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The first digit means the left most digit. 2410 is just an example and it can be replaced by any other numbers. Can any one help me to solve it?

daizuozhuo
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    What exactly are you allowed to use? – Igor Rivin Dec 13 '13 at 04:40
  • Any method. But computing the result obviously is the worst solution. – daizuozhuo Dec 13 '13 at 04:57
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    If you are allowed a calculator, you could write it as $10^{n\log(n)}$. See http://stackoverflow.com/questions/7613389/is-there-any-quick-way-to-determine-first-k-digits-on-nn – Prahlad Vaidyanathan Dec 13 '13 at 04:59
  • @PrahladVaidyanathan that's precisely why I was asking, it is not easier for me to compute decimal logs in my head than integer powers :( – Igor Rivin Dec 13 '13 at 05:06
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    $2410^{2410}=(2.41\cdot10^3)^{2410}=2.41^{2410}\cdot10^{3\times2410}$. Obviously, $10^{3\times2410}$ does not contribute to the value of the first digit. – Lucian Dec 13 '13 at 05:51

2 Answers2

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Using a calculatior we get that $\log_{10} (2410^{2410})=8150.66107260543188\ldots$.

Thus, $2410^{2410}$ has $8151$ digits, and the first one is $4$, since $10^{0.66107260543188}=4.582184853626742\ldots$.

In general, if we have a huge number $M$, then its first digit is the same as the first digit of $10^{\log_{10} M-\lfloor\log_{10} M\rfloor}$, where $\lfloor\cdot\rfloor$ is the integer part.

Incidentally, the second digit is 5, the third 8, the fourth 2 etc...

Yiorgos S. Smyrlis
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I'll try without computer as exposed here (the idea is a combination of Prahlad and Lucian's suggestiond) but it's not really easy... (with $\,\log(x)=\log_{10}(x)=\dfrac{\ln(x)}{\ln(10)}$ and since $\,\dfrac 1{\ln(10)}\approx 0.43$) : \begin{align} 2410\cdot\log(24.1)&\approx 2410\cdot(\log(3)+3\log(2)+\log\left(1+\frac1{240}\right))\\ &\approx 2410\cdot(0.47712+3\cdot 0.30103+0.43/240)\\ &\approx 2410\cdot (0.47712+0.90309+0.0018\\ &\approx 2410\cdot 1.382\\ &\approx 241\cdot 13.82\\ &\approx 241\cdot 13+241\cdot 0.82\\ &\approx \text{####}.62\\ \end{align} The first digit is obtained by computing $10^{\text{decimal part}}\approx 4$ (since $\log(4)\approx 0.60206$ and $\log(5)\approx 0.69897\approx 1-0.30103$).

(I had some luck here since the exact computation of $2410\log(24.1)= 3330.661072\cdots$ with $10^{0.661072\cdots}= 4.58218\cdots$)

Community
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Raymond Manzoni
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