HCF of 270,405

135
bcos 270 = 135*2
405 = 135*3

2008-12-18 23:45:36 補充：
Method of doing:

5 | 270 405
__________
9 | 54 81
__________
3 | 6 9
__________
2 3

HCF = 5*9*3 = 135

2008-12-18 23:56:41 補充：
有點兒移位 = =
但方法是兩數同時除它們的任意一個公因數 (240/5 & 405/5)
然後兩數被除後產生的商(54 & 81)，再用兩個商的任意一個公因數去除 (54/9 & 81/9)
所產的新的兩個商(6 & 9)，又再用新的兩個商的任意一個公因數去除(6/3 & 9/3)
如是者，直至最後的兩個商互質 (2 & 3，只可被 1 同時除盡)
最後將所有除數相乘 (5*9*3) 就是 HCF 了

2008-12-19 00:00:35 補充：
Method of doing:

5 |....270.......405
...____________
9 |......54.........81
...____________
3 |........6...........9
...____________
............2.......... 3

HCF = 5*9*3 = 135
看的時候把點踢走吧~

Firstly, factorize 270 and 405

270　=　2 ． 3^3 ． 5
405　=　3 ． 3^3 ． 5

If we want to find the HCF of them, we only need to take out their common factors (公因數）
=　3^3 ． 5　
=　27 ． 5　
=　135

H.C.F. of 270 and 405 is 135.Because 270 F. has 1,2,3,5,6,9,10,15,1

8,27,30,45,54,90,135,270.405 F. has 1,3,5,9,15,27,45,81,135,405.

They C.F. has 1,3,5,9,15,27,45,81,135.So,they H.C.F. is 135.

We can use the basic method:
wirte 270 and 405 into their prime factorizations

by short division

270= 2 x 3^3 x 5
405=3^4 x 5

hence HCF of 270 and 405 = 3^3 x 5
= 135