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How do we find the number of digits in the sum $1!+2!+\cdots+100!$ , Options are given as:

  1. $137$
  2. $283$
  3. $314$
  4. $189$
  5. none of these

So I tried to find individually number of digits but it wasn't effective.

miracle173
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Aditya Narayan Sharma
  • 2,896
  • Hint: take a decimal (base ten) logarithm. – Sean Roberson Oct 09 '16 at 17:43
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    Hint: since $100!$ has 2 digits more than $99!$ you need to estimate the former. The number of digits of a number $n$ written in decimal is about $\log_{10}(n)$, what can you say about $\log(100!)$? – Alessandro Codenotti Oct 09 '16 at 17:44
  • Here is a question about the number of digits of 100! http://math.stackexchange.com/questions/1075422/how-many-digits-are-there-in-100 –  Oct 09 '16 at 17:50

1 Answers1

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$$\log_{10} (100!) <\color{blue}{ \log_{10} (1!+2!...+100!)} < \log_{10} (99(99!)+100!)< \log_{10}(2(100!))=\log_{10} (2)+\log_{10} (100!)$$

$$157.97...<\color{blue}{ \log_{10} (1!+2!...+100!)}<158.37..$$

Ahmed S. Attaalla
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