Olympiad Maths Prep

Track / Stage 3 / 215 of 260 #215 of 2000

Problem 215

AMC 10/12, early questions
Algebra Difficulty 3.8 Find the answer UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica · Italy

Problem:
Qual è la seconda cifra (partendo da sinistra) del numero (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) ?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 .

Official solution

Solution:
La risposta è (B)\mathbf{( B )}. Si possono calcolare direttamente tutte le cifre del numero. Si ha 1016+1=1032−11016−110^{16}+1=\frac{10^{32}-1}{10^{16}-1} (è il prodotto notevole (a−1)(a+1)=a2−1(a-1)(a+1)=a^{2}-1 ), e scomposizioni analoghe per gli altri termini. Quindi
(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 cifre. \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{ cifre}} . \end{aligned}

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