1000! mod 10^250

1000! mod 10^250 

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the answer is 2

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Hi I'm trying to solve the above problem that was asked recently. Couldn't mod it because it was closed. 

So far I've found the 1000! has 249 zeros 
because there are 
200 multiples of 5 that will generate 200 zeros 
40 multiples of 25 that will generate an Additional 40 zeros 
8 multiples of 125 that will generate an Additional 8 zeros 
1 multiple of 625 that will generate an addition zero. 

So what I'm trying to find is what the last significant digit is. 
Now. 
1x2x3x4x1x6x7x8x9 
generates a value that ends in 6 
The same will apply to every other sequence ending in 
1,2,3,4,6,7,8,9 
of which there are 100 
6^100 conveniently also ends in 6 as does any power of 6. 

Multiplying of 10, 20,30,40, 60,70,80,90 will does the same thing for every set of 100 
as will 
the multiplying of 100, 200,300,400, 600,700,800,900 will does the same thing as well. 

However, I can't figure out how to deal with the multiples of 5 now that are not multiples of 10 and the multipliers that are multiples of 50 that aren't 100s 
and the 500. 

Any suggestions ?

Update: Actually, I just realised that I can bind every 5 to a 2; every 50 to a 20, and the 500 to a 200 
Leaving me with 111 sets of (1.3.4.6.7.8.9) 
which ends in an 8 
so 8^111 ends in a 2; because powers of 8 mod 10 repeat in sets of 4. 

So I'm guessing that the final answer is 2 
Anybody know if this would be correct ? 
Thanks.

Update 2: Thanks for the "can't bind the 5s" Forgot that that's kind of why they were left out to begin with. 
The 111 was from 
100 sequences of 1,2,3,...,9; 10 sequences of 10,20,30...,90; 1 sequence of 100,200,300... 
thanks for the Wolfram link - that's awesome.

Update 3: Primes seem to have been the way to go. 
1000! can be written as 
2^994.3^498.5^249.7^164.11^98.13^81.17^... etc 
which can written 
2^249.5^249.2^745.3^498. etc 

1000! can then also be thought of as Product(all non multiples of 5).5^160.Product(allnon multiples of 5 to 200).(5^2)^(40-8).Product(all non multiples of 5 to 40).(5^3)^(8-1)(Pupto8).(5^4)^1(1) 
which is 5^249.(product sequences with all least significant digits 1,2,3,4,6,7,8,9).product_sequence(1.2.3.... 
which is 5^249(sequence ending in 6)(sequence ending in 4) 
which is 5^249(sequence ending in 4) 

I already know that the (sequence ending in 4) has 2^249.2^745 as a factor. 
Taking out 2^249 from that (sequence ending in 4) will remove the issue with the 5s 
multiples of 2 end in the sequence 2,4,8,6, 2,4,8,6 etc. 
stepping back 249 times along this sequence starting at 4, we arrive at 2 
So I think that is a reasonable method and answer ? 
Thanks for all the help.

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 Best Answer:  Yes, it ends in 249 0's and the last significant digits are 10970027753472. I have a program that does the calculation. Results are in the image below, or athttp://i276.photobucket.com/albums/kk2/f... if YahooAnswers maintenance is preventing you from seeing it. 

I'm not quite sure of the details of what you did. You can't ignore the multiples of 5 just because they get matched up with a 2. E.g., take 30 and 40. Match up the 5's and you are left with factors of 6 and 8, which are different, and you have to account for those quotients after the 5's are out. You are looking at sets of 1x3x4x6x7x8x9 but when you take the 2 out to bind to a 5, what's left? 32x35 = 1120. 42*45 = 1890. So in one case you still have a 2 to deal with, in the other it's a 9. 

1*2*3*4*6*7*8*9*(5*10) = 72,576 x 50 = something ending in 6 x 50 = 3628800. Last s.d. is an 8. 
The product from 11 to 20 = something ending in 6 x (15x20) = something ending in 6 x 300. Last s.d. is again an 8 
But the product from, say, 31 to 40 = something ending in 6 x (35*40) = 6 x 1400 and it ends in a 4, not an 8. 

So you've matched up all the 5's, but you need to be concerned about what's left when you do that. I'm not quite sure if you've done that. 2 is the right answer, but I'm not sure that it's because 8^111 ends in 2. Maybe it is, but I don't see where you got 111. Is that from factoring out the 5's somehow? 

I think you have either figured out the right answer, or are on the right track. It looks like you may have a little more work to do to solve this analytically. 

Here's some add'l info, a table of the last 3 s.d.'s of n! 

100 864 
200 472 
300 496 
400 008 
500 864 
600 496 
700 384 
800 496 
900 432 
1000 472 

You can see how irregular it is. It's easy to count the factors of 5, but not so easy to determine that last digit of what you are left with after you factor them out. 

Another approach is to count all the prime factors of 1000!, toss out the 5's and 249 of the 2's, find p^e mod 1000, and then take the cumulative product mod 1000. Once again you get 472 as the last 3 s.d.'s. You get: 

P e p^e mod 1000 *** prod mod 1000 
2 745 832 832 
3 498 889 648 
5 0 1 648 
7 164 401 848 
11 98 281 288 
13 81 613 544 
17 61 617 648 
19 54 321 8 
23 44 241 928 
29 35 549 472 
31 33 191 152 
37 27 533 16 
41 24 561 976 
43 23 507 832 
47 21 847 704 
53 18 689 56 
59 16 41 296 
61 16 961 456 
67 14 329 24 
71 14 881 144 
73 13 33 752 
79 12 441 632 
83 12 161 752 
89 11 489 728 
97 10 49 672 
101 9 901 472 
103 9 583 176 
107 9 507 232 
109 9 389 248 
113 8 321 608 
127 7 503 824 
131 7 811 264 
137 7 433 312 
139 7 379 248 
149 6 601 48 
151 6 401 248 
157 6 449 352 
163 6 9 168 
167 5 607 976 
173 5 93 768 
179 5 899 432 
181 5 901 232 
191 5 951 632 
193 5 193 976 
197 5 757 832 
199 5 999 168 
211 4 441 88 
223 4 441 808 
227 4 841 528 
229 4 481 968 
233 4 521 328 
239 4 641 248 
241 4 561 128 
251 3 251 128 
257 3 593 904 
263 3 447 88 
269 3 109 592 
271 3 511 512 
277 3 933 696 
281 3 41 536 
283 3 187 232 
293 3 757 624 
307 3 443 432 
311 3 231 792 
313 3 297 224 
317 3 13 912 
331 3 691 192 
337 2 569 248 
347 2 409 432 
349 2 801 32 
353 2 609 488 
359 2 881 928 
367 2 689 392 
373 2 129 568 
379 2 641 88 
383 2 689 632 
389 2 321 872 
397 2 609 48 
401 2 801 448 
409 2 281 888 
419 2 561 168 
421 2 241 488 
431 2 761 368 
433 2 489 952 
439 2 721 392 
443 2 249 608 
449 2 601 408 
457 2 849 392 
461 2 521 232 
463 2 369 608 
467 2 89 112 
479 2 441 392 
487 2 169 248 
491 2 81 88 
499 2 1 88 
503 1 503 264 
509 1 509 376 
521 1 521 896 
523 1 523 608 
541 1 541 928 
547 1 547 616 
557 1 557 112 
563 1 563 56 
569 1 569 864 
571 1 571 344 
577 1 577 488 
587 1 587 456 
593 1 593 408 
599 1 599 392 
601 1 601 592 
607 1 607 344 
613 1 613 872 
617 1 617 24 
619 1 619 856 
631 1 631 136 
641 1 641 176 
643 1 643 168 
647 1 647 696 
653 1 653 488 
659 1 659 592 
661 1 661 312 
673 1 673 976 
677 1 677 752 
683 1 683 616 
691 1 691 656 
701 1 701 856 
709 1 709 904 
719 1 719 976 
727 1 727 552 
733 1 733 616 
739 1 739 224 
743 1 743 432 
751 1 751 432 
757 1 757 24 
761 1 761 264 
769 1 769 16 
773 1 773 368 
787 1 787 616 
797 1 797 952 
809 1 809 168 
811 1 811 248 
821 1 821 608 
823 1 823 384 
827 1 827 568 
829 1 829 872 
839 1 839 608 
853 1 853 624 
857 1 857 768 
859 1 859 712 
863 1 863 456 
877 1 877 912 
881 1 881 472 
883 1 883 776 
887 1 887 312 
907 1 907 984 
911 1 911 424 
919 1 919 656 
929 1 929 424 
937 1 937 288 
941 1 941 8 
947 1 947 576 
953 1 953 928 
967 1 967 376 
971 1 971 96 
977 1 977 792 
983 1 983 536 
991 1 991 176 
997 1 997 472

Source(s):http://www.wolframalpha.com/input/?i=100... gives you the answer too. Keep asking for "more digits."

来源： https://answers.yahoo.com/question/index?qid=20131127134508AAoHYGh&guccounter=1