Consider a product
P=8*(888...888), i.e.
8 times a number represented by a chain of
k eights.
For what value of k the sum of P's digits is exactly 1000?
8*8=64
8*88=704
8*888=7104
8*8888=71104
so it appears for k>2 the pattern is 7, followed by k2 1's, then a 0 and a 4.
To prove this assume the pattern holds the k'th value
let f(k) be the k'th value, then we have
f(k+1)=10f(k)+64
this can be seen by
f(k)=8*(8...8)
10f(k)=8*(8...80)
10f(k)+64=8*(8...80)+64=8*(8...80)+8*8=8*(8...88)=f(k+1)
if this is the case then
71...104*10+64=
71...1040+64=
71...1104
thus continuing the pattern.
So from this we get that for k>2 the sod is
k2+7+4=k+9
so if we want
k+9=1000
k=991

Posted by Daniel
on 20150715 10:36:48 