(20*2)分呀數學題小6-中一計L.C.M. H.C.F.

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。

sweets:27
chocolates:22