A B C D 是不同的数字（都不是0）,由它们组成的四位数：ABCD是13的倍数,BCDA是11的倍数,CDAB是9的倍数.求ABCD各是什么数字?_作业帮
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A B C D 是不同的数字（都不是0）,由它们组成的四位数：ABCD是13的倍数,BCDA是11的倍数,CDAB是9的倍数.求ABCD各是什么数字?
A B C D 是不同的数字（都不是0）,由它们组成的四位数：ABCD是13的倍数,BCDA是11的倍数,CDAB是9的倍数.求ABCD各是什么数字?
首先抱歉,系统可能会把我的加号改成空格,所以有莫名其妙空格的地方是加号!以下是过程.因为CDBA为9的倍数,所以A B C D为9的倍数,所以ABCD与BCDA都是9的倍数.而ABCD BCDA=1001*A 11*BCD=11*91*A 11*BCD为11的倍数,而BCDA也是11的倍数.所以ABCD为11的倍数.因此ABCD一定是9*11*13=1287的倍数.验证后你会发现,这些倍数大部分都是答案!61;
你好！有5个答案都满足条件，分别是：1，2，8，7
2，5，7，4
3，8，6，1
1287 13X11X9=1287
CDAB是9的倍数，所以A+B+C+D是9的倍数所以A B C D任意组合都是9的倍数10*ABCD=10*(1000*A+BCD)=10000*A+10*BCD
=(10*BCD+A)+9999*A
=BCDA+9999*A因为BCDA是11的倍数，9999*A也是11的倍数所以10*...您的访问出错了(404错误)
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3、感谢您使用本站，1秒后自动跳转public class MD5{/*
* A Java implementation of the RSA Data Security, Inc. MD5 Message
* Digest Algorithm, as defined in RFC 1321.
* Based on the JavaScript implementation of Paul Johnston
Copyright (C) Paul Johnston 1999 - 2000.
for details.
* Java Version by Thomas Weber (Orange Interactive GmbH)
* Convert a 32-bit number to a hex string with ls-byte first
*/String hex_chr = "abcdef";private String rhex(int num){
String str = "";
for(int j = 0; j &= 3; j++)
str = str + hex_chr.charAt((num &;&; (j * 8 + 4)) & 0x0F) + hex_chr.charAt((num &;&; (j * 8)) & 0x0F);}/*
* Convert a string to a sequence of 16-word blocks, stored as an array.
* Append padding bits and the length, as described in the MD5 standard.
*/private int[] str2blks_MD5(String str){
int nblk = ((str.length() + 8) &;&; 6) + 1;
int[] blks = new int[nblk * 16];
int i = 0;
for(i = 0; i & nblk * 16; i++) {
for(i = 0; i & str.length(); i++) {
blks[i &;&; 2] |= str.charAt(i) && ((i % 4) * 8);
blks[i &;&; 2] |= 0x80 && ((i % 4) * 8);
blks[nblk * 16 - 2] = str.length()*8;
* Add integers, wrapping at 2^32
*/private int add(int x, int y){
return ((x&0x7FFFFFFF) + (y&0x7FFFFFFF)) ^ (x&0x) ^ (y&0x);}/*
* Bitwise rotate a 32-bit number to the left
*/private int rol(int num, int cnt){
return (num && cnt) | (num &;&;&; (32 - cnt));}/*
* These functions implement the basic operation for each round of the
* algorithm.
*/private int cmn(int q, int a, int b, int x, int s, int t){
return add(rol(add(add(a, q), add(x, t)), s), b);}private int ff(int a, int b, int c, int d, int x, int s, int t){
return cmn((b & c) | ((~b) & d), a, b, x, s, t);}private int gg(int a, int b, int c, int d, int x, int s, int t){
return cmn((b & d) | (c & (~d)), a, b, x, s, t);}private int hh(int a, int b, int c, int d, int x, int s, int t){
return cmn(b ^ c ^ d, a, b, x, s, t);}private int ii(int a, int b, int c, int d, int x, int s, int t){
return cmn(c ^ (b | (~d)), a, b, x, s, t);}/*
* Take a string and return the hex representation of its MD5.
*/public String calcMD5(String str){
int[] x = str2blks_MD5(str);
int a = 0x;
int b = 0xEFCDAB89;
int c = 0x98BADCFE;
int d = 0x;
for(int i = 0; i & x. i += 16)
int olda =
int oldb =
int oldc =
int oldd =
a = ff(a, b, c, d, x[i+ 0], 7 , 0xD76AA478);
d = ff(d, a, b, c, x[i+ 1], 12, 0xE8C7B756);
c = ff(c, d, a, b, x[i+ 2], 17, 0x242070DB);
b = ff(b, c, d, a, x[i+ 3], 22, 0xC1BDCEEE);
a = ff(a, b, c, d, x[i+ 4], 7 , 0xF57C0FAF);
d = ff(d, a, b, c, x[i+ 5], 12, 0x4787C62A);
c = ff(c, d, a, b, x[i+ 6], 17, 0xA8304613);
b = ff(b, c, d, a, x[i+ 7], 22, 0xFD469501);
a = ff(a, b, c, d, x[i+ 8], 7 , 0x);
d = ff(d, a, b, c, x[i+ 9], 12, 0x8B44F7AF);
c = ff(c, d, a, b, x[i+10], 17, 0xFFFF5BB1);
b = ff(b, c, d, a, x[i+11], 22, 0x895CD7BE);
a = ff(a, b, c, d, x[i+12], 7 , 0x6B901122);
d = ff(d, a, b, c, x[i+13], 12, 0xFD987193);
c = ff(c, d, a, b, x[i+14], 17, 0xA679438E);
b = ff(b, c, d, a, x[i+15], 22, 0x49B40821);
a = gg(a, b, c, d, x[i+ 1], 5 , 0xF61E2562);
d = gg(d, a, b, c, x[i+ 6], 9 , 0xC040B340);
c = gg(c, d, a, b, x[i+11], 14, 0x265E5A51);
b = gg(b, c, d, a, x[i+ 0], 20, 0xE9B6C7AA);
a = gg(a, b, c, d, x[i+ 5], 5 , 0xD62F105D);
d = gg(d, a, b, c, x[i+10], 9 , 0x);
c = gg(c, d, a, b, x[i+15], 14, 0xD8A1E681);
b = gg(b, c, d, a, x[i+ 4], 20, 0xE7D3FBC8);
a = gg(a, b, c, d, x[i+ 9], 5 , 0x21E1CDE6);
d = gg(d, a, b, c, x[i+14], 9 , 0xC33707D6);
c = gg(c, d, a, b, x[i+ 3], 14, 0xF4D50D87);
b = gg(b, c, d, a, x[i+ 8], 20, 0x455A14ED);
a = gg(a, b, c, d, x[i+13], 5 , 0xA9E3E905);
d = gg(d, a, b, c, x[i+ 2], 9 , 0xFCEFA3F8);
c = gg(c, d, a, b, x[i+ 7], 14, 0x676F02D9);
b = gg(b, c, d, a, x[i+12], 20, 0x8D2A4C8A);
a = hh(a, b, c, d, x[i+ 5], 4 , 0xFFFA3942);
d = hh(d, a, b, c, x[i+ 8], 11, 0x);
c = hh(c, d, a, b, x[i+11], 16, 0x6D9D6122);
b = hh(b, c, d, a, x[i+14], 23, 0xFDE5380C);
a = hh(a, b, c, d, x[i+ 1], 4 , 0xA4BEEA44);
d = hh(d, a, b, c, x[i+ 4], 11, 0x4BDECFA9);
c = hh(c, d, a, b, x[i+ 7], 16, 0xF6BB4B60);
b = hh(b, c, d, a, x[i+10], 23, 0xBEBFBC70);
a = hh(a, b, c, d, x[i+13], 4 , 0x289B7EC6);
d = hh(d, a, b, c, x[i+ 0], 11, 0xEAA127FA);
c = hh(c, d, a, b, x[i+ 3], 16, 0xD4EF3085);
b = hh(b, c, d, a, x[i+ 6], 23, 0x04881D05);
a = hh(a, b, c, d, x[i+ 9], 4 , 0xD9D4D039);
d = hh(d, a, b, c, x[i+12], 11, 0xE6DB99E5);
c = hh(c, d, a, b, x[i+15], 16, 0x1FA27CF8);
b = hh(b, c, d, a, x[i+ 2], 23, 0xC4AC5665);
a = ii(a, b, c, d, x[i+ 0], 6 , 0xF4292244);
d = ii(d, a, b, c, x[i+ 7], 10, 0x432AFF97);
c = ii(c, d, a, b, x[i+14], 15, 0xAB9423A7);
b = ii(b, c, d, a, x[i+ 5], 21, 0xFC93A039);
a = ii(a, b, c, d, x[i+12], 6 , 0x655B59C3);
d = ii(d, a, b, c, x[i+ 3], 10, 0x8F0CCC92);
c = ii(c, d, a, b, x[i+10], 15, 0xFFEFF47D);
b = ii(b, c, d, a, x[i+ 1], 21, 0x85845DD1);
a = ii(a, b, c, d, x[i+ 8], 6 , 0x6FA87E4F);
d = ii(d, a, b, c, x[i+15], 10, 0xFE2CE6E0);
c = ii(c, d, a, b, x[i+ 6], 15, 0xA3014314);
b = ii(b, c, d, a, x[i+13], 21, 0x4E0811A1);
a = ii(a, b, c, d, x[i+ 4], 6 , 0xF7537E82);
d = ii(d, a, b, c, x[i+11], 10, 0xBD3AF235);
c = ii(c, d, a, b, x[i+ 2], 15, 0x2AD7D2BB);
b = ii(b, c, d, a, x[i+ 9], 21, 0xEB86D391);
a = add(a, olda);
b = add(b, oldb);
c = add(c, oldc);
d = add(d, oldd);
return rhex(a) + rhex(b) + rhex(c) + rhex(d);}}
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