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Let the number 75...57 denote a (r+2)-digit number whose first and last digits are both 7 and the remaining r digits are all 5. Consider the sum S = 77 + 757 + 7557 +... + 75...57 (with the last term having r fives). If S = (68 * 10^lambda + 11020) / 81, find the value of lambda.

1. lambda = r + 2
2. lambda = r + 1
3. lambda = r
4. lambda = 2r

Correct answer: lambda = r + 2

Solution

The term with k fives is Tₖ = 7*10^(k+1) + 5*(111...1 with k digits)*10 + 7... actually Tₖ = 7*10^(k+1) + (5/9)*(10^(k+1) - 10) + 7. Sum S = sumₖ₌₀^(r) Tₖ. After summing the geometric series and simplifying, the result contains 10^(r+2) in the numerator. Matching with (68*10^lambda + 11020)/81 gives lambda = r + 2.

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