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Find at least three numbers that satisfy all three conditions:

(1) there is a remainder of $1$ when the number is divided by $2$;

(2) there is a remainder of $2$ when the number is divided by $3$;

(3) there is a remainder of $3$ when the number is divided by $4$.

Since the LCM($2, 3, 4$) is $12$, all of them can divide $12$ without giving a remainder.

So, $12 - 1=11$ gives

a remainder of $1$ when divided by $2$,

a remainder of $2$ when divided by $3$,

a remainder of $3$ when divided by $4$.

Similarly $12*2-1=23$

and $12*3-1=35$

How would I explain to middle school students the "pulling out of a hat" of

first the $LCM-1$,

and secondly the $LCM*(n)-1$ parts?

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4 Answers 4

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Here’s the explanation that I’d probably use at that level.

Suppose that $n$ is a positive integer satisfying all three conditions. The first condition says that $n+1$ is even; the second says that $n+1$ is a multiple of $3$; and the third says that $n+1$ is a multiple of $4$. Any multiple of $4$ is automatically even, so the first condition is redundant, and we want to choose $n$ so that $n+1$ is a multiple of both $3$ and $4$. The least common multiple of $3$ and $4$ being $12$, this means that we want $n+1$ to be a multiple of $12$. If the concept of least common multiple hasn’t yet been introduced, you can just point out that since $12=3\cdot 4$, any multiple of $12$ is guaranteed to be a multiple of both $3$ and $4$. Either way, it’s now pretty clear that all numbers of the form $12n-1$ work, and it’s almost as easy to see that they’re the only ones that work.

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  • $\begingroup$ @skullpatrol: You’re welcome. $\endgroup$ Sep 5, 2013 at 18:22
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The simple answer could just be write out the sets; i.e,

$S_1=\{...,-3,\color{red}{-1},1,3,5,7,9,\color{red}{11},13,15,17,19,21,\color{red}{23},25...\}$

$S_2=\{...,-4,\color{red}{-1},2,5,8,\color{red}{11},14,17,20,\color{red}{23},26,...\}$

$S_3=\{...,-5,\color{red}{-1},3,7,\color{red}{11},15,19,\color{red}{23},27,...\}$.

As noted in the other answer, immediately we see that $-1,11$, and $23$ satisfy the conditions. From there, we can note that perhaps every $12$th number also satisfies. Then note that $LCM(2,3,4)=12$. Finally construct your formula as $LCM\cdot{n}-1$.

I tried not to use any modular arithmetic since they are middle schoolers, so I hope this helps.

$\mathbf{EDIT}$ After thinking about it, the last thing i would do is verify that the number $12n-1$ leaves the appropriate remainders when divided by $2,3,$ and $4$. Convert $12n-1$ to $12n+11$ for arithmetic's sake and show $$\frac{12n+11}{2}=6n+5+\frac{1}{2}$$ $$\frac{12n+11}{3}=4n+3+\frac{2}{3}$$ $$\frac{12n+11}{4}=3n+2+\frac{3}{4}$$

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    $\begingroup$ You're welcome. I felt like the tangibility of the sets themselves would sit easier with 8th graders. I don't know the curriculum of the middle schools but it's also a good introduction to the idea of a set as well. Glad I could help $\endgroup$ Sep 5, 2013 at 19:17
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(1) there is a remainder of 1 when the number is divided by 2;

(2) there is a remainder of 2 when the number is divided by 3;

(3) there is a remainder of 3 when the number is divided by 4.

Well, the first condition says that $n=2k+1$, the second one says that $n=3k'+1$ and the third one says that $n=4k''+3$. Note that all of them can be written as $n=mk-1$ for some k, where $m=2,3,4$. Now any number of the form $n=Lq-1$ where $L$ is a common multiple of $2,3,4$ will satisfy all the 3 conditions. Definitely, the least common multiple works, and since it's the least one all other common multiples of these numbers will be a multiple of it. So, any such number can be written as $n=12k-1$. This is how I would explain it to them.

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    $\begingroup$ I noticed that your user name is $\exists 1\nu4u$. $\endgroup$
    – dfeuer
    Sep 3, 2013 at 22:17
  • $\begingroup$ That's really cool. Thank you for pointing that out. I'll always use this one wherever they allow LaTeX from now on. $\endgroup$
    – user66733
    Sep 3, 2013 at 22:20
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    $\begingroup$ Even better: $\sum 1\nu4u$ :P $\endgroup$
    – dfeuer
    Sep 3, 2013 at 22:29
  • $\begingroup$ I suppose that doesn't work for Irish M Powers.... although I can get close with $\text{Irish}^{M}s$ $\endgroup$ Sep 3, 2013 at 22:34
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    $\begingroup$ @skullpatrol: well, by division algorithm it's obvious that you can write them as $n=2k+1$, $n=3k'+2$ and $n=4k''+3$. Now, you can write $n=2k+1=2k+(+2-1)=2(k+1)-1$, $n=3k'+2=3k'+(+3-1)=3(k'+1)-1$, $n=4k''+3=4k''+(4-1)=4(k''+1)-1$ setting $k+1=q$ and so forth you can show that all of them are of the form $n=mK-1$ for $m=2,3,4$. $\endgroup$
    – user66733
    Sep 5, 2013 at 13:14
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First, I'd likely flip these numbers a bit as your remainders are going to be "-1" modulo 2,3, and 4 which is something to note here about how you could get the "-1" magically.

Secondly, note that condition 3 implies condition 1. If there are going to be an odd remainder if divided by 4 this implies there would be 1 when divided by 2 which could be shown by looking at odd versus even and then dividing up the odds to being either 1 over a multiple of 4 or 1 under a multiple of 4. In university you could use algebra to show this but at lower grades it may be more important to get the links here.

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