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If \mathrm{r, s, t} are prime numbers and  \mathrm{p, q} are natural numbers such that LCM of \mathrm{p, q} is \mathrm{r^{2} s^{4} t^{2}}, then the number of ordered pairs \mathrm{(p, q)} is 

Option: 1

252


Option: 2

254


Option: 3

225


Option: 4

224


Answers (1)

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Numbers \mathrm{p} and \mathrm{q} must be of the form

\mathrm{\[ p=r^{a} s^{b} t^{c}, \quad q=r^{\alpha} s^{\beta} t^{\gamma} \]}

where           \quad 0 \leq a, \alpha \leq 2 \text{and at least one of a, }\alpha \, is\, 2
                     0 \leq b, \beta \leq 4 \text{\, and at least one of b,} \beta\, is \, 4
                      0 \leq c, \gamma \leq 2 \text{\, and at least one of c, }\gamma\, is\, 2

 Possible values of (a, \alpha)\: , and \: (c, \gamma)\: are
(0,2),(1,2),(2,2),(2,0),(2,1).

Possible values of (b, \beta)\: are
(0,4),(1,4),(2,4),(3,4),(4,4),(4,0),(4,1),(4,2),(4,3).

Thus, number of possible order pairs (p, q)$ is $5 \times 9 \times 5= 225.

Posted by

Rakesh

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