Find the shortest method! 2

Find the last 3 digits of 1 7 256 17^{256} .


The answer is 681.

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2 solutions

Rohit Ner
Feb 2, 2016

17 256 = ( 290 1 ) 128 = ( 128 0 ) . 290 128 ( 128 1 ) . 290 127 + + ( 128 126 ) . 290 2 ( 128 127 ) . 290 + 1 = + 683564800 37120 + 1 = + 683527681 \begin{aligned}{17}^{256}&={\left(290-1\right)}^{128}\\&=\binom{128}{0}.{290}^{128}-\binom{128}{1}.{290}^{127}+\cdots +\binom{128}{126}.{290}^{2}-\binom{128}{127}.290+1\\&=\cdots +683564800-37120+1\\&=\cdots+ 683527681 \end{aligned}

Good one! This was my intended solution!

Aditya Kumar - 5 years, 4 months ago
. .
Mar 16, 2021

1 7 256 = 988 379 827 094 910 507 716 745 743 102 107 480 127 824 727 816 322 428 606 347 790 683 579 951 095 596 498 669 067 626 734 803 713 141 036 057 094 787 944 698 968 645 135 679 565 717 661 558 158 816 617 770 802 898 563 279 790 220 892 557 595 023 518 649 428 021 494 387 820 796 115 301 188 899 585 376 100 742 208 114 172 011 847 942 342 894 726 894 890 539 279 217 291 395 388 121 713 690 717 038 112 599 784 263 681 17 ^ { 256 } = \text { 988 379 827 094 910 507 716 745 743 102 107 480 127 824 727 816 322 428 606 347 790 683 579 951 095 596 498 669 067 626 734 803 713 141 036 057 094 787 944 698 968 645 135 679 565 717 661 558 158 816 617 770 802 898 563 279 790 220 892 557 595 023 518 649 428 021 494 387 820 796 115 301 188 899 585 376 100 742 208 114 172 011 847 942 342 894 726 894 890 539 279 217 291 395 388 121 713 690 717 038 112 599 784 263 681 } .

Warning : the result is over one centillion, 1 0 303 10 ^ { 303 } in short scale .

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