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UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica

Algebra Difficulty 3.8 AMC 10/12 Find the answer Italy

Problem:
What is the second digit (starting from the left) of the number (1016+1)(108+1)(104+1)(102+1)(10+1)\left(10^{16}+1\right)\left(10^{8}+1\right)\left(10^{4}+1\right)\left(10^{2}+1\right)(10+1) ?

Pick one

Solution

Solution:
The answer is (B)\mathbf{( B )}. One can directly compute all the digits of the number. We have 1016+1=1032−11016−110^{16}+1=\frac{10^{32}-1}{10^{16}-1} (this is the well-known identity (a−1)(a+1)=a2−1(a-1)(a+1)=a^{2}-1 ), and analogous decompositions for the other terms. Hence
(1016+1)(108+1)(104+1)(102+1)(10+1)=1032−11016−11016−1108−1108−1104−1104−1102−1102−110−1=1032−19=399…99⏞9=111…1⏞32 digits. \begin{aligned} & \left(10^{16}+1\right)\left(10^{8}+1\right)\left(10^{4}+1\right)\left(10^{2}+1\right)(10+1) \\ = & \frac{10^{32}-1}{10^{16}-1} \frac{10^{16}-1}{10^{8}-1} \frac{10^{8}-1}{10^{4}-1} \frac{10^{4}-1}{10^{2}-1} \frac{10^{2}-1}{10-1} \\ = & \frac{10^{32}-1}{9}=\overbrace{\frac{399\ldots 9}{9}}^{9}=\overbrace{111\ldots 1}^{32 \text{ digits}} . \end{aligned}

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.