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Given that the divisor n = 3p5q7r of the form 4𝛌+1 , 𝛌≥0 then

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p + r is always even

p + q+r is always odd

q can he any integer

if p is odd then r is even

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Detailed Solution

n = 3^{p} 5^{q} 7^{r}

Hence, any +ve integer power of 5 will be in the form 4𝛌2+1
Even power of 3 and 7 will be in the form of

and odd power of 3 and 7 will be form of

Hence, both p and r must be even
Thus p + r is always even q can be any integral value
Hece a and c are correct option

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