Two dice A and B are rolled. Let the numbers obtained on A and B be and respectively. If the variance of is , where and are co-prime, then the sum of the positive divisors of is equal to [2023]

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Solution

Q.
Two dice A and B are rolled. Let the numbers obtained on A and B be $α$ and $β$ respectively. If the variance of $α - β$ is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to [2023]

1 72

2 48

3 36

4 31

Ans.
(2)

Cases $α - β$ $P$

(6,1) 5 1/36

(6,2), (5,1) 4 2/36

(6,3), (5,2), (4,1) 3 3/36

(6,4), (5,3), (4,2), (3,1) 2 4/36

(6,5), (5,4), (4,3), (3,2), (2,1) 1 5/36

(6,6), (5,5), (4,4), (3,3), (2,2), (1,1) 0 6/36

(1,2), (2,3), (3,4), (4,5), (5,6) −1 5/36

(1,3), (2,4), (3,5), (4,6) −2 4/36

(1,4), (2,5), (3,6) −3 3/36

(1,5), (2,6) −4 2/36

(1,6) −5 1/36

$E (x) = \sum x_{i} P (x_{i}) = 0$

$E (x^{2}) = \sum x_{i}^{2} P (x_{i}) = 2 [\frac{25}{36} + \frac{32}{36} + \frac{27}{36} + \frac{16}{36} + \frac{5}{36}]$

$= 2 [\frac{105}{36}] = \frac{35}{6}$

$Variance ρ^{2} = E (x^{2}) - {[E (x)]}^{2} = \frac{35}{6} = \frac{p}{q} \Rightarrow p = 35 = 7 \times 5$

$∴$ $Sum of divisors = (5^{0} + 5^{1}) (7^{0} + 7^{1}) = 6 \times 8 = 48$

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