Solution plz

Let

x = 2l + 1, y = 2m + 1, z = 2n + 1

Where l ≥ 0, m ≥ 1, n ≥ 2

Given that y - x = 2, z - y = 2

Put

m - l = 1, n - m = 1

x + z = 4m + 2

So m ≥ 1

x + z can be 6, 10, 14, 18, 22, .....

So not divisible by 4

A

aaaaaa is divisible by 13

Till 42nd digit it is divisible by 13.

After 44th digit(let say x) remaining digits = 36(multiple of 6) means they are also divisible by 13.

So 2x(43rd and 44th digit) must be divisible by 13.

26 is divisible by 13.

So x = 6

@Railway_employee18_Pradeep_ku @S_nEha_1 @pandamic2020 @ChiragNagpal

6kx² + 12kx - 24x + 16 = (ax + b)²

Compare

a² = 6k, b² = 16

ab = 6(k - 2)

Square and put values

3k² - 20k + 12 = 0

k = 6, 2/3

Please iska batado ans iska nhi samajh ara

Complex Number 🙆‍♂

Puma

(x² + 13x + 42)(x² + 13x + 40) = 840

Let x² + 13x = p

(p + 42)(p + 40) = 840

p² + 82p + 840 = 0

(p + 12)(p + 70) = 0

p = -12, -70

x² + 13x + 12 = 0

x = -1, -12

x² + 13x + 70 = 0

D < 0

No real solutions.

C