MHB What Integers Between -100 and 400 Satisfy Specific Modular Conditions?

[evinda]

Hello! :)
I am given the following exercise:
Which integers of the interval: $[-100,400]$ have the identity: divided by $11$,the remainder is $2$ and divided by $13$,the remainder is $3$.

It is like that:
$[x]_{11}=[2]_{11} \Rightarrow x \equiv 2(\mod 11) \Rightarrow 11 \mid x-2 \Rightarrow x=11k+2, k \in \mathbb{Z} (*) $
Also, $[x]_{13}=[3]_{13} \Rightarrow x \equiv 3(mod 13) $
$(*) \Rightarrow 11k+2 \equiv 3(\mod 13) \Rightarrow 11k=1 (\mod 13) \Rightarrow [11] [k]=[1] \text{ in } \mathbb{Z}_{13} \Rightarrow [k]=[6] \Rightarrow k \equiv 6(\mod 13) \Rightarrow k=13l+6, l \in \mathbb{Z}$
Then, from the relation $(*)$, $x=11(13l+6)+2$ and we find the values of $x$ from the relation: $-100 \leq x \leq 400$.Instead of doing it like that:
$[x]_{13}=[3]_{13} \Rightarrow x \equiv 3(mod 13) $
$(*) \Rightarrow 11k+2 \equiv 3(\mod 13) \Rightarrow 11k=1 (\mod 13) \Rightarrow [11] [k]=[1] \text{ in } \mathbb{Z}_{13} \Rightarrow [k]=[6] \Rightarrow k \equiv 6(\mod 13) \Rightarrow k=13l+6, l \in \mathbb{Z}$

Could I do it also like that:
$x=3+13l,l \in \mathbb{Z}$,and find also the values of $x$ for which the relation $-100 \leq x \leq 400$ stand and then the solution will be the common $x$s ?

 

[I like Serena]

Hello! :)
I am given the following exercise:
Which integers of the interval: $[-100,400]$ have the identity: divided by $11$,the remainder is $2$ and divided by $13$,the remainder is $3$.

It is like that:
$[x]_{11}=[2]_{11} \Rightarrow x \equiv 2(\mod 11) \Rightarrow 11 \mid x-2 \Rightarrow x=11k+2, k \in \mathbb{Z} (*) $
Also, $[x]_{13}=[3]_{13} \Rightarrow x \equiv 3(mod 13) $
$(*) \Rightarrow 11k+2 \equiv 3(\mod 13) \Rightarrow 11k=1 (\mod 13) \Rightarrow [11] [k]=[1] \text{ in } \mathbb{Z}_{13} \Rightarrow [k]=[6] \Rightarrow k \equiv 6(\mod 13) \Rightarrow k=13l+6, l \in \mathbb{Z}$
Then, from the relation $(*)$, $x=11(13l+6)+2$ and we find the values of $x$ from the relation: $-100 \leq x \leq 400$.Instead of doing it like that:
$[x]_{13}=[3]_{13} \Rightarrow x \equiv 3(mod 13) $
$(*) \Rightarrow 11k+2 \equiv 3(\mod 13) \Rightarrow 11k=1 (\mod 13) \Rightarrow [11] [k]=[1] \text{ in } \mathbb{Z}_{13} \Rightarrow [k]=[6] \Rightarrow k \equiv 6(\mod 13) \Rightarrow k=13l+6, l \in \mathbb{Z}$

Could I do it also like that:
$x=3+13l,l \in \mathbb{Z}$,and find also the values of $x$ for which the relation $-100 \leq x \leq 400$ stand and then the solution will be the common $x$s ?

Hi!

Yes. You can also do it like that (the second way)... for this problem.

It should be clear that it is a lot more work.
And it only works because the integers have been limited to an interval that is not too big.

Are you perchance working on a chapter titled Chinese Remainder Theorem?

I suspect you will get more problems that are more complicated, so that you will have to do it the first way.

 

[evinda]

Hi!

Yes. You can also do it like that (the second way)... for this problem.

It should be clear that it is a lot more work.
And it only works because the integers have been limited to an interval that is not too big.

Are you perchance working on a chapter titled Chinese Remainder Theorem?

I suspect you will get more problems that are more complicated, so that you will have to do it the first way.

I understand!Thank you very much! ;)
Yes,I am working on a chapter with this title.. (Nod)

 

[chisigma]

Hello! :)
I am given the following exercise:
Which integers of the interval: $[-100,400]$ have the identity: divided by $11$,the remainder is $2$ and divided by $13$,the remainder is $3$.

It is like that:
$[x]_{11}=[2]_{11} \Rightarrow x \equiv 2(\mod 11) \Rightarrow 11 \mid x-2 \Rightarrow x=11k+2, k \in \mathbb{Z} (*) $
Also, $[x]_{13}=[3]_{13} \Rightarrow x \equiv 3(mod 13) $
$(*) \Rightarrow 11k+2 \equiv 3(\mod 13) \Rightarrow 11k=1 (\mod 13) \Rightarrow [11] [k]=[1] \text{ in } \mathbb{Z}_{13} \Rightarrow [k]=[6] \Rightarrow k \equiv 6(\mod 13) \Rightarrow k=13l+6, l \in \mathbb{Z}$
Then, from the relation $(*)$, $x=11(13l+6)+2$ and we find the values of $x$ from the relation: $-100 \leq x \leq 400$.Instead of doing it like that:
$[x]_{13}=[3]_{13} \Rightarrow x \equiv 3(mod 13) $
$(*) \Rightarrow 11k+2 \equiv 3(\mod 13) \Rightarrow 11k=1 (\mod 13) \Rightarrow [11] [k]=[1] \text{ in } \mathbb{Z}_{13} \Rightarrow [k]=[6] \Rightarrow k \equiv 6(\mod 13) \Rightarrow k=13l+6, l \in \mathbb{Z}$

Could I do it also like that:
$x=3+13l,l \in \mathbb{Z}$,and find also the values of $x$ for which the relation $-100 \leq x \leq 400$ stand and then the solution will be the common $x$s ?

The diophantine equation You have to solve is...

$\displaystyle x \equiv 2\ \text{mod}\ 11$

$\displaystyle x \equiv 3\ \text{mod}\ 13\ (1)$

The solving procedure is illustrated in...

http://mathhelpboards.com/number-theory-27/applications-diophantine-equations-6029-post28283.html#post28283

Here is $N = n_{1}\ n_{2} = 143$ so that $N_{1}=13 \implies \lambda_{1} = 13^{- 1} \text{mod}\ 11 = 6$ and $N_{2}=11 \implies \lambda_{2} = 11^{- 1} \text{mod}\ 13 = 6$. The solution is...

$\displaystyle x = (2 \cdot 6 \cdot 13 + 3 \cdot 6 \cdot 11)\ \text{mod}\ 143 = 68\ \text{mod}\ 143\ (2)$

... so that the requested numbers are -75, 68, 211, 354...

Kind regards

$\chi$ $\sigma$

 

[soroban]

Hello, evinda!

I solved it with basic algebra.

Which integers on the interval: [-100,400] have the identity:
divided by 11,the remainder is 2,
and divided by 13,the remainder is 3.

We have: .\begin{Bmatrix}N &=& 11a + 2 & [1] \\ N &=& 13b + 3 & [2]\end{Bmatrix}

Equate [1] and [2]: .11a + 2 \:=\:13b + 3

. . a \:=\:\frac{13b+1}{11} \quad\Rightarrow\quad a \:=\:b + \frac{2b+1}{11}\;\;[3]

Since a is an integer, 2b+1 must be a multiple of 11.

This first happens when b = 5
. . and in general when b = 5+11k.

Sustitute into [3]: .a \:=\: (5+11k) + \frac{2(5+11k) + 1}{11}
. . which simplifies to: .a \:=\:13k+6

Substitute into [1]:
. . N \:=\:11(13k+6) + 2 \quad\Rightarrow\quad N \:=\:143k + 68

For k = 1,2, we have: .N \:=\:211,\,354