Option (c) is correct

Let the three consecutive odd numbers be – (x), (x + 2), (x + 4)

So,

(x)² + (x + 2) ² + (x + 4) ² = 2531

x² + x² + 4x + 4 + x² + 8x + 16 = 2531

3x² + 12x – 2511 = 0

x² + 4x – 837 = 0

x²+ (31 – 27) x – 837 = 0

(x + 31) (x – 27) = 0

x = –31, x = 27

Since numbers are positive

So, 27, 29, 31 are the number in which two (29, 31) are prime.

विकल्प (c) सही है

मान लीजिए कि तीन क्रमागत विषम संख्याएँ

=(x), (x + 2), (x + 4) हैं

इसलिए,

(x)² + (x + 2) ² + (x + 4) ² = 2531

x² + x² + 4x + 4 + x² + 8x + 16 = 2531

3x² + 12x – 2511 = 0

x² + 4x – 837 = 0

x²+ (31 – 27) x – 837 = 0

(x + 31) (x – 27) = 0

x = –31, x = 27

चूंकि संख्याएँ धनात्मक हैं

अतः, 27, 29, 31 वे संख्याएँ हैं जिनमें दो (29, 31) अभाज्य हैं।