCF(p, q) = HCF(x + y, x − y)

Let x = Ha, y = Hb where HCF(a, b) = 1

⇒ p = H(a + b)

⇒ q = H(a − b)

HCF(p, q) = H × HCF(a + b, a − b)

If a and b both odd ⇒ a + b and a − b are even

⇒ HCF(a + b, a − b) = 2

⇒ HCF(p, q) = 2H

If one of a, b is even ⇒ HCF(a + b, a − b) = 1

⇒ HCF(p, q) = H

∴ Both statements I and II are correct.

Given:

x⁴ = x² + 1 , x > 0

Formula used:

Put y = x² ⇒ x⁴ = y²

Calculation:

⇒ y² = y + 1

⇒ y² − y − 1 = 0

⇒ y = (1 + √5) / 2 (positive root)

⇒ x² = (1 + √5)/2

⇒ x⁴ = y² = ((1 + √5)/2)²

⇒ x⁴ = (6 + 2√5)/4 = (3 + √5)/2

⇒ 2x⁴ = 3 + √5

∴ The correct answer is 3 + √5.

Given:

$\dfrac{p+q}{q+r}=\dfrac{r+s}{s+p}$

Formula used:

Cross multiplication: (p+q)(s+p) = (r+s)(q+r)

(A−B)(A+B) = A² − B²

Calculation:

(p + q)(s + p) = (r + s)(q + r)

⇒ ps + p² + qs + pq = rq + r² + sq + sr

⇒ (p² − r²) + (ps − sr) + (pq − rq) = 0

⇒ (p − r)(p + r) + s(p − r) + q(p − r) = 0

⇒ (p − r)(p + r + s + q) = 0

⇒ (p − r) = 0 or (p + r + s + q) = 0

⇒ p = r or (p + r + s + q) = 0.

∴ The correct answer is: Either p + q + r + s = 0 or p = r.

Given:

n ∈ N, n < 7

Find number of n for which gcd(12n + 2, 8n + 1) = 1

Formula used:

Use Euclid: gcd(a, b) = gcd(b, a − q×b)

Calculations:

⇒ gcd(12n + 2, 8n + 1) = gcd(8n + 1, (12n + 2) − (8n + 1))

⇒ gcd(8n + 1, 4n + 1)

⇒ gcd(4n + 1, (8n + 1) − 2×(4n + 1))

⇒ gcd(4n + 1, −1) = 1

⇒ gcd(12n + 2, 8n + 1) = 1 for every n

n values = 1, 2, 3, 4, 5, 6 ⇒ count = 6

∴ Number of values = 6.

Given:

f = x³ + y³ + 3xy − 1

g = (x − y)⁴ − 1

Formula used:

HCF(f, g) = greatest common divisor of polynomials. Factor both and compare common factors.

Calculation:

⇒ Factor g: (x + y)⁴ − 1 = (x + y − 1)(x + y + 1)(x² + 2xy + y² + 1)

⇒ Factor f: x³ + y³ + 3xy − 1 = (x + y − 1)(x² − x y + x + y² + y + 1)

⇒ Common factor: (x + y - 1)

∴ HCF = (x + y - 1).

Given:

x = n(n + 1)(n + 2)

n is an even natural number

Formula used:

Three consecutive numbers always include:

• one multiple of 3

• at least two even numbers if n is even

Calculation:

Let n = 2k

⇒ x = 2k(2k + 1)(2k + 2)

⇒ x = 2k(2k + 1)2(k + 1)

⇒ x = 4k(k + 1)(2k + 1)

From k and (k + 1), one is even 

⇒ x has factor 8

Among k, (k + 1), (2k + 1) one is divisible by 3 

⇒ x is always divisible by 24

Check Statement I:

48 = 16 × 3

x has only 8 × 3

Example: n = 2 ⇒ x = 2 × 3 × 4 = 24

⇒ Not divisible by 48 

Statement I is false

Check Statement II:

x is divisible by 24

⇒ x² is divisible by 24² = 576

144 = 16 × 9 ⊂ 576

⇒ x² is always divisible by 144 

∴ Only Statement II is correct.

Correct option: II only.

Given:

Find LCM of A = x⁴ + x²y² + y⁴, B = x³y + y⁴, C = x⁴y² − x³y³

Formula used:

LCM = product of distinct irreducible factors with highest powers present in A, B, C

Calculations:

⇒ A = x⁴ + x²y² + y⁴ = (x² + xy + y²)(x² − xy + y²)

⇒ B = x³y + y⁴ = y(x³ + y³) = y(x + y)(x² − xy + y²)

⇒ C = x⁴y² − x³y³ = x³y²(x − y)

Highest powers needed: x³ (from C), y² (from C), factors (x − y), (x + y), (x² + xy + y²), (x² − xy + y²)

⇒ Combine: LCM = x³ y² (x − y)(x + y)(x² + xy + y²)(x² − xy + y²)

⇒ (x − y)(x + y)(x² + xy + y²)(x² − xy + y²) = x⁶ − y⁶

⇒ LCM = x³ y² (x⁶ − y⁶)

The correct answer is Option (2).

Given:

Let XYZ be a 3-digit number. Let D be the difference between XYZ and ZYX.

We need to find the remainder when D is divided by 99.

Calculation:

Let XYZ = 100X + 10Y + Z and ZYX = 100Z + 10Y + X

D = (100X + 10Y + Z) - (100Z + 10Y + X)

⇒ D = 100X - X + 10Y - 10Y + Z - 100Z

⇒ D = 99X - 99Z

⇒ D = 99(X - Z)

When D is divided by 99:

⇒ Remainder = 0

∴ The correct answer is option (1).

Given:

p, q are natural numbers

(p + q)^p+q is divisible by 512

Formula used:

512 = 2⁹

If a = 2^k × odd, then aⁿ contains 2^k×n

Calculation:

Let s = p + q

Then (p + q)^p+q = s^s

Check options:

s = 4

4 = 2²

⇒ s^s = 2^2×4 = 2⁸ < 2⁹ 

s = 6

6 = 2 × 3

⇒ s^s contains 2⁶ < 2⁹ 

s = 8

8 = 2³

⇒ s^s = 2^3×8 = 2²⁴ ≥ 2⁹ 

s = 12

12 > 8 but not least

∴ The least value of (p + q) is 8.