Literally stripping 1 from 100000
(in binary)
2006-07-08 02:54:40
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answer #1
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answered by 11:11 3
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10+10+20+10-20+1 = 31 (and 5 zeros also)
2006-07-08 00:57:44
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answer #2
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answered by yourownlove 3
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How about this? Using the inverse base-10 logarithmic function and 0!=1:
31 = invLog(0!)*(0!+0!+0!) + 0!
= invLog(1)*(1+1+1) + 1
= (10)*(3) + 1
= 31 Q.E.D.
2006-07-08 03:22:49
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answer #3
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answered by Ѕємι~Мαđ ŠçїєŋŧιѕТ 6
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its easy
take 5 zeros and add 31 to them
2006-07-08 00:55:51
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answer #4
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answered by Ivanhoe Fats 6
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I know how to do with 8 zeros: (as 0!=1)
31 = (0!+0!)^((0!+0!)^(0!+0!) +0!) - 0!
= 2^(2^2 + 1) - 1
or even with 7 zeros:
31 = (0!+0!)^((0!+0!+0!)!-0!) - 0!
=2^(3! - 1) - 1
2006-07-08 02:44:26
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answer #5
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answered by love.wisdom 2
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if you mean as in binary representation ...
the corresponding binary number is 11111 which is 1+2+4+8+16 = 31
2006-07-08 00:58:38
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answer #6
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answered by Luay14 6
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31+0+0+0+0+0=31
Done :)
2006-07-08 00:57:12
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answer #7
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answered by AnswerMachine 2
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00000+31=31
2006-07-08 00:58:49
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answer #8
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answered by San 2
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Impossible. All you have to work with is 5 zeroes. How can you get even ONE when all you have are 5 zeroes.
Speaking of zeroes, here comes one now!
TAAAAAAAAAKE COVERRRRRR!
2006-07-08 01:54:46
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answer #9
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answered by Thomas C 4
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00000 > 31
the magic!
00-convert binary to decimal get your 3
take number 0,0 and put it to the power of remaining 0 > get 1
write these two next to each other and get
31
2006-07-08 01:47:21
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answer #10
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answered by ash_wiz4rd 2
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