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1 solution
so many dights! Great job!
What is the meaning of N, [.] and 100?
This is a direct quote from Wolfram Mathematica. N is a function that gives a n -digit approximation to an expression. The square brackets are Wolfram Mathematica notation of a function call, in this case, the N function. The function call has two arguments: the expression to be evaluated and the number of digits desired. Playfully, this time I requested 100 digits.
Wolfram Mathematica return exact expressions, e. g., π , when possible; rather than just giving a numerical approximation, 3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982 14808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881 09756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066 06315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759 59195309218611738193261179310511854807446237996274956735188575272489122793818301194912983367336244065664 30860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277 85771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034 41815981362977477130996051870721134999999837297804995105973173281609631859502445945534690830264252230825 33446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823 53787593751957781857780532171226806613001927876611195909216420199.
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Did you just used a robot to do this?
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In the sense that that question about using robots is normally used, no. Wolfram Mathematica is a computer algebra system. I used it as a calculator for this problem. Just now, I repeated the calculation using my PC's builtin calculator program: 0.00157773065680813584112427000935. Do you consider that to be using a robot.?
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@A Former Brilliant Member – yes because computer systems are bots.
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@I changed My Name. – Doing problem yet again, since even using a US$1 scientific calculator offends you.
Using a printed booklet of 6-digit logarithms from 1961 published by Dennison, which predates the general availability of calculators, although to be honest I do own a Friden Model 4 mechanical calculator that probably would offend your sensibilities and which is just a year younger than I. Printed tables of logarithms have existed since the early 1600s and were hand computed.
All logarithms and antilogarithms below are base 10
The Brilliant LaTeX processor is not working. I believe the demonstration can be read with some difficulty.
0.301030+1 0.260206+0 0.278754+1 0.106787+0 0.385541+1 0.385541-1 0.242963+0 0.260206+0 0.242963+0 0.503169+0 0.305136+3 0.198033-3 0 . 0 0 1 5 7 7 7 3 Log 20 above divided by 5 Log 19 Log(Log 19) the sum of the above 2 values subtracting 2 for the division by 100 the antilogarithm of the number above copy of first result copy of second result sum of the above two values Log 2019 Subtract above two logarithms antilogarithm of above value and this is the answer
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@A Former Brilliant Member – \begin{array}{ll} \text{0.301030+1} & \text{Log 20} \ \text{0.260206+0} & \text{above divided by 5} \ \ \text{0.278754+1} & \text{Log 19} \ \text{0.106787+0} & \text{Log(Log 19)} \ \text{0.385541+1} & \text{the sum of the above 2 values} \ \text{0.385541-1} & \text{subtracting 2 for the division by 100} \ \text{0.242963+0} & \text{the antilogarithm of the number above} \ \ \text{0.260206+0} & \text{copy of first result} \ \text{0.242963+0} & \text{copy of second result} \ \text{0.503169+0} & \text{sum of the above two values} \ \text{0.305136+3} & \text{Log 2019} \ \text{0.198033-3} & \text{Subtract above two logarithms} \ \text{0.00157773} & \text{antilogarithm of above value and this is the answer}.\ \end{array}}
N [ 2 0 1 9 5 2 0 × 1 9 1 9 / 1 0 0 , 1 0 0 ] ⇒ 0 . 0 0 1 5 7 7 7 3 0 6 5 6 8 0 8 1 3 5 8 4 1 1 2 4 2 7 0 0 0 9 3 5 1 2 0 2 3 4 8 0 9 1 9 5 3 0 4 7 7 0 0 2 8 3 4 9 5 0 0 8 8 5 3 2 3 2 4 8 9 1 6 7 1 8 7 8 4 0 8 0 0 9 7 8 9 5 9 7 0 0 8 4 9 5 4 0 0 4 8 3 0 0
Because of complaints of using a computer to do the work, here is a computer transcription of a pencil and paper version of the computation using a Dennison 1961 booklet of common 6-place logarithms and antilogarithms. This method was available by the middle 1600s and well predates modern computers, even the steam powered (potentially) Analytical Engine .
0.301030+1 0.260206+0 0.278754+1 0.106787+0 0.385541+1 0.385541-1 0.242963+0 0.260206+0 0.242963+0 0.503169+0 0.305136+3 0.198033-3 0 . 0 0 1 5 7 7 7 3 Log 20 above divided by 5 Log 19 Log(Log 19) the sum of the above 2 values subtracting 2 for the division by 100 the antilogarithm of the number above copy of first result copy of second result sum of the above two values Log 2019 Subtract above two logarithms antilogarithm of above value and this is the answer