正整数集合Ak最小元素为1,最大元素为2007,并且各元素可以从小到大排列成一个公差为k的等差数列,求并集A17 A59 中有几个元素

正整数集合Ak最小元素为1,最大元素为2007,并且各元素可以从小到大排列成一个公差为k的等差数列,求并集A17 A59 中有几个元素
正整数集合Ak最小元素为1,最大元素为2007,并且各元素可以从小到大排列成一个公差为k的等差数列,求并集A17 A59 中有几个元素

正整数集合Ak最小元素为1,最大元素为2007,并且各元素可以从小到大排列成一个公差为k的等差数列,求并集A17 A59 中有几个元素
2007-1=2006,
2006/17=118
2006/59=34,
所以A17中含有118+1=119个元素,
A59中含有34+1=35个元素,
因为17和59都是素数所以最小共倍数是17*59=1003,
2006/1003=2,
所以最多有3个数字相同,
先找到第一个元素1,
再加1003n,即得其它两个元素1004,2007,
具体为
于是并集元素的个数为119+35-3=151.
完毕.
附:
A17=
{1,18,35,52,69,86,103,120,137,154,171,188,205,222,239,256,273,290,307,324,341,358,375,392,409,426,443,460,477,494,511,528,545,562,579,596,613,630,647,664,681,698,715,732,749,766,783,800,817,834,851,868,885,902,919,936,953,970,987,1004,1021,1038,1055,1072,1089,1106,1123,1140,1157,1174,1191,1208,1225,1242,1259,1276,1293,1310,1327,1344,1361,1378,1395,1412,1429,1446,1463,1480,1497,1514,1531,1548,1565,1582,1599,1616,1633,1650,1667,1684,1701,1718,1735,1752,1769,1786,1803,1820,1837,1854,1871,1888,1905,1922,1939,1956,1973,1990,2007}
A59=
{1,60,119,178,237,296,355,414,473,532,591,650,709,768,827,886,945,1004,1063,1122,1181,1240,1299,1358,1417,1476,1535,1594,1653,1712,1771,1830,1889,1948,2007}
A17∪A59=
{1,18,35,52,60,69,86,103,119,120,137,154,171,178,188,205,222,237,239,256,273,290,296,307,324,341,355,358,375,392,409,414,426,443,460,473,477,494,511,528,532,545,562,579,591,596,613,630,647,650,664,681,698,709,715,732,749,766,768,783,800,817,827,834,851,868,885,886,902,919,936,945,953,970,987,1004,1021,1038,1055,1063,1072,1089,1106,1122,1123,1140,1157,1174,1181,1191,1208,1225,1240,1242,1259,1276,1293,1299,1310,1327,1344,1358,1361,1378,1395,1412,1417,1429,1446,1463,1476,1480,1497,1514,1531,1535,1548,1565,1582,1594,1599,1616,1633,1650,1653,1667,1684,1701,1712,1718,1735,1752,1769,1771,1786,1803,1820,1830,1837,1854,1871,1888,1889,1905,1922,1939,1948,1956,1973,1990,2007}

由于n-1和k都是整数，而且乘积=2006，由于：2006=17*59，所以当公差为17时，结果为：
A17={1，18，35，……，2007}
共有项数为：（2007-1）/17+1=60项
同理：A59共有：(2007-1)/59+1=18项
假设A17和A59中有相同的数，则：1+(n-1)17=1+(m-1)59
17（n-1）=59(m-1)由于1...

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由于n-1和k都是整数，而且乘积=2006，由于：2006=17*59，所以当公差为17时，结果为：
A17={1，18，35，……，2007}
共有项数为：（2007-1）/17+1=60项
同理：A59共有：(2007-1)/59+1=18项
假设A17和A59中有相同的数，则：1+(n-1)17=1+(m-1)59
17（n-1）=59(m-1)由于17和59没有公约数，最小公倍数为2006，因此
满足该条件的集合为：U={1,2007}其项数为：2
所以，并集的项数=18+60-2=76

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