標題:
(20*2)分呀數學題小6-中一計L.C.M. H.C.F.
發問:
108 sweets and 88 chocolates are to be shared among some children . if each child gets the same numbers of sweets and chocolates each child gets at least?
最佳解答:
This is a question which consist of the knowledge of HCF (Highest Common Factor) but is your question wroong?? at least??? should be at most??? if at least, the number is 1. HCF of 108 and 88 divided by 2 , 54 and44 divided by 2 , 27 and 22 so HCF = 4, so each child can get 27 sweets and 22 chocloates... 這是一條有關最大公因數的問題，但你的題目是否有錯？？「每人最少可分多少朱古力及糖果？？」 應該是「每人最多可分多少？？」 108及88的HCF 先將兩數除2，得54及44 再除2，得27及22 不能找另一個因素同時相除兩數 所以最大公因數係4。 而每位小朋友可得27粒糖及22朱古力。
其他解答:
sweets:27 chocolates:22
(20*2)分呀數學題小6-中一計L.C.M. H.C.F.
發問:
108 sweets and 88 chocolates are to be shared among some children . if each child gets the same numbers of sweets and chocolates each child gets at least?
最佳解答:
This is a question which consist of the knowledge of HCF (Highest Common Factor) but is your question wroong?? at least??? should be at most??? if at least, the number is 1. HCF of 108 and 88 divided by 2 , 54 and44 divided by 2 , 27 and 22 so HCF = 4, so each child can get 27 sweets and 22 chocloates... 這是一條有關最大公因數的問題，但你的題目是否有錯？？「每人最少可分多少朱古力及糖果？？」 應該是「每人最多可分多少？？」 108及88的HCF 先將兩數除2，得54及44 再除2，得27及22 不能找另一個因素同時相除兩數 所以最大公因數係4。 而每位小朋友可得27粒糖及22朱古力。
其他解答:
sweets:27 chocolates:22
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