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anonymous

  • 5 years ago

If my age is divided by 5, the remainder is 4. If it’s divided by 3, the remainder is 2. If it’s divided by 7, the remainder is 5. How old am I?

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  1. anonymous
    • 5 years ago
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    x/5 =4 x/3 =2 x/7=5 x/5 + x/3 + x/7 =4+2+5 x(1/5 + 1/3 + 1/7)=11 x=11(1/5 + 1/3 + 1/7)

  2. anonymous
    • 5 years ago
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    what is the value of x?

  3. anonymous
    • 5 years ago
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    you can do what remains

  4. anonymous
    • 5 years ago
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    on no. try chinese remainder theorem

  5. anonymous
    • 5 years ago
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    will work it out if you like

  6. anonymous
    • 5 years ago
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    I agree with satellite; the working above in nonsense.

  7. anonymous
    • 5 years ago
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    is*

  8. dumbcow
    • 5 years ago
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    i believe its 89

  9. anonymous
    • 5 years ago
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    That is meaningless; like above, use the chinese remainder theorem.

  10. anonymous
    • 5 years ago
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    i believe that it is early in the morning to use crt, but it is actually interesting to list. remainder of 4 when divided by 5 9, 14, 19, 24, 29, ... and so see what happens when you consider the remainders when dividing by 3. the patter in 0, 2, 1, 0, 2 , 1,

  11. anonymous
    • 5 years ago
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    in any case method is described clearly here http://marauder.millersville.edu/~bikenaga/numbertheory/chinese-remainder/chinese-remainder.html

  12. anonymous
    • 5 years ago
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    for a really nice description of the history of the chinese remainder theorem see "the mathematical experience"

  13. anonymous
    • 5 years ago
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    so is the answer 89 ?

  14. anonymous
    • 5 years ago
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    \[x = 89 + 105k \] But realistically, k =0 is the only solution.

  15. anonymous
    • 5 years ago
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    @INewton : what is k ?

  16. anonymous
    • 5 years ago
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    k is an integer where \[k \geq 0 \]

  17. anonymous
    • 5 years ago
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    \[x \equiv \begin{cases} 2\mod\ 3 \\ 4\mod\ 5 \\ 5\mod\ 7\end{cases}\] \[\text{Let } x_1 \equiv \begin{cases} 1\mod\ 3 \\ 0\mod\ 5 \\ 0\mod\ 7\end{cases} \implies x_1 \equiv \begin{cases} 1\mod\ 3 \\ 0\mod\ 35 \end{cases} \implies x_1=35x_1' \text{ for some integer } x_1'\] \[\text{Therefore } 35x_1' \equiv 1 \mod 3 \implies 2x_1' \equiv 1\mod 3 \] \[\text{Let's take } x_1' = 2 \implies x_1 = 70\] Following the obvious notation we find similarly: \[x_2 = 21 \text{ and } x_3 = 15 \] \[\text{Clearly, } 2x_1 + 4x_1 + 5x_3 = x_{min} + 105k \implies x_{min} = 299 - 105k\] For the smallest positive integer x we have x = 89 (k=2), be we also have solutions of the form \[x = 89 + 105k\ \forall k \in \mathbb{N} \]

  18. anonymous
    • 5 years ago
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    Ignore the typos ¬_¬ (e.g. it should be 2x_1 + 4x_2 + 5x_3)

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