,196,227,344,198,
203,247,116,19,8,212,230,31,6,328,65,48,52,59,41,
122,33,117,11,18,25,71,36,45,83,76,89,92,31,65,70,
83,96,27,33,44,50,61,24,112,136,149,176,180,194,143,
171,205,296,87,12,44,51,89,98,34,41,208,173,66,9,35,
16,95,8,113,175,90,56,203,19,177,183,206,157,200,
218,260,291,305,618,951,320,18,124,78,65,19,32,121,
18,53,57,84,96,207,244,66,82,119,71,11,86,77,213,54,
82,316,245,303,86,97,106,212,18,37,15,81,89,16,7,81,
39,96,14,43,216,118,29,55,109,136,172,213,64,8,227,
304,611,221,364,819,375,128,296,11,18,53,76,10,15,
23,19,71,84,120,134,66,73,89,96,230,48,77,26,101,
127,936,218,439,178,171,61,226,313,215,102,18,167,
262,114,218,66,59,48,27,19,13,82,48,162,119,34,127,
139,34,128,129,74,63,120,11,54,61,73,92,180,66,75,
101,124,265,89,96,126,274,896,917,434,461,235,890,
312,413,328,381,96,105,217,66,118,22,77,64,12,12,7,
55,24,83,67,97,109,121,135,181,203,219,228,256,21,
34,77,319,374,382,675,684,717,864,203,4,18,92,16,63,
82,22,46,55,69,74,112,135,186,175,119,213,116,312,
343,264,119,186,218,343,417,845,951,124,209,49,617,
856,924,936,72,19,29,11,35,42,40,66,85,94,112,65,82,
115,119,236,244,186,172,112,85,6,56,38,44,85,72,32,
47,73,96,124,217,314,319,221,644,817,821,934,922,
416,975,10,22,18,46,137,181,101,39,86,103,116,138,
164,212,218,296,815,380,412,460,495,675,820,952。
沃德是如何设法破译出第二页的呢？密码文中数字的数目大
大超过了26个（字母表中字母的数目），沃德想，既然如此，这些数字是不是有可能与比尔曾依次编号的文件中的单词相对应呢？考虑到这一点，沃德试着对许多著名文件中单词的字母进行编号并用那些字母代替密码文中的数字。“这全都是徒劳无益的，”沃德写道，“直到后来，《独立宣言》为其中一张纸的数字提供了线索而重新激发了我的希望。”沃德的做法是给《独立宣言》中每个单词的第一个字母进行编号。例如，他这样给前9个词进行编号：
他从这些单词中发现1＝，2＝1，3＝t，4＝C，5＝0，6＝h，7＝E，8＝1，9＝B。你已经可以看到比尔有两种办法给字母I加密：2或 8。等到他给整个《独立宣言》编号之后，他对许多字母无疑就有了众多的选择。通过自由运用所有这些选择，他借助频率分析法破译难以译出的密码文。这样，由于沃德碰巧发现了适当的密钥——《独立宣言》——而破译了这段密码，他运用这一密钥而推断出下列一段文字：
“我在离布法德约4英里处的贝德福德县里的一个离地面6英尺深的洞穴或地窖中贮藏了下列物品，这些物品为各队员——他们的名字在后面第三张纸上——公有