我想問下有咩快既方法可以計到E條數...!?

2007-09-16 2:32 am
a)11111X999=??
b)100+99-98-97+96+95......-2-1=??
c)59994+498+301+22=??

回答 (2)

2007-09-16 4:30 am
✔ 最佳答案
a)11111X999=??
change 999 to 1000 - 1
Thus, 11111 * 999
= 11111 * 1000 - 11111
= 11111000 - 11111
= 11099889 (ans.)

b)100+99-98-97+96+95......-2-1=??
For 100+99-98-97+96+95......-2-1,
the sum each pair of 1st no. and 3rd no.,………98th no. and 100th no. = 2
also, no, of pairs = 100 / 2 = 50
thus, 100+99-98-97+96+95......-2-1
= 2 * 50
= 100 (ans.)

c)59994+498+301+22=??
Change 59994 to 60000 - 6
Change 498 to 500 - 2
Change 301 to 300 + 1
Change 22 to 20 + 1
Thus, 59994+498+301+22
= (60000 + 500 + 300 + 20) + (-6 - 2 + 1 + 2)
= 60820 - 5
= 60815 (ans.)

希望幫到你.
2007-09-16 2:43 am
用計數機篤......好似有一條式計到..(不過唔記得左^o^~)
參考: me


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