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 Pieces of eight (Posted on 2015-07-15)
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?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 analytical solution | Comment 1 of 2
8*8=64
8*88=704
8*888=7104
8*8888=71104

so it appears for k>2 the pattern is 7, followed by k-2 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
k-2+7+4=k+9

so if we want
k+9=1000
k=991

 Posted by Daniel on 2015-07-15 10:36:48

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