## Saturday, 22 January 2011

### Enigma Number 1628

Was reading through New Scientist and thought I'd have a go at their "Enigma" puzzle, which I've always glossed over since they're pretty difficult maths. However, since I've got a computer I thought I'd see what the answer is anyway.

The question is this: raising a number to the power of itself is like 5 to the power of 5, or 5**5 in Python; a 'reverse' number is a number with its digits backwards, so 45 is the reverse of 54; a number that ends with another means that the least significant digits of the former are the same as all of the digits of the latter in the same order. Given this, what is the 2 digit number x such that x raised to the power of x ends with x, the reverse of x raised to the reverse of x ends with the reverse of x, the reverse of x raised to the power of x ends with the reverse of x and the digits in x aren't the same?

Translating this into Python is quite simple. We start with all of the possible answers, ie. the numbers 00-99, which we get via range(100). Next we can split them into a pair of the "tens and units" as they used to say in primary school. We do this with [(a/10,a%10) for a in range(100)]. Next we can put these back into a list of numbers by undoing the separation of the digits, like this [x*10+y for x,y in [(a/10,a%10) for a in range(100)]]. Next we want to put our filters on this. Namely, the 2 digits x and y cannot be equal, so x!=y; the number raised to itself, (x*10+y)**(x*10+y), must ends with itself, ie. ((x*10+y)**(x*10+y))%100==x*10+y; the same must be true for the reverse, where we just swap x and y; and the reverse to the power of the number, (y*10+x)**(x*10+y), must end in the reverse, y*10+x, so ((y*10+x)**(x*10+y))%100==y*10+x. Putting it all together we get a list of answers:

[x*10+y for x,y in [(a/10,a%10) for a in range(100)] if x!=y and ((x*10+y)**(x*10+y))%100==(x*10+y) and ((y*10+x)**(y*10+x))%100==(y*10+x) and ((y*10+x)**(x*10+y))%100==(y*10+x)]

Easy :)

PS: The answers are 16, 61 and 57