2007/02/12
Riddle
/**
* This solves the riddle:<BR><BR>
*
* If you have a 10 digit number using all the digits 0-9, find a number
* where the leftmost value is divisible by 1, the two leftmost values are
* divisible by 2, etc. The full number will eventually be divisible by 10.
*
* The answer to this riddle is 3816547290.
*/
private static void solve() {
for (int x0 = 0; x0 <= 9; x0++) {
final long t0 = x0;
for (int x1 = 0; x1 <= 9; x1++) {
if (x1 == x0) continue;
final long t1 = 10 * t0 + x1;
if (t1 % 2 != 0) continue;
for (int x2 = 0; x2 <= 9; x2++) {
if (x2 == x1 || x2 == x0) continue;
final long t2 = 10 * t1 + x2;
if (t2 % 3 != 0) continue;
for (int x3 = 0; x3 <= 9; x3++) {
if (x3 == x2 || x3 == x1 || x3 == x0) continue;
final long t3 = 10 * t2 + x3;
if (t3 % 4 != 0) continue;
for (int x4 = 0; x4 <= 9; x4++) {
if (x4 == x3 || x4 == x2 || x4 == x1 || x4 == x0) continue;
final long t4 = 10 * t3 + x4;
if (t4 % 5 != 0) continue;
for (int x5 = 0; x5 <= 9; x5++) {
if (x5 == x4 || x5 == x3 || x5 == x2
|| x5 == x1 || x5 == x0) continue;
final long t5 = 10 * t4 + x5;
if (t5 % 6 != 0) continue;
for (int x6 = 0; x6 <= 9; x6++) {
if (x6 == x5 || x6 == x4 || x6 == x3
|| x6 == x2 || x6 == x1 || x6 == x0) continue;
final long t6 = 10 * t5 + x6;
if (t6 % 7 != 0) continue;
for (int x7 = 0; x7 <= 9; x7++) {
if (x7 == x6 || x7 == x5 || x7 == x4
|| x7 == x3 || x7 == x2 || x7 == x1 || x7 == x0) continue;
final long t7 = 10 * t6 + x7;
if (t7 % 8 != 0) continue;
for (int x8 = 0; x8 <= 9; x8++) {
if (x8 == x7 || x8 == x6 || x8 == x5
|| x8 == x4 || x8 == x3 || x8 == x2
|| x8 == x1 || x8 == x0) continue;
final long t8 = 10 * t7 + x8;
if (t8 % 9 != 0) continue;
for (int x9 = 0; x9 <= 9; x9++) {
if (x9 == x8 || x9 == x7 || x9 == x6
|| x9 == x5 || x9 == x4 || x9 == x3
|| x9 == x2 || x9 == x1 || x9 == x0) continue;
final long t9 = 10 * t8 + x9;
if (t9 % 10 != 0) continue;
System.out.println(x0 + "" + x1 + "" + x2 + ""
+ x3 + "" + x4 + "" + x5 + ""
+ x6 + "" + x7 + "" + x8 + "" + x9);
return;
}
}
}
}
}
}
}
}
}
}
}
Copyright © 2006 Stein Gunnar Bakkeby
