algebra

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a farmer is taking her eggs to market in her cart, but she hits a pothole, whick knocks over all the containers of eggs. Though she herself is unhurt, every egg is broken. So she goes to the insurance agent who asks her how many eggs she had. She says she doesnt know, but she remembers some things from various ways she tried packing the eggs. She knows that when she put the eggs in groups of two, there was one egg left over. When she out them in groups of three there was also one egg left over. The same thing happened when she put them in groups of four, groups of five, or groups of six. But when she put them in groups of seven, she ended up withcomplete groups of seven with no eggs left over. How many eggs did she have?

  • Chinese remainder theorem -

    This 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 -

    you 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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