Precompute the 279 6-digit palindromic numbers that are the product of 2 3-digit numbers.

使用双轨搜索,或者由于该清单太小,只是一带。

名单上:

[99999, 101101, 102201, 105501, 106601, 108801, 110011, 111111, 117711, 119911, 121121, 122221, 123321, 127721, 128821, 129921, 131131, 133331, 135531, 137731, 138831, 140041, 141141, 142241, 143341, 147741, 149941, 154451, 155551, 156651, 159951, 161161, 162261, 165561, 168861, 171171, 174471, 178871, 180081, 182281, 184481, 187781, 188881, 189981, 198891, 201102, 202202, 204402, 209902, 210012, 212212, 213312, 214412, 215512, 216612, 219912, 220022, 221122, 222222, 225522, 227722, 231132, 232232, 234432, 235532, 238832, 239932, 242242, 244442, 246642, 249942, 252252, 255552, 256652, 257752, 258852, 259952, 262262, 266662, 270072, 272272, 273372, 276672, 277772, 279972, 280082, 282282, 284482, 286682, 289982, 290092, 292292, 294492, 296692, 297792, 299992, 301103, 302203, 303303, 306603, 308803, 320023, 321123, 324423, 329923, 330033, 333333, 335533, 343343, 345543, 348843, 354453, 357753, 359953, 363363, 366663, 367763, 369963, 371173, 372273, 374473, 375573, 377773, 378873, 384483, 391193, 393393, 397793, 399993, 401104, 402204, 404404, 405504, 407704, 408804, 409904, 412214, 414414, 416614, 420024, 421124, 424424, 425524, 426624, 428824, 432234, 434434, 436634, 438834, 440044, 441144, 442244, 443344, 444444, 445544, 447744, 452254, 456654, 459954, 461164, 462264, 464464, 468864, 469964, 470074, 471174, 474474, 476674, 477774, 484484, 485584, 487784, 488884, 489984, 491194, 493394, 505505, 506605, 507705, 508805, 509905, 510015, 512215, 513315, 514415, 515515, 519915, 520025, 522225, 523325, 525525, 528825, 531135, 534435, 535535, 536635, 543345, 545545, 548845, 549945, 550055, 551155, 554455, 555555, 560065, 561165, 564465, 565565, 570075, 571175, 573375, 575575, 576675, 577775, 579975, 580085, 585585, 588885, 589985, 592295, 595595, 601106, 602206, 603306, 604406, 606606, 611116, 612216, 616616, 618816, 619916, 623326, 630036, 631136, 636636, 639936, 642246, 648846, 649946, 650056, 652256, 653356, 654456, 656656, 657756, 660066, 661166, 663366, 666666, 672276, 675576, 678876, 689986, 693396, 696696, 698896, 723327, 729927, 737737, 747747, 749947, 770077, 780087, 793397, 801108, 802208, 804408, 807708, 809908, 819918, 821128, 824428, 828828, 840048, 853358, 855558, 861168, 886688, 888888, 906609]

Here s something to play with. It tries to pick options based on restrictions in the multiplicand digits. We re multiplying

  a1a2a3
*
  b1b2b3

为棕榈地选择一位外部数字限制了3和3的备选办法。 我没有为挑选外部目标数字或1和b1进行自动化,尽管我们能够做到。 这里就是一个例子,就是那些人都到9岁,导致棕榈地最多不足1 000 000 000。

Java印法:

var outer = 9;
var a1 = 9;
var b1 = 9;

var modHash = new Array(10);
var iterations = 0;
var highest = 0;

for (var i = 1; i < 10; i++){
  modHash[i] = {0: [0]}
  for (var j = 1; j < 10; j++){
    iterations ++;
    var r = i * j % 10;
    if (modHash[i][r])
      modHash[i][r].push(j);
    else 
      modHash[i][r] = [j];
  }
}

function multiples(x, y, carry, mod){
  for (var i in modHash[x]){
    var m = (10 + mod - i - carry) % 10;
    
    if (modHash[y][m]){
      for (var j in modHash[x][i]){
        for (var k in modHash[y][m]){
          iterations ++;
          var palindrome = num(
            a1,modHash[y][m][k],x,
            b1,modHash[x][i][j],y);
          
          var str = String(palindrome);
          
          if (str == str.split("").reverse().join("") && palindrome > highest){
            console.log(palindrome);
            highest = palindrome;
          }
        }
      }
    }
  }
}

function num(a1, a2, a3, b1, b2, b3){
  return (100*a1 + 10*a2 + a3)
       * (100*b1 + 10*b2 + b3);
}

var a3b3s = [];

for (let x = 1; x < 10; x++) {
  for (let y = x; y < 10; y++) {
    iterations++;
    const xy = x * y;
    if (xy % 10 == outer) {
      a3b3s.push(
        [x, y, Math.floor(xy / 10)]);
    }
  }
}

console.log("[[a3,b3,carry]]: " + JSON.stringify(a3b3s));

for (var i in a3b3s){
  for (var mod = 0; mod < 10; mod++){
    let x = a3b3s[i][0],
        y = a3b3s[i][1],
        carry = a3b3s[i][2];
    multiples(x, y, carry, mod);
  }
}

console.log("Highest: " + highest);
console.log("Iterations: " + iterations);