IQ數學題

2010-11-29 9:32 pm
1500最少要減去多少,才能同時被4、6和11整除?

回答 (6)

2010-11-30 1:20 am
✔ 最佳答案
LCM ( 4, 6, 11 ) = LCM ( 12, 11 ) = 132

132 x 11 = 1452 and 132 x 12 = 1584

Hence, we only need to consider 1452.

Since the difference between 1500 and 1452 is 48,

you need to reduce the number by 48 for divisible.
參考: Mathematics Teacher Mr. Ip
2010-11-30 12:46 am
1500最少要減去多少,才能同時被4、6和11整除?
Sol
[4,6,11]=2[2,3,11]=2*2*3*11=132
132p<1500
p<=11.36
p=11
1500-132*11=48
2010-11-30 12:10 am
48~~~~~~~~~~~
2010-11-29 11:11 pm
首先要揾出 4,6,11 的最小公倍数(LCM) = 132
又132最大倍数且少于1500 =1452 1500 - 1452 = 48答
1500最少要減去48 ,才能同時被4、6和11整除?
2010-11-29 11:00 pm
首先要揾出 4,6,11 o既公倍数

4 x 6 x 11 = 264

再乘到最接近 1500...

264 x 2 = 528
264 x 3 = 792
264 x 4 = 1056
264 x 5 = 1320
264 x 6 = 1584

o甘姐系 1500 -1320 = 180
答案系 180

2010-12-01 09:13:51 補充:
唔好意思... 搞错o左...
唔咪公倍数, 系最小公倍数(LCM)

姐系 : 4, 6, 11 o既(LCM)

2 / 4, 6, 11
________
2, 3, 11

2 x 2 x 3 x 11 = 132

再乘到最接近 1500...
132 x 2 = 264
132 x 3 = 396
132 x 4 = 528
...
132 x 11 = 1452 <<<
132 x 12 = 1584

o甘姐系 1500 -1452 = 48
答案系 48 先o岩...

^^
參考: 希望可以帮到你丫... ^^
2010-11-29 10:09 pm
1500最少要減去180=1320 1320÷11=120 1320÷6=220 1320÷4=330

1500減去444=1056 1056÷11=96 1056÷6=176 1056÷4=264

1500減去708=792 792÷11=72 792÷6=132 792÷4=198

1500減去972=528 528÷11=48 528÷6=88 528÷4=132

1500減去1236=264264÷11=24 264÷6=44 264÷4=66


希望幫到你
參考: OK?自己?


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