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For all positive integers  k, if the greatest divisor of $25^{k}+12k-1$  is d , then 4$\sqrt{d}$=

Answer:

Explanation:

We have

 $25^{k}+12k-1$ , for all positive integer k

when k=1, $25^{1}+12(1)-1=36$

when k=2 , $25^{2}+12(2)-1$

=625+24-1=648

HCF of 36 and 648 is 36

So, greatest divisor is 36

$\therefore$ d=36

Now. 44 $\sqrt{d}=4\sqrt{36}=4 \times 6=24$