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

Question

The number of values of $a \in \mathbb{N}$ such that the variance of $3, 7, 12, a, 43-a$ is a natural number is:

Accepted Answer

Given,

5 numbers are 3, 7, 12, a, 43 - a

$$\therefore$$ Mean $$(\overline x ) = {{3 + 7 + 12 + a + 43 - a} \over 5}$$

$$ = {{65} \over 5}$$

$= 13$

We know,

Variance $$({\sigma ^2}) = {{\sum {x_1^2} } \over n} - {(\overline x )^2}$$

$$ = {{{3^2} + {7^2} + {{12}^2} + {a^2} + {{(43 - a)}^2}} \over 5} - {(13)^2}$$

$$ = {{9 + 49 + 144 + {a^2} + 1849 + {a^2} - 86a} \over 5} - 169$$

$$ = {{2{a^2} - 86a + 2051} \over 5} - 169$$

$$ = {{2{a^2} - 86a + 2051 - 845} \over 5}$$

$$ = {{2{a^2} - 86a + 1206} \over 5}$$

$$ = {2 \over 5}({a^2} - 43a + 603)$$

Variance will be natural number if ${a^2} - 43a + 603$ in multiple of 5.

$$\therefore$$ ${a^2} - 43a + 603 = 5n$

$$ \Rightarrow {a^2} - 43a + 603 - 5n = 0$$ ..... (1)

$$\therefore$$ $$a = {{43\, \pm \,\sqrt D } \over 2}$$

Now "a" will be natural number if

(1) D is a perfect square and

(2) $$43\, \pm \,\sqrt D $$ is multiple of 2

From equation (1),

$$D = {( - 43)^2} - 4\,.\,1\,.\,(603 - 5n)$$

$= 1849 - 2412 + 20n$

$= 20n - 563$

For any value of n, unit digit of 20n is always 0. Then 20n - 563 will give a number whose unit digit is 7.

For perfect square numbers ex : 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, 102 = 100

So, unit digit is either 1, 4, 6, 5, 6, 9, 0 it can't be 7. So D can't be perfect square.

Solutions · 1