Least Common Multiple of 3 modulo equations

Question

i%7=1   ->  i=1+7n
i%9=2   ->  i=2+9n
i%11=3  ->  i=3+11n

Besides writing out everything:

1,8,15,22...
2,11,20,29...
3,14,25,36...

to find the LCM. Is there a better way to do this?

Answer 1

In general, you should use the Chinese Remainder Theorem.

However, for these particular numbers, there is a trick! Note that the congruences are equivalent to $2x+5\equiv 0\pmod{7}$, $2x+5\equiv 0\pmod{9}$, and $2x+5\equiv 0\pmod{11}$. So we want $2x+5$ to be divisible by $7$, $9$, and $11$, and therefore by $(7)(9)(11)$, which is $693$. So $2x+5=693$ will work. That gives $x=344$.

Remark: If you are not familiar with congruence notation, look at the sequences of numbers you produced. We will concentrate on the first one, which was $1,8,15,22,\dots$ (there was a mild typo).

Double these, and add $5$. We get $7,21, 35, 49,\dots$, the odd multiples of $7$. Similarly, when you double the numbers in your second sequence, and add $5$, you get the odd multiples of $9$. Similar treatment to the third sequence gets you the odd multiples of $11$. So we want $2x+5$ to be an odd multiple of $(7)(9)(11)$. The smallest positive one is $693$. So now solve $2x+5=693$.

You can also carry through the same argument using the language of remainders. We want $(2x+5)\%7=0$, $(2x+5)\%9=0$, and $(2x+5)\%11=0$.

Answer 2

$\begin{eqnarray}\rm {\bf Hint} &&\rm\qquad\ x &\equiv&\rm n\pmod{2n\!+\!5},\ \ n = 1,2,3\\ \Rightarrow &&\rm\ 2x\!+\!5 &\equiv&\rm 0\pmod{2n\!+\!5},\ \ n = 1,2,3 \end{eqnarray}$

Thus $\rm\:7,9,11\: |\: 2x\!+\!5\:\Rightarrow\: lcm(7,9,11) = 693\:|\:2x\!+\!5.\:$

Therefore $\rm\ mod\ 693\!:\,\ x \,\equiv\, \dfrac{-5}2\,\equiv\, \dfrac{693-5}2 \,\equiv\, 344.$

Remark $ $ The idea generalizes: see CCRT for congruences in A.P. (Arithmetic Progression).