"How many integers exist such that not only are they multiples of 904 to the Power 2008 but also are factors of 904 to the power 2015 ?"
Answer (1)
12th Mar, 2018
Hi!
lets understand with first term. 904^2015 is a factor of itself and you need to multiply 904^2008 by 904^7 to achieve that. Similarly you can do it for the second term and hence you'll be able to get it for the nth term. Then you can finally solve it considering it as an infinite series and get 'n' in the end
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