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Title: Many sevens Post by BNC on Jul 9th, 2003, 9:51am Look at this addition: 7 + 72 + 73 + .... + 749 Without calculating, what are the two LSB digits? |
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Title: Re: Many sevens Post by NickH on Jul 9th, 2003, 4:44pm I assume LSB means Least Significant Binary? If so, then... ::[hide] Since 7 = -1 (mod 8), 7odd power = 1112 (mod 8) and 7even power = 12 (mod 8) Therefore 7 + ... + 748 = 02 (mod 8) Hence 7 + ... + 749 = 1112 (mod 8) So the last three LSB digits are 111. [/hide]:: |
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Title: Re: Many sevens Post by BNC on Jul 9th, 2003, 9:20pm I actually meant the 2 least significant decimal figures :-[ but [hide]NuckH's method works there as well[/hide] |
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Title: Re: Many sevens Post by wowbagger on Jul 10th, 2003, 2:32am [quote author=BNC link=board=riddles_easy;num=1057769477;start=0#2 date=07/09/03 at 21:20:15]I actually meant the 2 least significant decimal figures :-[/quote] As far as I know, LSB stands for "least significant bit", so strictly speaking there's no such thing as a decimal LSB digit. Perhaps "least significant decimal digit" would be more appropriate. I thought that your intention was to aim at decimal digits, but your original statement can be misleading - as was proven by "NuckH"'s answer. |
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Title: Re: Many sevens Post by Sameer on Aug 20th, 2003, 10:30am "LSB" has been loosely used to signify the "last" digits in whatever "base" system you are talking. Anyways in decimal system we will notice that 7 + 7^2 + 7^3 + 7^4 mod 100 comes to zero. This is true for subsequent quadruples too (following NickH's solution) Subsequently we are left with 7^49 which belongs to 7^(4n-3) group or whose first element is 7 hence giving the answer to be 7^49 mod 100 = 7 so last two decimals ditis are "07" This process made me think: Can you generalise on last two digits or at least last digit for the sum (where n is any number): n^1 + n^2 + ... + n^k for a base "b" ? |
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