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What is the number of zeros in the decimal expansion of $11^{100}-1$?

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According to python 12. Your number is 13780612339822270184118337172089636776264331200038466433146477552154985209552307‌​6769401159497458526446000. –  P.. Oct 25 '12 at 18:44
    
how can it be solved using some theorems? –  Vaolter Oct 25 '12 at 18:46
    
Or if that's binary, you have 3^4 - 1 = 80, so the answer is 1 zero in the decimal expansion. –  Patrick87 Oct 25 '12 at 18:46
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You can easily show that $11^{100} - 1$ has exactly $3$ trailing $0$'s. However, the $0$'s in the middle of the number are not really very amenable to analysis. –  Robert Israel Oct 25 '12 at 19:05
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2 Answers

We ca see that,

$$11^{100}-1=(10+1)^{100}-1=(10^{100}+100\cdot10^{99}+99\cdot50\cdot 10^{98}+...+1)-1$$

Now, try to investigate the above.

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11^100 - 1 = 137806123398222701841183371720896367762643312000384664331464775521549852095523076769401159497458526446000 which has 12 zeros.

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