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Question 25 of the Australian Mathematics Competition, Junior Level, Year 2002:

What are the last 5 digits of this sum? $$1+11+111+\cdots+\underbrace{11111...1}_{\text{$2002$ "1"s}}$$ Note, the last number a.k.a $11111...$ contains 2002 digits of 1.

I've tried to solve this problem but could only get through half of the question.

My method:

So firstly, let's try to solve for the last digit. The one digit can be obtained by adding all 2002 1's together. This ends with a 2. Thus the last digit is 2. Using the same logic for the second to last and 3rd to last digits, they should be 1 and 0 respectively. This means that the answer should end in XX012. I was correct so far but got stuck on the next few steps.

You help would be greatly appreciated.

Thank you.

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1 Answer 1

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Define $n_i$ in the obvious manner. Then for $i \geq 5, n_i \equiv 11111 \pmod{100000}$. Thus, you're looking for:

\begin{align}\sum_{i=1}^{2002} n_i \pmod {100000} &\equiv 1234 + \sum_{i=5}^{2002} 11111 \pmod {100000} \\ &\equiv 1234 + 1998 \cdot 11111 \pmod {100000} \\ &\equiv 1234 + 99778 \pmod {100000} \\ & \equiv 1012 \pmod {100000}.\end{align}

The last five digits of the sum, therefore, are $01012$.

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    $\begingroup$ Very nice proof. $\endgroup$
    – KaleBhodre
    Commented Aug 2, 2023 at 0:32

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