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Five Fives (Posted on 2005-09-18) Difficulty: 4 of 5
A mathematician who was exceedingly fond of the number five set to work trying to express as many consecutive integers using no numerals besides '5', and only up to five of them. She allowed herself to use any standard mathematical notation she knew, as long as it didn't contain any numerals. For example, she could use the symbol for 'square root', but not 'cube root' (because it contains a '3'). She determined that the highest consecutive integer she could express this way was 36. Her last few calculations were as follows:
  • 31 = 5*5 + 5 + (5/5)
  • 32 = 55*.5 + 5 - .5
  • 33 = (55 + 5) * .55
  • 34 = 5!/5 + 5/.5
  • 35 = (5 + (5+5)/5) * 5
  • 36 = 5*5 + 55/5
  • 37 = ?
Was she correct in thinking 36 was the highest consecutive integer she could express this way? Can you express 37 using only up to five 5's?

Note: The intention here is to find an exact expression, so rounding expressions like [] "greatest integer" are not allowed.
Note: Can you do it without using letters of any kind (x, log, lim, sum, etc.)?

See The Solution Submitted by Josh70679    
Rating: 4.4737 (19 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution eureca ! ! 308 in five fives | Comment 254 of 391 |
I am not introducing a new symbol but rather expanding the scope of using the "!"; (exclamation mark) . In combinatorics n! is number of permutations of n objects (e.g ; 5!=120) while !n denotes number of derangments (e.g.!5=44) n! is called factorial; n!=1*2*3*4...*n !n is called subfactoriall (LOOK IT UP ) !n=n!*(1/2! -!/3! +1/4!-....+- 1/n! ) !3=2 ; !4=9; !5=44;!6 =265; !7=1854 For our purposes; 265+44-1=308 !(SIX)+!5-5/5=308

If a one-time use of a subfactorial is not accepted by "the solving team", so be it ! At least use this opportunity to learn about derangments, the Gamma function, latin squares etc Anyhow it was fun, breaking Josh daVinci's code..... !Ady
  Posted by Ady TZIDON on 2005-10-13 20:49:10
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