(數論)數學問題,求教please!!!!!!:(((((

2013-04-12 6:24 am
Let a and b be a positive integers. Show that the number of multiples of b in the sequence

a,2a,3a......,ab

is gcd(a,b)
Category
Science & Mathematics > Mathematics
更新1:

請問可以解釋一下, The multiples of b = Bd in the sequence are BdA , 2BdA , 3BdA , .... , (d-1)BdA , dBdA respectively 為什麼是BdA 開頭???>< 感謝萬分,想不明白

回答 (2)

2013-04-12 9:55 am
✔ 最佳答案
Let gcd(a,b) = d , then a = dA , b = dB were (A,B) = 1. The sequence a , 2a , 3a ,..., ab becomes dA , 2dA, 3dA , .... , dBdA.The multiples of b = Bd in the sequence are
BdA , 2BdA , 3BdA , .... , (d-1)BdA , dBdA respectively.Total d terms , the result follow.

2013-04-13 14:18:11 補充:
每項都有因數A, 所以BdA是sequence中最小能被Bd整除的一項。
2013-04-15 11:03 pm
THX!!!!!!!!!!!!


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