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 Five Fives (Posted on 2005-09-18)
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.)
 more 3 fivers | Comment 360 of 391 |
(In reply to re: no 778, but here's some u.q.'s like 477.. by Ady TZIDON)

Ady, thanks for catching my error, but it really should be:

477 = !((sqrt(5/.5`)!)/.5`
because: !((sqrt(5/.5`)!)/.5 = !(sqrt(9)!)/.5` = !(3!)/.5` = !6/.5` = 265*1.8 = 477

I've put (n!) in parenathes because of order of opperations. Factorial first then subfactorial. Or does it matter? Is factorial always done first?

172 = 5!/.5` - !5
176 = !5/.5/.5
309 = 265 + 44 = !(sqrt(5/.5`)!) + !5
380 = 76*5 = (5! - !5)*5
1200 = (5 + 5)*5!
1325 = 265*5 = !(sqrt(5/.5`)!)*5
1408 = !5*.5^-5
1440 = sqrt(5/.5`)!!/.5
...

Edited on October 24, 2005, 5:19 pm
 Posted by brad on 2005-10-24 17:16:01

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