anonymous
  • anonymous
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?
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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)
anonymous
  • anonymous
what is the value of x?
anonymous
  • anonymous
you can do what remains

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anonymous
  • anonymous
on no. try chinese remainder theorem
anonymous
  • anonymous
will work it out if you like
anonymous
  • anonymous
I agree with satellite; the working above in nonsense.
anonymous
  • anonymous
is*
dumbcow
  • dumbcow
i believe its 89
anonymous
  • anonymous
That is meaningless; like above, use the chinese remainder theorem.
anonymous
  • anonymous
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,
anonymous
  • anonymous
in any case method is described clearly here http://marauder.millersville.edu/~bikenaga/numbertheory/chinese-remainder/chinese-remainder.html
anonymous
  • anonymous
for a really nice description of the history of the chinese remainder theorem see "the mathematical experience"
anonymous
  • anonymous
so is the answer 89 ?
anonymous
  • anonymous
\[x = 89 + 105k \] But realistically, k =0 is the only solution.
anonymous
  • anonymous
@INewton : what is k ?
anonymous
  • anonymous
k is an integer where \[k \geq 0 \]
anonymous
  • anonymous
\[x \equiv \begin{cases} 2\mod\ 3 \\ 4\mod\ 5 \\ 5\mod\ 7\end{cases}\] \[\text{Let } x_1 \equiv \begin{cases} 1\mod\ 3 \\ 0\mod\ 5 \\ 0\mod\ 7\end{cases} \implies x_1 \equiv \begin{cases} 1\mod\ 3 \\ 0\mod\ 35 \end{cases} \implies x_1=35x_1' \text{ for some integer } x_1'\] \[\text{Therefore } 35x_1' \equiv 1 \mod 3 \implies 2x_1' \equiv 1\mod 3 \] \[\text{Let's take } x_1' = 2 \implies x_1 = 70\] 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 \[x = 89 + 105k\ \forall k \in \mathbb{N} \]
anonymous
  • anonymous
Ignore the typos ¬_¬ (e.g. it should be 2x_1 + 4x_2 + 5x_3)

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