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Math Help - Farmer's Story Problem

  1. #1
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    Exclamation Farmer's Story Problem

    A cowboy was asked how many cows were on the ranch. He replied that he was unsure, but he knew that when he ocunted them by twos, threes, fours, fives, or sixes, he always had one left over. The only way he could acoid this was to count by sevens; he then had none left over. What is the smallest number of cows on the ranch?

    Please help me answer this! Thank you!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by swimalot View Post
    A cowboy was asked how many cows were on the ranch. He replied that he was unsure, but he knew that when he ocunted them by twos, threes, fours, fives, or sixes, he always had one left over. The only way he could acoid this was to count by sevens; he then had none left over. What is the smallest number of cows on the ranch?

    Please help me answer this! Thank you!
    we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.

    here is our list,

    56,63,70,77,84,91

    91 is the answer,

    If you check all the other conditions you will see that they hold.

    I hope this helps.

    edit: 4|91 12(3)+3
    so 91 is out.

    Sorry.
    Last edited by TheEmptySet; May 3rd 2008 at 10:27 AM. Reason: incorrect
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  3. #3
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    farmers story

    posted by swimalot


    count by sixes until you find a number divisible by each counting multiple.this is 60 so the farmer has 61 cows
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    Behold, the power of SARDINES!
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    Quote Originally Posted by bjhopper View Post
    posted by swimalot


    count by sixes until you find a number divisible by each counting multiple.this is 60 so the farmer has 61 cows
    Quote Originally Posted by swimalot View Post
    The only way he could acoid this was to count by sevens; he then had none left over.
    I am sorry this is incorrect becuse of the above statement 7 must divide the number of cows

    7|61 gives 61=7\cdot 8 +5

    He would have 5 cows left over. So 61 can't be an answer
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  5. #5
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    Hello, swimalot!

    A cowboy was asked how many cows were on the ranch.
    He replied that he was unsure, but he knew that when he counted them
    by twos, threes, fours, fives, or sixes, he always had one left over.
    When counted by sevens, he had none left over.
    What is the smallest number of cows on the ranch?
    Let N = number of cows on the ranch.

    The LCM of 2, 3, 4, 5, 6 is 60
    . . Hence: . N \:=\:60a + 1 ... for some integer a.

    Since N is divisible by 7: . 60a + 1 \:=\:7b ... for some integer b.

    Solve for b\!:\;\;b \:=\:\frac{60a+1}{7} \;=\;8a + \frac{4a+1}{7}

    Since b is an integer, 4a + 1 must be divisible by 7.

    The first time this happens is: a = 5


    Therefore: . N \;=\;60(5)+1 \;=\;\boxed{301}

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  6. #6
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    Quote Originally Posted by TheEmptySet View Post
    we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.

    here is our list,

    56,63,70,77,84,91

    91 is the answer,

    If you check all the other conditions you will see that they hold.

    I hope this helps.
    Hello Tessy

    91 doesn't work for ... four

    91=88+3
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  7. #7
    Super Member angel.white's Avatar
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    You could use the Chinese Remainder Theorem, Topsquark has a lovely example in post #5 here http://www.mathhelpforum.com/math-he...nt-modulo.html
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  8. #8
    Super Member angel.white's Avatar
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    I will do the Chinese Remainder Theorem for you:

    we know that if we divide it by 1, 2, 3, 4, 5, 6 that we have a remainder of 1, and when divided by 7, there is a remainder of 0. So let x be the number of cows, then we have:

    x \equiv 1 (mod2)
    x \equiv 1 (mod3)
    x \equiv 1 (mod4)
    x \equiv 1 (mod5)
    x \equiv 1 (mod6)
    x \equiv 0 (mod7)

    Now, 2, 4, 6 are not pairwise coprime, I am not sure how to handle this, so I will remove the 4 and the 6, and then test them afterwards.

    This gives us:
    x \equiv 1 (mod2)
    x \equiv 1 (mod3)
    x \equiv 1 (mod5)
    x \equiv 0 (mod7)

    So because 2*3*5*7 = 210, we need to find a pair of integers (r_i, s_i) such that:

    2r_1 + \left ( \frac{210}{2} \right )s_1 = 1

    3r_2 + \left ( \frac{210}{3} \right )s_2 = 1

    5r_3 + \left ( \frac{210}{5} \right )s_3 = 1

    7r_4 + \left ( \frac{210}{7} \right )s_4 = 1

    Using Topsquark's numbers (I got different ones the last time I did it, but it doesn't matter, because any solution will work)
    (r_1, s_1) = -51, 1)
    (r_2, s_2) = -23, 1)
    (r_3, s_3) = -25, 3)
    (r_4, s_4) = -17, 4)

    Then
    x \equiv 1 \cdot \left ( \frac{210}{2} \right )s_1 + 1 \cdot \left ( \frac{210}{3} \right )s_2 + 1 \cdot \left ( \frac{210}{5} \right )s_3 + 0 \cdot \left ( \frac{210}{7} \right )s_4 ~~~~(mod 210)

    Simplify and substitute in the values
    x \equiv \left ( \frac{210}{2} \right ) \cdot 1 + \left ( \frac{210}{3} \right )\cdot 1+ \left ( \frac{210}{5} \right )\cdot 3 ~~~~(mod 210)

    Simplify
    x \equiv 301 (mod 210)

    And because 301 \equiv 91 (mod 210)
    x \equiv 91 (mod 210)
    We choose this one because it is the smallest positive number which is congruent to 301 (mod210).

    Now, I omitted the 4 and 6 because I didn't know how to deal with them as they were not pairwise coprime, but we should test each of them to make sure it works:  1 \equiv 91 (mod 4) this is false (1=0*4+1, and 91=22*4+3) so we know that x\not{=}91 So we try the next number that satisfies x \equiv 91 (mod 210)

    x = 91 + 1*210 = 301

    and we test this one
    1 \equiv 301 (mod 4) This is true (1=0*4+1, and 301=75*4+1)
    1 \equiv 301 (mod 6) This is true (1=0*6+1, and 301=50*6+1)

    So the answer is 301 cows.
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