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教育王國 討論區 小學雜談 math question for help
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math question for help [複製鏈接]

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10
1#
發表於 08-12-23 11:43 |只看該作者 |倒序瀏覽 |打印
Can anybody help?

A pack of biscuits to be shared equally among
3 children --> 2 biscuits left
5 children --> 3 biscuits left
7 children --> 4 biscuits left

What is the minimum number of biscuits in that pack of biscuits
   0    0    0    0

Rank: 5Rank: 5


1889
2#
發表於 08-12-23 11:54 |只看該作者
answer: 53

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10
3#
發表於 08-12-23 11:57 |只看該作者

回覆 # 的文章

Thank you.
Can you show us the steps in reaching the answer.

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1889
4#
發表於 08-12-23 14:44 |只看該作者
N = 2(mod3)
N = 3(mod5)
N = 4(mod7)

5x7 = 35 is the smallest no. which is divisible by 5 and 7 and has remainder of 2 when divided by 3

3x7 = 21 is the smallest no. which is divisible by 3 and 7 and has a remainder of 1 when divided by 5
therefore 21x3 = 63 is divisible by 3 and 7 and has a remainder of 3 when divided by 5

3x5 = 15 is the smallest no. which is divisible by 3 and 5 and has a remainder of 1 when divided by 7
therefore 15x4 = 60 is divisible by 3 and 5 and has a remainder of 4 when divided by 7

LCM 3x5x7 = 105

N = 35+63+60 + 105t where t is integer
N = 158 + 105t (general solution)
N = 53 is smallest when (t = -1)

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10
5#
發表於 08-12-23 15:39 |只看該作者

回覆 # 的文章

Thanks very much.

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3060
6#
發表於 08-12-24 11:41 |只看該作者
原帖由 chauch 於 08-12-23 15:39 發表
Thanks very much.


幾年班maths 題目

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10
7#
發表於 08-12-24 13:23 |只看該作者

回覆 # 的文章

It's just an I.Q. math question for a P 5 student.  We can get to the answer by simple logical thinking (trial by error), but the approach posted by baekdoosan is obviously beyond primary or even secondary school level.

Rank: 3Rank: 3


218
8#
發表於 08-12-25 00:02 |只看該作者
三人同行七十稀
五樹梅花廿一枝
七子團圓正半月
餘百零五便得知

三三數之,餘數乘以七十
五五數之,餘數乘以廿一
七七數之,餘數乘以十五
三者相加,如不大於一百零五,即為答案;否則須減去一百零五或其倍數

(金庸.射鵰英雄傳)

原帖由 chauch 於 08-12-23 11:43 發表
Can anybody help?

A pack of biscuits to be shared equally among
3 children --> 2 biscuits left
5 children --> 3 biscuits left
7 children --> 4 biscuits left

What is the minimum number of biscuits i ...
別人笑我太瘋癲我笑他人看不穿
不見五陵豪傑墓無花無酒鋤作田
Heidi's bookself

Rank: 3Rank: 3


367
9#
發表於 08-12-25 00:46 |只看該作者

回覆 # 的文章

wa, hohowanwoo.....

不如用個最原始的方法:

分三份餘二, 即:
5, 8, 11, 14... 47, 50, 53

分五份餘三, 即:
8, 13, 18, 23.. 43, 48, 53

分七份餘四, 即:
11, 18, 25, 31... 39, 46, 53

Rank: 3Rank: 3


138
10#
發表於 09-1-20 21:08 |只看該作者
I prefer 最原始的方法!
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