More fun in 2016, Part 15

Let A(n) = n + S(n) + S(S(n)) = 2016

First, we will find a lower boundary for n.

As the biggest 3-digit number is 999, the maximum for S(n) is 9×3 =27, and the biggest value for S(S(n)) is 10 (when S(n)=19), the biggest value which we can achieve, cannot be bigger, than

999 + 27 + 10 = 1036 (and it is also easy to see, that the actual maximum value is 1035) , which is much smaller, than 2016.

Hence, we have to look for a 4-digit number.

As the smallest 4 digit numbers start with 1 and 2016 is very close to 1999, we will look for numbers between 1000 and 1999 first.

Now, we can find a lower boundary for n.

n is minimal, when S(n) and S(S(n)) are maximal (as their sum with n is constant).

The maximum value of S(n) is 1+ 9 + 9 + 9 = 28, and the maximum value of S(S(n)) is 10 (both at 28 and 19).

Therefore, the minimum value for n is not smaller, than

2016 - 28 - 10=1978

While A(1978)= 1978 + 25 + 7 = 2010 is not the solution (too small), by examining the values of A(n) for values of n slightly greater, than 1978, we will get our solution ( $n= \boxed {1986}$ ) quickly.

A(1986)= 1986 + 24 + 6 = 2016

def d(n):

s=0

x=str(n)

for i in range(len(x)):

    s=s+int(x[i])

return s

for n in range(2016):

x=n+d(n)+d(d(n))

if x==2016:

    print n