Author Topic: The Poetry of Logarithms (or, Be Drunk on Math)  (Read 1606 times)

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Pascal Program: General Triangle Solver
« Reply #15 on: August 18, 2018, 09:47:57 am »
This Pascal Code is what motivated me to create the Area Program above.   You will see that the same menu is displayed, but that more must be considered in each case, since in some situations we will need to use the Law of Cosines or the Law of Sines to find lengths of sides, and in other situations we need to use the Law of Cosines or the Law of Sines to compute angle measures. 

While it is important to apply these methods by hand with pencil and paper, drawing a figure, labeling the given data and unknowns, and showing yourself how to find altitude and base, basically reconstructing the formulas in a natural, non-memorized manner, writing code and running various scenarios forced me to develop a systematic strategy.  It forced me to really study the various cases, especially the ambiguous case where two different triangles are possible from the data.   There was certain data which do not determine a triangle because the sum of two side lengths were greater than the length of the third side.   Other data would provide inverse trig functions with values which were not defined for the domain, for instance, feeding arc cosine values less than -1 or greater than 1, so, in these cases the data does not determine a triangle either.  These values had to be "caught" so that they were not stupidly fed to the inverse function,  Then I used a goto and label so as to exit the program gracefully, rather than allow it to produce garbled erroneous output or just crash and burn.

In the C program I did not have to use any goto/label strategy, but, rather, I just made the functions return integer (rather than void).  This way, when I wanted to exit gracefully, I just had to "return 0;".

Hence, another example of just how greatly computer programming can complement mathematical understanding in the process of solving the problem methodically.

So, as I noticed there is not much code available on the Internet which handles all the cases, I am including this as well as the above C code.  Maybe someone will be pleased to stumble upon it in our little shop of philosophical horrors.  i am sure that if you do a comparison, you will find the code in these last two posts is of a different caliber than those clogging up the ZooTubes with programs that handle only one scenario.  It's kind of pitiful, actually.  It makes me wonder about how the search engines out there have gone down the tubes over the years, most likely a consequence of advertisements and money which bump certain sites up to the top of the hit list.   It's a shame that, for instance, you can't just type C Program for finding Area of a Triangle to find the code I just wrote here.   What's going on with the search engines?  Even explicitly searching for whybother.freeboards.org "C Program for finding Area of a Triangle" does not find us here.  This leads me to expect that our message board is for all intents and purposes off the grid.   Maybe it's not such a bad thing, but it makes me feel so utterly invisible ... Fuuck it.  Two tears in a bucket.

Oh well.  Maybe someone besides myself will appreciate the thoroughness.

This next one can be compiled with Lazarus.

(*********************************************************************)
program solve_triangle (INPUT, OUTPUT);

type
  flag = 1..2;

var
  a, b, c, b2, c2 : real;
  angle_a, angle_b, angle_c, angle_b2, angle_c2 : real;
  pi : real;
  option : integer;
  second_solution : boolean;
  no_solution : boolean;

(**************************************************************************)
procedure display_menu (var option : integer);

begin
  writeln('What is your option?');
  writeln('<1> 2 sides and included angle');
  writeln('<2> 2 angles and included side');
  writeln('<3> 3 sides');
  writeln('<4> 2 sides and angle opposite one of them');
  write('ENTER: ');
  readln(option);
  writeln;
end;

(**************************************************************************)
procedure get_data (option : integer);

begin
  case option of
       1:  begin
              write('Enter the length of a side: ');
              readln(a);
              write('Enter the length of the other side: ');
              readln(b);
              write('Enter the measure of the included angle: ');
              readln(angle_c);
           end;

       2: begin
            write('Enter the measure of an angle: ');
            readln(angle_a);
            write('Enter the measure of the other angle: ');
            readln(angle_b);
            angle_c := 180 - (angle_a + angle_b);
            write('Enter the length of the included side: ');
            readln(c);
          end;

       3: begin
            write('Enter the length of a side: ');
            readln(a);
            write('Enter the length of another side: ');
            readln(b);
            write('Enter the length of the third side: ');
            readln(c);
          end;

       4: begin
            write('Enter the measure of the angle: ');
            readln(angle_a);
            write('Enter the length of the side opposite this angle: ');
            readln(a);
            write('Enter the length of the other side: ');
            readln(b);
          end;
  end;  (* case *)
  writeln;
end;

(* The flag tells the procedure whether to solve for an angle *)
(* or a side using the Law of Cosines *)
(***************************************************************)
procedure law_of_cosines (var c, ang_c : real; a, b : real; choice : flag);

label
  Exit;

var
  cosine : real;

begin
  if choice = 1
    then
    begin
      c := sqrt(a * a + b * b - 2 * a * b * cos(pi * ang_c/180));
      if (a + b < c) then begin
        writeln('NO SOLUTION');
        no_solution := TRUE;
        writeln('Since ', a:1:3, ' + ', b:1:3, ' = ', (a+b):1:3, ' < ', c:1:3, ',');
        writeln('there is no triangle with the given sides.');
        end;
      end
    else
      begin
        cosine := (c * c - a * a - b * b)/(-2 * a * b);
        if (a + b < c) then begin
        writeln('NO SOLUTION');
        no_solution := TRUE;
        writeln('Since ', a:1:3, ' + ', b:1:3, ' = ', (a+b):1:3, ' < ', c:1:3, ',');
        writeln('there is no triangle with the given sides.');
        end;
        if (no_solution) then begin
              writeln('We will procede to exit immediately if not sooner.');
              GoTo Exit;
        end;

        if cosine > 0
          then begin
            ang_c := arctan(sqrt(1 - cosine * cosine)/cosine);
            ang_c := 180 * ang_c/pi;
          end;

        if cosine < 0
          then begin
            cosine := -cosine;
            ang_c := arctan(sqrt(1 - cosine * cosine)/cosine);
            ang_c := 180 - 180 * ang_c/pi;
          end;

        if cosine = 0
          then ang_c := 90;
      end;
 Exit:
end;

(* The flag tells the procedure whether to solve for an angle *)
(* or a side using the Law of Sines *)
(***************************************************************)
procedure law_of_sines (var a, ang_a : real; b, ang_b : real;
                              choice : flag);

var
  sine : real;

begin
  if choice = 2
    then begin
           sine := a * sin(pi * ang_b/180)/b;
           if sine > 0
             then begin
                    ang_a := arctan(sine/sqrt(1 - sine * sine));
                    ang_a := 180 * ang_a/pi;
             end;
           if sine < 0
             then begin
                    sine := -sine;
                    ang_a := arctan(sine/sqrt(1 - sine * sine));
                    ang_a := 180 - 180 * ang_a/pi;
             end;
    end
  else a := b * sin(pi * ang_a/180)/(sin(pi * ang_b/180));
end;

(*************************************************************************)
procedure display_results;

begin

if (no_solution)
  then  writeln('No results to display.')
  else
begin
  write('The lengths of the sides are: ');
  writeln(a:1:3, ' ', b:1:3, ' ', c:1:3);
  write('The measures of the angles are: ');
  writeln(angle_a:1:3, ' ', angle_b:1:3, ' ', angle_c:1:3);
  writeln;
  if second_solution
    then begin
           writeln('OR');
           writeln;
           write('The lengths of the sides are: ');
           writeln(a:1:3, ' ', b2:1:3, ' ', c2:1:3);
           write('The measures of the angles are: ');
           write(angle_a:1:3, ' ', angle_b2:1:3, ' ');
           writeln(angle_c2:1:3);
    end;
 end;
end;

(************************************************************************)
begin (* main *)
  display_menu(option);
  get_data(option);
  pi := 4 * arctan(1);
  second_solution := FALSE;
  no_solution := FALSE;
  case option of
       1: begin
            law_of_cosines(c, angle_c, a, b, 1);
            law_of_sines(a, angle_a, c, angle_c, 2);
            angle_b := 180 - (angle_a + angle_c);
       end;
       2: begin
            law_of_sines(a, angle_a, c, angle_c, 1);
            law_of_sines(b, angle_b, c, angle_c, 1);
       end;
       3: begin
            law_of_cosines(c, angle_c, a, b, 2);
            law_of_sines(a, angle_a, c, angle_c, 2);
            angle_b := 180 - (angle_a + angle_c);
       end;
       4: begin
            if angle_a < 90
              then begin
                     if a < b * sin(pi * angle_a/180)
                       then begin
                              writeln('NO SOLUTION');
                              no_solution := TRUE;
                              writeln('angle < 90 AND length ', a:1:3, ' < ', b:1:3, '*sin(', (angle_a):1:3, ')');
                            end
                       else begin
                              law_of_sines(b, angle_b, a, angle_a, 2);
                              angle_c := 180 - (angle_a + angle_b);
                              law_of_sines(c, angle_c, a, angle_a, 1);
                       end;
                     if (b * sin(pi * angle_a/180) < a) and (a < b)
                       then begin
                              angle_b2 := 180 - angle_b;
                              angle_c2 := 180 - (angle_a + angle_b2);
                              law_of_sines(b2, angle_b2, a, angle_a, 1);
                              law_of_sines(c2, angle_c2, a, angle_a, 1);
                              second_solution := TRUE;
                       end;
              end;
            if (90 <= angle_a) and (angle_a < 180)
              then if a <= b
                then begin
                       writeln('NO SOLUTION');
                       writeln('90 <= angle < 180 AND ', a:1:3, ' <= ', b:1:3);
                       no_solution := TRUE;
                     end
                else begin
                       law_of_sines(b, angle_b, a, angle_a, 2);
                       angle_c := 180 - (angle_a + angle_b);
                       law_of_sines(c, angle_c, a, angle_a, 1);
                end;
       end;
  end;  (* case *)
  display_results;
  end.

« Last Edit: August 18, 2018, 05:09:08 pm by Henry [He Not Rich] »
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

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Holden

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Something Interesting for Herr Hauser
« Reply #16 on: August 22, 2018, 02:04:00 am »
(2222^5555+5555^2222)/7.

Find the remainder-without using computer-only pen and paper maybe used.
La Tristesse Durera Toujours                                  (The Sadness Lasts Forever ...)
-van Gogh.

Nation of One

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Without a computer?  EEEEK!

This is a problem in Number Theory since it involves integers.  You would need to apply a combination of modular arithmetic (finding the remainder) and the Remainder Theorem.  I love problems having to do with Number Theory.

I will assume you do not want to apply anything too fancy, such as Fermat's little theorem

My strategy would be to first rewrite the problem as  [(2222^5555) mod 7 + (5555^2222) mod 7] mod 7.   (see below)

I think that I can explain my strategy from the get go most easily by showing you an example with smaller numbers so that you can prove to yourself why the remainder theorem works.  By now, one thing you should know about me is that I am not impressed with mysterious tricks absent of meaningful explanation.  Often people write complete nonsense thinking that they are writing true statements.  For example, they will write  (2222^5555) / 7 = (3^5555)/ 7, which is clearly false.  What they mean is (2222^5555) mod 7 = (3^5555) mod 7. 

They could at least write (2222^5555) % 7 = (3^5555) % 7, as this is the same as what I wrote using the symbol "mod".   I urge you to avoid the lazy habit of writing clearly false statements such as (2222^5555) / 7 = (3^5555)/ 7 when what is meant is that the remainders of the divisions are equal.  There are many who understand how to get a result, but are terrible at communicating why the result is correct.  Perhaps they don't really understand why, only understand how, and therefore are incapable of explaining why the answer is such and such.

Forgive me for going about this in a round about manner, but I think it is valuable in the long run.  I am not so much explaining this procedure to YOU, but explaining the procedure to MYSELF - and, in turn, hopefully, explaining the procedure to YOU, coming around full circle.  My trigonometry exercises can wait.  The electrical problems with the automobile can wait.  This is more interesting to me.

Let's look at a simpler problem first so you will see exactly why I choose this strategy.  I ask that you be patient with this long-winded explanation.  It will benefit the both of us in the long run.

Suppose the problem asked to find the remainder of (2^5 + 5^2)/7.

Using the notation of modular arithmetic, the problem of finding the remainder after dividing an integer by the integer, 7,  can be stated as (2^5 + 5^2) mod 7.  For future reference, sometimes modular arithmetic uses the symbol %, where a mod n is written as a%n.

I am going to use the fact that (2^5 + 5^2) mod 7 = [(2^5 mod 7) + (5^2 mod 7)]  mod 7.

Notice that the left hand member of the above equation does in fact equal the right hand member:

(2^5 + 5^2) mod 7 = (32 + 25) mod 7 = 57 mod 7 = 1 because 56/7 = 8 and 57 - 56 = 1, or, look at it this way:  57 = 56 + 1 = (8 * 7) + 1.   This is "modular arithmetic".

Now the right hand side:

[(2^5 mod 7) + (5^2 mod 7)]  mod 7 = [(32 mod 7) + (25 mod 7)] mod 7 = [4 + 4] mod 7 = 8 mod 7 = 1.

Do you see that 32 mod 7 = 4 since 28/7 = 4 and 32 - 28 = 4?  [32 = (4 * 7) + 4]

You see that 25 mod 7 = 4 because 21/7 = 3 and 25 - 21 = 4.  [25 = (3 * 7) + 4]

I just want you to appreciate how modular arithmetic answers the question "find the remainder when one integer is divided by another".   

So, clearly, (2^5 + 5^2) mod 7 = 1; that is, 57%7 = 1.

Before considering the original problem, next consider what the remainder of (22^55 + 55^22)/7 would be.   Remember, first we write this as:

(22^55 + 55^22) mod 7 = [(22^55 mod 7) + (55^22 mod 7)]  mod 7

22^55 :  22 mod 7 = 1, so (22^55 mod 7) = (1^55 mod 7) = 1

55^22:  55 mod 7 = 6, so (55^22 mod 7) = (6^22 mod 7)

6^22 = 6^(2 * 11) = (6^2)^11 = 36^11

36 mod 7 = 1, so (6^22) mod 7 = 1, so (55^22) mod 7 = 1

Therefore,

(22^55 + 55^22) mod 7 = [(22^55 mod 7) + (55^22 mod 7)]  mod 7 = (1 + 1) mod 7 = 2

Do you see why 2 mod 7 = 2?   

7 = (0 * 7) + 2

______________________________________________________________________________
Now, BEGIN SOLUTION
______________________________________________________________________________

I will now proceed in a similar manner with the problem you present, and this way, you can refer to the above example.

Quote
(2222^5555+5555^2222)/7.

Find the remainder-without using computer-only pen and paper maybe used.

First rewrite the problem in terms of modular arithmetic.

The number you are looking for can be expressed as:

(2222^5555 + 5555^2222) mod 7 =  [(2222^5555) mod 7 + (5555^2222) mod 7] mod 7.

Now, let us break this down into parts we can chew on.


What is 2222 mod 7?

Do the long division.  You find 7 divides 2222 317 times with a remainder 3.  That is, (317 * 7) + 3 = 2222.

You may now rest assured that (2222^5555) mod 7 names the same number as (3^5555) mod 7.

Breaking it down into parts further still, note that 5555 has prime factors 5 * 11 * 101 or 5 * 1111 for now.

3^5555 = 3^(5 * 1111) = (3^5)^1111 = 243^1111

OK, so 243 mod 7 = 5 because (34 * 7) = 238 and 238 + 5 = 243

By the logic followed in the above example,
(2222^5555) mod 7 = (3^5555) mod 7 = [(3^5)^1111 mod 7] = (243^1111) mod 7,
and, since 247 mod 7 = 5, we conclude that (243^1111) mod 7 = 5^1111 mod 7.

Notice that 5^1111 = 5^(11 * 101) = (5^11)^101

*******************************************************************************************************************
At this point, if you insist on the restriction of using no sophisticated instruments for calculating, you are stuck computing 5 * 5 * 5 * 5 * 5 * 5* 5* 5 * 5 * 5 * 5 with paper and pencil, or just pulling out of your ass the conclusion that (5^1111) mod 7 is also 5.   Or they will run to the aid of a fancy theorem to avoid this Devil in the Details.  In a jail cell, one would have to use logarithms:  11 log(5) = log(x), so x = antilog(11 * 0.6889) , and then we lose precision, so our integer arithmetic goes out the window. 
*******************************************************************************************************************

Rather than going on such a tangent, I have no other choice than apply Fermat's Little Theorem:



Since 11 is prime, (5^11) mod 7 = 5 mod 11 = 5

That's not much help to us, but it is helpful to see that 5^x mod 7 = 5^(x + 6) mod 7;
hence, (5^5) mod 7 = (5^11) mod 7, so (3125 mod 7) = 3

How?  Using long division, 3125/7 = 446 + 3/3125; that is, 3125 = (446 * 7) + 3 = 3122 + 3.

So, now we know that (5^1111) mod 7 = 5^(11 * 101) mod 7 = (5^11)^101 mod 7 = 3^101 mod 7

Again, we apply some reasoning about the nature of modular arithmetic.  I'm not going to call it Fermat's Little theorem since this makes it too mysterious or even mystical.

Let's look at the nature of the numbers 3^n mod 7.

I want to break it down to this level so that you might even be inspired to print a hard copy of this post and carry it with you into the office to look at during lunch break.

For any integer x. there are at most 7 possible numbers for  x^n mod 7.

3^0 mod 7 = 1 mod 7 = 1
3^1 mod 7 = 3 mod 7 = 3
3^2 mod 7 = 9 mod 7 = 2
3^3 mod 7 = 27 mod 7 = 6
3^4 mod 7 = 81 mod 7 = 4    [because 81 - 77 = 4]
3^5 mod 7 = 243 mod 7 = 5  [because 243 - (34 * 7) = 243 - 238 = 5]
3^6 mod 7 = 729 mod 7 = 1  [because 729 - (104 * 7) = 729 - 728 = 1]

Now the numbers will repeat:

3^7 mod 7 = 2187 mod 7 = 3
3^8 mod 7 = 3^(8-6) mod 7 = 3^2 mod 7 = 2;

You will be delighted to know that we do not have to write out all the 3^n up to and including 101.

Every six terms repeat.   101%6 = 5; that is, 101 mod 6 = 5.

I feel obligated to point out that, even when using such utterly brute force, we are still applying the same idea known as Fermat's Little Theorem.  If p is prime and n is any integer, then n^p is congruent to n mod p.   In particular, if p does not divide n, then n^(p-1) is congruent to 1 mod p.

So, you see, with 5^11, since 7 is prime and does not divide 5, 5^6 is congruent to 1 mod 7, so 5^11 is congruent to 5^(11 mod 6) mod 7 = 5^5 mod 7 = 3, as we found.

Likewise, since 3^6 mod 7 = 1, (3^101 mod 7) = 3^(101 mod 6) mod 7 = 3^5 mod 7 = 5

How?  Well, 101 mod 6 = 5 since 101 = (16 * 6) + 5 = 96 + 5

So, believe it or not, we have arrived at the solution for the first part in this brute force manner without any mysticism or Visions from "God" or resorting to typing 2222^5555%7 into Sage CAS, which is kind of like a supernatural power, I'll be the first to concede to that.

 We shown that 2222^5555 mod 7 = 5

_______________________________________________________________________________
A cigarette and hot dog are in order before showing 5555^2222 mod 7, but for that part, we can breeze through it, although hopefully not quite  as mystically as the "hot shots" who post their cryptic responses on "Beat the GMAT", where one Birottam Dutta declares "And remainder of the second part will be 2^1111/7 which will be 2," where, while the answer is correct, his statement, 2^1111/7 will be 2, makes no sense whatsoever.  I mean, he leaves it to the reader to assume and interpret what he actually means.   I find that more than a little irritating; and so, I bend over backwards to say what I mean and mean what I say.

He means 2^1111 mod 7 = 2, but, even so, he states the result with a wave of his hands, as though it ought to be obvious.   >:(

I would hope to never leave such a terse response. 





« Last Edit: August 22, 2018, 09:15:37 pm by Henry [He Not Rich] »
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

~ Tabak und Kaffee Süchtigen ~

Nation of One

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5555^2222 mod 7

Using long division you find 5555/7 = 793 + 4/5555, or 5555 = (793 * 7) + 4 = 5551 + 4

So, 5555^2222 mod 7 = 4^2222 mod 7

4^2222 = 4^(2 * 1111) = (4^2)^1111 = 16^1111

(16 mod 7)^1111 mod 7 = 2^1111 mod 7

2^1111 = 2^(11 * 101) = (2^11)^101

Note that:

2^1 mod 7 = 2 mod 7 = 2
2^2 mod 7 = 4 mod 7 = 4
2^3 mod 7 = 8 mod 7 = 1

So, 2^11 mod 7 = 2^(11 mod 3) mod 7 = 2^2 mod 7 = 4

Hence, (2^11)^101 mod 7 = 4^101 mod 7

Now, 4^(101) = (2^2)^101,
so 4^101 mod 7 = (2^2)^101 mod 7 =  2^[2^(101 mod 3)] mod 7
= 2^(2^2) mod 7 = 2^4 mod 7 = 16 mod 7 = 2.

Therefore, 5555^2222 mod 7 = 2.

So, we can now easily find the remainder of (2222^5555+5555^2222)/7 without using a computer.

(2222^5555+5555^2222) mod 7 = [(2222^5555) mod 7 + (5555^2222) mod 7] mod 7

Above, we showed that (2222^5555 mod 7) = 5

So,  [(2222^5555) mod 7 + (5555^2222) mod 7] mod 7 = (5 + 2) mod 7 = 7 mod 7 = 0.
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

~ Tabak und Kaffee Süchtigen ~

Holden

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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #19 on: August 24, 2018, 10:09:47 am »
You got it! Here is my stab at it(please let me know if I  make a conceptual mistake  or if any  step is insufficiently lucid):

When 2222  is  divided by 7,it leaves a remainder  of  3.Hence, for the remainder purpose
(2222^5555)/7 gives us (3^5555/7)
=(3*3^5554)/7=[3(3^2)^2777]/7 =[3* (7+2)^2777]/7 gives us  (3*2^2777)/7
=(3*2^2*2^2775)/7=[3*2^2*(2^3)^  925]/7

[3*2 *   ( 8  ) ^  925]/7 gives us  (12/7) Remainder=5

Similarly,(5555^2222)/7 gives   us (4^2222)/7 =[(2^2)^2222]/7
=(2)^4444/7=(2*2^4443)/7 =[2*(2^3) ^   1481]/7 gives us
[2*  (  8  )   ^   1481]/7 gives us[2*(1)^1481]/7  gives   us  2   (remainder)

Hence, (2222^5555)/7+(5555^2222)/7  gives  us  (5+2)/7 gives  us
Remainder=0
« Last Edit: August 24, 2018, 10:34:51 am by Holden »
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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #20 on: August 24, 2018, 11:52:38 am »
Holden,

I see how you are subtracting powers from the exponent and then multiplying by factors subtracted.  That's not something I even considered.  Interesting ...

The part where I see an error may have just been an oversight on your part, towards the end of the first part.   I will try to be clear as to where it is unclear to me.   In order to communicate this, I will have to break it down and "think out loud":

You state:

(3^5555/7)
=(3*3^5554)/7

This is clear to me.  You subtract 1 from 5555 and multiply by three.  OK

Then you equate this to:

=[3(3^2)^2777]/7

Again, this is clear since 3^5554 = 3^(2 * 2777) = (3^2)^2777

Then you equate this to:

=[3* (7+2)^2777]/7 gives us  (3*2^2777)/7

This I get, since 3^2 = 7 + 2 and 7 mod 7 = 0.  So far, so good.

Then you equate this to:

=(3*2^2*2^2775)/7=[3*2^2*(2^3)^  925]/7

Some added parenthesis would have made that less ambiguous, but I am able to follow your logic.
This is how I would have written it.  Notice the use of parenthesis and added steps.  Also note that I do not include the "/7" in these equations so as to make it more clear:

3 * 2^2777 = 3 * 2^(2775 + 2) = 3 * (2^2) * 2^2775 = 3 * (2^2) * 2^(3 * 925)
= 3 * (2^2) * (2^3)^925

Other than added parenthesis and removing the /7, what I have written is in total agreement with what you wrote thus far.  Now, here is where I detect something is not clear:

From [3*2^2*(2^3)^  925]/7 , you then state:

[3*2 *   ( 8  ) ^  925]/7 gives us  (12/7) Remainder=5

But [3*(2^2)*(2^3)^925]/7 = [3*4*8^925]/7 = [12*8^925]/7

Now, 12*8^925 mod 7 = 5, true, but 3*2*8^925 mod 7 = 6, so you may have just transposed it incorrectly at that point.   What is more alarming to me is the statement:

[3*2 *   ( 8  ) ^  925]/7 gives us  (12/7) Remainder=5

I see that "12 mod 7 = 5" and that "(12/7) gives us a remainder of 5" are equivalent statements.  What I do not see is how "[3*2 *   ( 8  ) ^  925]/7" gives us this.

8 mod 7 = 1, so [3*2*8^925] mod 7 = [3*2*1^925] mod 7 = 6 mod 7 = 6.

I suppose you meant  [12*8^925] mod 7 = 12*1^925 mod 7 = 12 mod 7 = 5.  Maybe you just forgot to type the ^2.

You asked me to let you know if any step could be more lucid, and that is what I did.  Conceptually, I think you have the hang of it, for sure.   I still think you would do well to use the modular arithmetic notation rather than awkwardly state "leaves the remainder".  It is much smoother.

If you don't like typing mod for some reason, you could always write with function notation, as in:

R([12*8^925]/7) = R(12/7) = 5, where the function R() returns the remainder after integer division, which is the same meaning of 12 mod 7.

Peace.   Rest assured that communicating these things is not easy, and it is a great testimony to our similar temperaments that I am able to critique your work without your taking offense.  I am no master, that is for sure.  I am just stating the truth from my perspective. 

I am glad we can discuss such things with a high degree of emotional detachment. 

On a positive note, I concede that I am impressed with the creative approach you took, as far as manipulating the exponent with subtraction.  That is something I will have to add to my own bag of tricks.  I do appreciate the creativity you display in your approach.

Also, understand that I go out of my way to be polite when discussing such things since I know how sensitive I am to criticism, I have to assume others are equally sensitive.
« Last Edit: August 24, 2018, 12:37:04 pm by Henry [He Not Rich] »
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Holden

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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #21 on: August 25, 2018, 01:52:30 pm »
Thanks for your response.Much appreciated.
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Drunk on modular arithmetic
« Reply #22 on: January 20, 2022, 03:38:31 pm »
A cool 6 page presentation of modular arithmetic by Micguela A. Lerma.

Group theoretic view of modulo arithmetic by S. Parthasarathy

an aside:  There are some interesting resources, such as review sheets, for a 2011 course in Mathematical Reasoning by a certain Professor Loveless (Dr. Loveless?)   :D

It was in one of the "final exam review sheets" that I found the following statement made by Andrew Wiles.   While just a technicality, many sources seem to have mistakenly typed "proceed" rather than "precede".  The word, proceed, does not make sense in the context, so I have decided to use the correct word, precede.

check source:  Aleksandar Nikolov

Quote from: Andrew Wiles
Perhaps I can best describe my experience of doing mathematics in terms of a journey through a dark unexplored mansion. You enter the first room of the mansion and it’s completely dark. You stumble around bumping into the furniture, but gradually you learn where each piece of furniture is.  Finally, after six months or so, you find the light switch, you turn it on, and suddenly  it’s all illuminated.   You can see exactly where you were.   Then you move into the next room and spend another six months in the dark.  So each of these breakthroughs, while sometimes they’re momentary, sometimes over a period of a day or two, they are the culmination of-and couldn’t exist without the many months of stumbling around in the dark that precede them.
« Last Edit: January 21, 2022, 08:39:59 pm by ... »
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

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