× 0 . 0 . 0 . 2 5 1 2 5 2 2 5 3 2 … 5 … 4 …
The above shows a long multiplication, where only the first 4 decimal places are shown.
As you can see, the first 4 digits after the decimal place of the final product is 1234.
But what is the first 8 digits after the decimal place of the final product?
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We know 0 . 2 2 2 2 ⋯ = 2 × 0 . 1 1 1 1 ⋯ = 2 × 9 1 = 9 2 & 0 . 5 5 5 5 ⋯ = 5 × 0 . 1 1 1 1 ⋯ = 5 × 9 1 = 9 5
So the answer is 9 2 × 9 5 = 8 1 1 0 ≈ 0 . 1 2 3 4 5 6 7 9
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Note that ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 0 . 2 2 2 ⋯ = 0 . 2 + 0 . 0 2 + 0 . 0 0 2 + ⋯ = n = 1 ∑ ∞ 1 0 n 2 0 . 5 5 5 ⋯ = 0 . 5 + 0 . 0 5 + 0 . 0 0 5 + ⋯ = n = 1 ∑ ∞ 1 0 n 5 Thus above result gives us n = 1 ∑ ∞ 1 0 n 2 = 9 2 n = 1 ∑ ∞ 1 0 n 5 = 9 5 Making the answer n = 1 ∑ ∞ 1 0 n 2 ⋅ n = 1 ∑ ∞ 1 0 n 5 = 8 1 1 0 = 0 . 1 2 3 4 5 6 7 9 0 ⋯ Thus, our required answer is 1 2 3 4 5 6 7 9 .