Let the number of digits of #2^2005# be x and that of #5^2005# be y then the possible value of x+y=?

1 Answer
George C.
Aug 5, 2018

#2005#

Explanation:

Note that "number of digits" of #n# is essentially #floor(log_10 n) + 1#

So:

#x = floor(log_10 2^2005) + 1 = floor(2005 log_10 2) + 1#

#y = floor(log_10 5^2005) + 1#

#color(white)(y) = floor(2005 log_10 5) + 1#

#color(white)(y) = floor(2005(1 - log_10 2)) + 1#

#color(white)(y) = floor(-2005(log_10 2)) + 2006#

#color(white)(y) = -ceil(2005 log_10 2) + 2006#

So:

#x + y = floor(2005 log_10 2) - ceil(2005 log_10 2) + 2006 = 2005#

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