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In the courtyard of a school, more than 100 children were playing and less than 200. If they were grouped in groups of 7, there would be 2 children left. If they were put together in groups of 11, there would be 3 children left. How many children were there in the courtyard?(Answer:$135$)

I try

$n=7q+2$

$n=11q_1+3∴7q+2=11q_1+3⟹7q−11q_1=1$

Is there a relationship between the quotientes?

peta arantes
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  • @user2661923 "n" is the total number of children in the garden. – peta arantes Jan 05 '23 at 15:23
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    If you have never worked with problems like this before, and if you have received no Number Theory training, then I advise brute force. From the constraints, $(n-2) \equiv 0 \pmod{7}$ and $(n-2) \equiv 1 \pmod{11}.$ Why not manually check which element $r$ in ${0,1,2,\cdots,10}$ is such that $7r \equiv 1\pmod{11}.$ Then, you will be guaranteed that $(n-2)$ must have form $7 \times (r + 11k)$, where $k$ is some integer. – user2661923 Jan 05 '23 at 15:30
  • @user2661923 It's the basics of division. The dividend(n) is equal to the quotient(q,q1) times the divisor(7,11)plus the remainder(2,3). – peta arantes Jan 05 '23 at 15:31
  • I understand that. The point of my now deleted comments was for you to proofread and edit your posting, which you have done. – user2661923 Jan 05 '23 at 15:32
  • Use CRT to solve $,n\equiv 2\pmod{7},\ n\equiv 3\pmod{11},,$ e.g. via Easy CRT in the linked dupe. – Bill Dubuque Jan 05 '23 at 15:48

2 Answers2

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Assume your number to be $k$ where $k∈(100,200)$,

As you said,

$$k=7n_1+2=11n_2+3$$ Where $n_1,n_2∈\mathbb{N}$

So, by above equation$$ 7n_1-11n_2=1$$

As there is constraint for $k$, $n_2∈[9,17]$, on checking we get $n_2=12$, as $n_1$ is also an integer.

So, $k=135, n_1=19, n_2=12$

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Let number of students is n; the number theory equalaent to youquestion is:

Remainder of number n when divided by 7 is 2 and when it is divided by 11 the remainder is 3, find n.

We can write:

$n=7t+2=11t_1+3\Rightarrow -7 t+11 t_1=-1\space\space\space\space\space\space (1)$

one solution to this equation is $(t, t_1)=(8, 5)$

The homogeneous equation of (1) is:

$11 t_1-7 t=0$

and it's solution is: $t_1=7$ and $t=11$ and we can write the general form of solutions as:

$t=11\alpha+8$ and $t_1=7\beta +5$

So we have (1) in form of:

$n=7(11\alpha+8)+2=11(7\beta +5)+3$

Or:

$77\alpha+58=77\beta+58$

For $\alpha=\beta=1$ equation gives $n=135$.

sirous
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