Consider N=2004^2004.
1) What are the first 3 digits of N?
2) What are the last 3 digits of N?
(In reply to
alternate method (not tested) by Larry)
Your alternate method for part 2 is basically the same as mine, except broken down into smaller steps--really small steps in this case.
The alternate method for part 1 won't work, as inaccuracies will creep up from lower order positions, similar to the way when memory sizes are expressed in K, the difference between a factor of 1000 and a factor of 1024 creeps up into the very first digits when you get into G's.
Here's the method carried out, with the method in the left side and the actual powers in the right (when large enough, expressed in scientific notation):
4016 4016016
8048 8048096064
1612 16128384512256
3230 3.232128256256102D+16
6472 6.477185025537229D+19
1296 1.298027879117661D+23
2597 2.601247869751792D+26
5204 5.212900730982591D+29
1042 1.044665306488911D+33
2088 2.093509274203778D+36
4184 4.195392585504371D+39
8384 8.40756674135076D+42
1680 1.684876374966692D+46
3366 3.376492255433251D+49
6745 6.766490479888235D+52
1351 1.356004692169602D+56
2707 2.717433403107883D+59
5424 5.445736539828198D+62
1086 1.091325602581571D+66
2176 2.187016507573468D+69
4360 4.38278108117723D+72
8737 8.783093286679169D+75
1750 1.760131894650506D+79
3507 3.527304316879613D+82
7028 7.068717851026745D+85
1408 1.41657105734576D+89
2821 2.838808398920903D+92
5653 5.688972031437488D+95
1132 1.140069995100073D+99
2268 2.284700270180545D+102
4545 4.578539341441813D+105
9108 9.175392840249393D+108
1825 1.838748725185978D+112
3657 3.684852445272701D+115
7328 7.384444300326492D+118
1468 1.479842637785429D+122
2941 2.965604646122D+125
5893 5.943071710828488D+128
1180 1.190991570850029D+132
2364 2.386747107983458D+135
Note that already by 2004^5 = 3.232128256256102D+16, the alternate method yields 3230 -- incorrect in the 4th digit. The last line above represents 2004^41, and by then the 3rd digit is incorrect.
For 2004^2004, the alternate method produces 6757 as the last 4-digit number, indicating 675 as the first 3 digits, which is not correct.
The following table, at intervals of 50 in the power, show the proposed alternate method followed by the actual value expressed in scientific notation:
power alt. method 2004^power
50 1231 1.244188 x10^ 165
100 1518 1.548003 x10^ 330
150 1871 1.926006 x10^ 495
200 2308 2.396312 x10^ 660
250 2841 2.981462 x10^ 825
300 3494 3.709498 x10^ 990
350 4304 4.615312 x10^ 1155
400 5308 5.742313 x10^ 1320
450 6537 7.144515 x10^ 1485
500 8056 8.889116 x10^ 1650
550 9933 1.105973 x10^ 1816
600 1223 1.376038 x10^ 1981
650 1508 1.712049 x10^ 2146
700 1858 2.130110 x10^ 2311
750 2284 2.650256 x10^ 2476
800 2811 3.297416 x10^ 2641
850 3462 4.102604 x10^ 2806
900 4268 5.104409 x10^ 2971
950 5260 6.350842 x10^ 3136
1000 6480 7.901639 x10^ 3301
1050 7989 9.831121 x10^ 3466
1100 9837 1.223176 x10^ 3632
1150 1213 1.521860 x10^ 3797
1200 1494 1.893480 x10^ 3962
1250 1838 2.355844 x10^ 4127
1300 2260 2.931112 x10^ 4292
1350 2785 3.646853 x10^ 4457
1400 3434 4.537369 x10^ 4622
1450 4232 5.645338 x10^ 4787
1500 5212 7.023859 x10^ 4952
1550 6424 8.738998 x10^ 5117
1600 7913 1.087295 x10^ 5283
1650 9757 1.352799 x10^ 5448
1700 1201 1.683136 x10^ 5613
1750 1478 2.094137 x10^ 5778
1800 1818 2.605499 x10^ 5943
1850 2240 3.241730 x10^ 6108
1900 2761 4.033320 x10^ 6273
1950 3404 5.018207 x10^ 6438
2000 4192 6.243590 x10^ 6603
2004 6757 1.006990 x10^ 6617
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Posted by Charlie
on 2004-11-23 16:28:43 |
re: alternate method (not tested)
