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.)?
My latest work (though admittedly lazy) has found solutions for 956978. Here they are:
956 = (.5^5)*5#5!/5#
957 = (.5^5)*5#�ã(5/.5`)
958 = (.5^5)*5#�ã(5!/5#)
959 = (.5^5)*5#5/5
960 = (.5^5)*5#
961 = .5�ã(5#+5/5) < .5th root
963 = (.5^5)*5#+�ã(5/.5`)
964 = (.5^5)*5#+5!/5#
965 = (.5^5)*5#+5
966 = (.5^5)*5#+5#/5
967 = (.5^5)*5#+�ã(!5+5)
968 = .5*!5*!5
969 = .5*!5*!5+5/5
970 = .5*!5*!5+�ã(5!/5#)
971 = .5*!5*!5+�ã(5/.5`)
972 = .5*!5*!5+5!/5#
973 = .5*!5*!5+5
974 = .5*!5*!5+5#/5
975 = .5*!5*!5+�ã(!5+5)
976 = (.5^5)*(5#+.5)
977 = .5*!5*!5+5/.5`
978 = .5*!5*!5+5/.5
So far that's what I've gotten to. I might try a bit harder now that we're so close to 1000. Might get there tonight.
Edited on February 17, 2006, 10:06 pm

Posted by Justin
on 20060208 10:17:34 