The number of values of a ∈ N such that the variance of 3, 7, 12, a, 43 - a is a natural number is :

Solution

Maths sir · 4 steps

The mean does not depend on a, so test the variance's numerator for divisibility by 5.

1. Find the mean. The sum is 3 + 7 + 12 + a + (43 - a) = 65, so the mean is 13 for every value of a.
2. Square the deviations. The deviations from 13 are -10, -6, -1, a - 13, 30 - a. Their squares add up to 100 + 36 + 1 + (a-13)² + (30-a)² = 2a² - 86a + 1206.

   σ² = (2a² - 86a + 1206)/5

3. Work mod 5. Since 86 ≡ 1 and 1206 ≡ 1 (mod 5), the numerator leaves the same remainder as 2a² - a + 1. Only the remainder of a on division by 5 matters, so five cases cover every a.

   a mod 5	2a² - a + 1	Remainder mod 5
   0	1	1
   1	2	2
   2	7	2
   3	16	1
   4	29	4

4. So. The remainder is never 0, so 5 never divides the numerator and the variance is never an integer. No natural number a works, and the answer is 0, option (a).

The idea to keep. When a question asks for which integers an expression is an integer, reduce it modulo the denominator. Checking every possible remainder settles all cases at once.

Answer from the NTA final answer key. Solution by PrepWiser.