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

In summary: For k = -1, we have: .N \:=\:-75So, the integers on the interval [-100,400] are 211, 354 and -75.In summary, the integers in the interval [-100,400] that have the identity of being divided by 11 with a remainder of 2 and divided by 13 with a remainder of 3 are 211, 354, and -75. These values can be found by solving the diophantine equation and using basic algebraic techniques.
  • #1
evinda
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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 ?
 
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  • #2
evinda said:
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! :eek:

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.
 
  • #3
I like Serena said:
Hi! :eek:

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)
 
  • #4
evinda said:
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

$\chi$ $\sigma$
 
  • #5
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: .[tex]\begin{Bmatrix}N &=& 11a + 2 & [1] \\ N &=& 13b + 3 & [2]\end{Bmatrix}[/tex]

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

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

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

This first happens when [tex]b = 5[/tex]
. . and in general when [tex]b = 5+11k.[/tex]

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

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

For [tex]k = 1,2[/tex], we have: .[tex]N \:=\:211,\,354[/tex]
 

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

1. What does it mean to find integers of an interval?

Finding integers of an interval refers to identifying all the whole numbers (positive, negative, and zero) within a given range or span of values. This can also be referred to as finding the set of integers within a specific interval.

2. How do you find integers of an interval?

To find integers of an interval, you can start by identifying the beginning and end points of the interval. Then, you can count up or down in whole numbers until you reach the end point. The numbers you count will be the integers within the interval. Alternatively, you can use a number line to visualize and identify the integers within the interval.

3. Can an interval have negative integers?

Yes, an interval can have negative integers. An interval can span across both positive and negative numbers, so it is possible to have negative integers within the interval. For example, the interval from -5 to 5 would include the integers -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, and 5.

4. What is the difference between an open and closed interval?

An open interval does not include the endpoints, while a closed interval includes the endpoints. For example, the interval (1, 5) would include all integers between 1 and 5, but not including 1 and 5. The interval [1, 5] would include all integers between 1 and 5, including 1 and 5.

5. Why is finding integers of an interval important?

Identifying integers within an interval is important because it allows us to understand and work with a specific set of numbers. This can be useful in mathematical equations, data analysis, and various scientific calculations. It also helps us to visualize and comprehend the relationship between numbers within a given range.

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