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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x/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
 5 years ago
Best ResponseYou've already chosen the best response.0what is the value of x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do what remains

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on no. try chinese remainder theorem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0will work it out if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I agree with satellite; the working above in nonsense.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is meaningless; like above, use the chinese remainder theorem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0in any case method is described clearly here http://marauder.millersville.edu/~bikenaga/numbertheory/chineseremainder/chineseremainder.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for a really nice description of the history of the chinese remainder theorem see "the mathematical experience"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so is the answer 89 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x = 89 + 105k \] But realistically, k =0 is the only solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@INewton : what is k ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k is an integer where \[k \geq 0 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x \equiv \begin{cases} 2\mod\ 3 \\ 4\mod\ 5 \\ 5\mod\ 7\end{cases}\] \[\text{Let } x_1 \equiv \begin{cases} 1\mod\ 3 \\ 0\mod\ 5 \\ 0\mod\ 7\end{cases} \implies x_1 \equiv \begin{cases} 1\mod\ 3 \\ 0\mod\ 35 \end{cases} \implies x_1=35x_1' \text{ for some integer } x_1'\] \[\text{Therefore } 35x_1' \equiv 1 \mod 3 \implies 2x_1' \equiv 1\mod 3 \] \[\text{Let's take } x_1' = 2 \implies x_1 = 70\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ignore the typos ¬_¬ (e.g. it should be 2x_1 + 4x_2 + 5x_3)
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