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December 20, 2014

December 20, 2014

Posted by **Jean** on Monday, August 23, 2010 at 12:12am.

- Chinese remainder theorem -
**MathMate**, Monday, August 23, 2010 at 8:17amThis is a particular case of a class of problems that can be solved by the Chinese remainder theorem.

In this particular case, it can be solved relatively easily without the theorem, since all the left-overs are one, except the last.

Since there was one egg left when packed in 2,3,4,5,6, we need to find the LCM (lowest common multiple) of 2,3,4,5,6 and add one to get the*least*possible answer, namely 60+1=61.

However, we recognize that 60k+1 will always satisfy the first 5 conditions, where k is a positive integer.

To satisfy the 7th condition, we need to find the*least*value of k such that 60k+1 is divisible by 7.

61 mod 7 = 5

121 mod 7 = 2

181 mod 7 = 6

241 mod 7 = 3

301 mod 7 = 0

So 301 is the*least*number of eggs the farmer had. She could have had 721 eggs (=301+420), but with the proof that she gave her insurance, she is likely to get paid for 301!

- algebra -
**jai**, Tuesday, August 24, 2010 at 9:31pmyou have to find the least common multiple of 2, 3, 4, 5 and 6.

thus, LCM = 60

now, add 1 to 60, and check if the new number is divisible by 7,,

60 + 1 = 61

61/7 = 8 remainder 5

61 is not divisible by 7,, thus repeat this process using multiples of 60,, try 120,,

121/7 = 17 remainder 2

*after trial and error, i found 301,, just check if there is a number less than this which satisfies the conditions given.

so there,, =)

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