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Topic: Many sevens (Read 479 times) |
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BNC
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Look at this addition:
7 + 72 + 73 + .... + 749
Without calculating, what are the two LSB digits?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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NickH
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Re: Many sevens
« Reply #1 on: Jul 9th, 2003, 4:44pm » |
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I assume LSB means Least Significant Binary? If so, then...
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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.
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Nick's Mathematical Puzzles
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BNC
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Re: Many sevens
« Reply #2 on: Jul 9th, 2003, 9:20pm » |
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I actually meant the 2 least significant decimal figures  but NuckH's method works there as well
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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wowbagger
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Re: Many sevens
« Reply #3 on: Jul 10th, 2003, 2:32am » |
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on Jul 9th, 2003, 9:20pm, BNC wrote:| I actually meant the 2 least significant decimal figures |
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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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Sameer
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Re: Many sevens
« Reply #4 on: Aug 20th, 2003, 10:30am » |
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"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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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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