Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Given m, which one of the following is necessarily true?
Explanation:
a + b + c + d = 4m + 1
a2 + b2 = (a + b)2 – 2ab
a2 + b2 is minimum when 2ab is maximum.
The product of two numbers is maximum when the numbers are equal.
∴ a² + b² is minimum when a = b
Similarly, c² + d² is minimum when c = d
∴ a² + b² + c² + d² is minimum when a = b and c = d
∴ (a² + b² + c² + d²)min = 2(a² + c²)
But, a² + c² is minimum when a = c
∴ a² + b² + c² + d² is minimum when a = b = c = d
When a = b = c = d, a + b + c + d is a multiple of 4.
But, a + b + c + d = 4m + 1
So, one out of a, b, c, d must be one greater than the other three.
∴ a = b = c = m and d = m + 1
∴ a² + b² + c² + d² = m² + m² + m² + (m + 1)² = 4m² + 2m + 1
Hence, option (b).
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.