Question

say a function f(num) returns the no. of digits in a number
then f(num) + f(num^2) = 1000
is this possible??

Answer 1

here num is an integer...

Answer 2

no if you have num =31 then 31+ 961 = 992
if num =32 then 32 + 1024 > 1000

Answer 3

not possible

Answer 4

case I
f(num) = k, f(num^2) = 2k


case II
f(num) = k, f(num^2) = 2k-1

Answer 5

if you solve for k in both cases, you wont get an integer solution so its not possible

Answer 6

its equal to solving this equation

\[\left[ x \right]+\left[ 2x \right]=1000\] where x=log(num)

there is no solution for this equation