Math Problem Solved in Python
I use the Python programming language, and came across a mathematics problem in an oldish (July 2005) copy of News Scientist.
The problem (number 1343 in their regular enigma column) was this: Find the largest whole number, whose digits are different, and do not include 0, that is divisible by all of its individual digits.
After playing with it, and reading around a little, I came up with a small python script which could solve the problem in 10 to 15 seconds on my machine - and that is not on an unused machine.
Some notes on my assumptions
As any even number x 5 make a 10, which would not be allowable because of the 0 rule, the number probably does not include a 5. This assumption, which reduces the possibilities vastly is used throughout the code.
The code uses a string form of the number quite extensively since the digits are so important here. The skipnumberatindex function is not as clean as I would like - but it does what it is supposed to well.
So - here is my code:
def has_repeating_number(numstr): for digit in numstr: first = numstr.find(digit) next = numstr.find(digit,first+1) if(next != -1): return (True, next) return (False, -1) def can_be_divided_by_digits(number,numstr): for digstr in numstr: if number % int(digstr) > 0: return False; return True; #This base number saves going through the repeat and #other tests when you have subtracted one from an inner #number. An earlier version set the digits after to 9's basenumberstr = "98764321" def skipnumberatindex(index, number, numstr): #skip the complex stuff if it is merely enough to subtract 1 if index == len(numstr) - 1: return number - 1 digit = numstr[index] while digit=='0': index -= 1 digit = numstr[index] digitminusone = str(int(digit)-1) rindex = (len(numstr) - index) - 1 outnumstr = (numstr[:index], digitminusone, basenumberstr[:rindex]) outnumstr = ''.join(outnumstr) return int(outnumstr) #I have started at the following number for some good #reasons. Since we can only have one of each digit, and #that none of them are 5, or 0, then we can only have 8 #digits. Since we are going for the largest number, #then the digits are present in descending order currentnumber = 98764321 while currentnumber > 0: currentnumber -= 1 numstr = str(currentnumber) #Skip zeros if (currentnumber % 10 == 0) or (currentnumber % 5 == 0): currentnumber -= 1; elif '0' in numstr: #Get the place of zero in the number, from the right currentnumber = skipnumberatindex(numstr.find('0'), currentnumber, numstr) elif '5' in numstr: #Get the place of a five in the number currentnumber = skipnumberatindex(numstr.find('5'), currentnumber, numstr) else: (hasrepeat, where) = has_repeating_number(numstr) if hasrepeat: #Okay - reverse the where a little currentnumber = skipnumberatindex(where, currentnumber, numstr) elif not can_be_divided_by_digits(currentnumber,numstr): currentnumber -= 1 else: break print "Number should be %s\n" % (str(currentnumber))
If you have any comments or further optimisations, please let me know!
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