With the pocket calculator:
123456789012345678901234567890 + 678901234567890123456789012345 = 8.023580235802358e+29
With this I have the exact value:
S = 123456789012345678901234567890 + 678901234567890123456789012345 = 802358023580235802358023580235

With the pocket calculator:   123456789123456789 * 8001 = 987777769776777900000
With this I have the exact value:
P = 123456789123456789 x 8001 = 987777769776777768789

With the pocket calculator:   123456789123456789 ^ 3 = 1.8816763774341882e+51
With this (x^3 -> x*x*x) I have the exact value:
P = 123456789123456789 x 123456789123456789 = 15241578780673678515622620750190521
P = P x 15241578780673678515622620750190521 = 1881676377434183981909562699940347954480361860897069

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1881676377434183981909562699940347954480361860897069 / 5 ?
/ 5  -->  x 2 / 10
1881676377434183981909562699940347954480361860897069 x 2 = 3763352754868367963819125399880695908960723721794138
376335275486836796381912539988069590896072372179413.8

1881676377434183981909562699940347954480361860897069 / 4 ?
/ 4  -->  x 25 / 100
1881676377434183981909562699940347954480361860897069 x 25 = 47041909435854599547739067498508698862009046522426725
470419094358545995477390674985086988620090465224267.25

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An example of the properties that can be conjectured: by operating in the following way we obtain integer numerical values gradually constituted by an increasing quantity of "1" (observation made a few centuries ago by the Arab author Ibn-Al-Banna):

0*9 + 1 = 1,  1*9 + 2 = 11,  12*9 + 3 = 111,  123*9 + 4 = 1111, 

     n = 0; m = 0
     m = m*10 + n    <---- 
     n = n + 1            |
     print  m*9 + n  ----> 

...
  m = 1234567  n = 8
  P = 1234567 x 9 = 11111103
  S = P + 8 = 11111111  
  m = 1234567*10 + 8 = 12345678  n = 9
  P = 12345678 x 9 = 111111102  
  S = P + 9 = 111111111  
  m = 12345678*10 + 9 = 123456789  n = 10
  P = 123456789 x 9 = 1111111101
  S = P + 10 = 1111111111  
  m = 123456789*10 + 10 = 1234567900  n = 11
  P = 1234567900 x 9 = 11111111100
  S = P + 11 = 11111111111
...
  m = 1234567901234567901234567898*10 + 29  n = 30
  P = 12345679012345679012345679009 x 9 = 111111111111111111111111111081
  S = P + 30 = 111111111111111111111111111111
...

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30! = 265252859812191058636308480000000
I can get 30! directly with factorial 2. In any case:
P = 2 x 1 = 2
P = 3 x P = 6
P = 4 x P = 24
...
P = 29 x P = 8841761993739701954543616000000
P = 30 x P = 265252859812191058636308480000000

factorial 2 works up to 170!. For larger values I can use this script.
With "factorial 2": 170! = 725741561530799896739672821112926311471699168129645137654357798
798900561843401706157852350749242617459511490991237838520776666022565442753025328900773207
510902400430280058295603966612599658257104355829425756896631343961226257109494680671120556
8880457193340212661452800000000000000000000000000000000000000000
For 171! I cas multiply this value by 171, and so on.

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With pocket calculator I have:
2 ^ 1023 = 8.98846567431158e+307     2 ^ 1024 = Infinity
I can deduce that 2^ 1024 = 17.9769313486232e307. To get all the figures ...
P = 2 x 1 = 2
P = p x p = 4
P = p x p = 16
...
P = p x p = 1797693134862315907729305190789024733617976978942306572734300811577326758055009
6313270847732240753602112011387987139335765878976881441662249284743063947412437776789342486
5485276302219601246094119453082952085005768838150682342462881473913110540827237163350510684 586298239947245938479716304835356329624224137216

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Euler conjectured that there were no positive integers x, y, z and w such that x^4 = y^4 + z^4 + w^4.
In 1988 this conjecture was shown to be false by taking x = 422481, y = 95800, z = 217519 and w = 414560:

P = 422481 x 422481 = 178490195361
P = P x P = 31858749840007945920321     x^4
P = 95800 x 95800 = 9177640000
P = P x P = 84229075969600000000     y^4
P = 217519 x 217519 = 47314515361
P = P x P = 2238663363846304960321     z^4
P = 414560 x 414560 = 171859993600
P = P x P = 29535857400192040960000     w^4
S = 84229075969600000000 + 2238663363846304960321 = 2322892439815904960321
S = S + 29535857400192040960000 = 31858749840007945920321     y^4+z^4+w^4

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If I know that √2 = 1.41421356237309504880168872420969807..., how can I find one more digit of √2?
1414213562373095048801688724209698078 x 1414213562373095048801688724209698078 = 1999999999999999999999999999999999998388724615478808833198118227916894084
1414213562373095048801688724209698079 x 1414213562373095048801688724209698079 = 2000000000000000000000000000000000001217151740224998930801495676336290241
So:  √2 = 1.414213562373095048801688724209698078...

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x*(10+x)+5*x^3 where x = 123456789012345:  9408381861768148891412848380186556157340600
Indeed:

10+x
123456789012345 + 10 = 123456789012355

x*(10+x)
123456789012355 x 123456789012345 = 15241578753239903688452522475

x^2
123456789012345 x 123456789012345 = 15241578753238669120562399025

x^3
15241578753238669120562399025 x 123456789012345 = 1881676372353626729966819028056573540963625

5*x^3
1881676372353626729966819028056573540963625 x 5 = 9408381861768133649834095140282867704818125

x*(10+x)+5*x^3
9408381861768133649834095140282867704818125 + 15241578753239903688452522475 = 9408381861768148891412848380186556157340600

With the pocket calculator:   9.40838186176815e+42

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I can check the calculations with the WolframAlpha online software. Example: