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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

1919.0. "Concours General, France, 1992, problem 5" by EVTSG8::ESANU (Au temps pour moi) Thu Dec 15 1994 11:22

What is the last digit of the floor of

           1992
	 10
	--------    ?
	  83
	10   + 7

T.R Title User Personal
Name Date Lines

1919.1 that's (10^83 + 7)q + r with 0 <= r < 10^83 + 7 CSC32::D_DERAMO Dan D'Eramo, Customer Support Center Thu Dec 15 1994 11:35 13

T.R	Title	User	Personal Name	Date	Lines
1919.1	that's (10^83 + 7)q + r with 0 <= r < 10^83 + 7	CSC32::D_DERAMO	Dan D'Eramo, Customer Support Center	`Thu Dec 15 1994 11:35`	13
	Spoiler follows... Write 10^1992 = (10^83 + 7)q + r. In decimal, the remainder r ends in a 1, so (10^83 + 7)q must end in a 9, which means q must end in a 7. Why does r end in a 1? Well, 1192 = 24 * 83, so 10^1992 is (10^83)^24, which has the same residue as (-7)^24 mod (10^83 + 7). In decimal (-7)^24 has the same last digit as 7^24, and the last digits 7^(4n+k) are 1,7,9,3 for k = 0,1,2,3. 24 corresponds to k=0, with last digit 1. Dan

        Spoiler follows...
        
        Write 10^1992 = (10^83 + 7)q + r.  In decimal, the remainder r
        ends in a 1, so (10^83 + 7)q must end in a 9, which means q
        must end in a 7.
        
        Why does r end in a 1?  Well, 1192 = 24 * 83, so 10^1992 is
        (10^83)^24, which has the same residue as (-7)^24 mod (10^83 + 7).
        In decimal (-7)^24 has the same last digit as 7^24, and the last
        digits 7^(4n+k) are 1,7,9,3 for k = 0,1,2,3.  24 corresponds
        to k=0, with last digit 1.
        
        Dan