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///////////////////////////////////////////////////////////////////////////////
// Reverse division
///////////////////////////////////////////////////////////////////////////////
// Enigma 1557 Richard England, New Scientist magazine, August 8, 2009
///////////////////////////////////////////////////////////////////////////////
//
// 1. What is the four-digit integer divisible by 11 and 13 whose reverse
//    is a larger integer also divisible by 11 and 13?
//
// 2. What is the four-digit integer divisible by 17 and 19 whose reverse
//    is a larger integer also divisible by 17 and 19?
//
//
///////////////////////////////////////////////////////////////////////////////
//
// Solve the problem by running the query:
//
//          all Enigma_1557(n1,n2)
//
///////////////////////////////////////////////////////////////////////////////
//
// Results:
//
//
// n1 = 1859
// n2 = 3876
// ___ Solution: 1 ___ [00:00:00] __ [Backtracks: 20991] ____
//
// Number of solutions: 1   Number of backtracks: 28987
// Elapsed time: 00:00:00
//
///////////////////////////////////////////////////////////////////////////////

local Digit = L[0..9]

pred Enigma_1557(n1::L,n2::L) iff
a1::Digit & b1::Digit & c1::Digit & d1::Digit &
n1  = a1*1000 + b1*100 + c1*10 + d1 & a1 <> 0 & n1 mod 11 = 0 & n1 mod 13 = 0 &
n1r = d1*1000 + c1*100 + b1*10 + a1 & d1 <> 0 & n1r mod 11 = 0 & n1r mod 13 = 0 &
n1r > n1 &
a2::Digit & b2::Digit & c2::Digit & d2::Digit &
n2  = a2*1000 + b2*100 + c2*10 + d2 & a2 <> 0 & n2 mod 17 = 0 & n2 mod 19 = 0 &
n2r = d2*1000 + c2*100 + b2*10 + a2 & d2 <> 0 & n2r mod 17 = 0 & n2r mod 19 = 0 &
n2r > n2

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