(2^2007+5)^2=225N find sum of digits in no. N?

Question

^ MEANS POWER
ANS 24061,24073,24089,24097

Answer 1

Expanding we can get 2^4014+ 5^2 =225N.

Now divide both sides by 225.So

(2^4014 + 5^2 ) /225 =N

=> (2^4014 + 5^2 ) /15^2 =N

=> 2^4014/15^2 + 5^2/15^2 =N

=> 2^4014/225 + (5/15)^2 =N

=> 2^4014/225 + (1/3)^2 =N

=> 2^4014/225 + 1/9 =N

so . there it is

N = 2^4014/225 + 1/9

Do the calculation on a UNIX system and you will get the sum of digits. Don't use a windows machine. You will end up with exponents.

Answer 2

Well, I'd start by expanding the expression (2^2007 + 5)^2 = 2^4014 + 25*2^2007 + 25. I'd rewrite it as 2^4014 + 25*(2^2007 + 1). Then, not seeing anything useful, I'd rewrite the original problem as (2^2007 + 5) = 15*sqrt(N). I can rewrite the power of 2 to make it (16*2^2003 + 5) = 15*sqrt(N). Now, I can further rewrite this as (15*2^2003 + 2^2003 + 5) = 15*sqrt(N). Dividing through by 15 gives me sqrt(N) = 2^2003 + (2^2003)/15 + 1/3. And then, frankly, I remember why I hated the AIME. This is an AIME problem, right? In any event, I think I'm on the right track. The bottom line is you just need to keep trying different approaches, different ways of breaking down the original problem, until you get a form that works.

Answer 3

9.......225 2+2+5=9

Answer 4

24061