Hint: Only $\frac{1}{m_{i}}$ of the integers are $\equiv a_{i}\mod m_{i}$. Now add the $\frac{1}{m_{i}}$, and voila.

It is not true. Must be $a_{i}=a_{j}(mod(gcd(m_{i},m_{j}))$.

Rust wrote: It is not true. Must be $a_{i}=a_{j}(mod(gcd(m_{i},m_{j}))$. ¿¿¿

For example $k=2, a_{1}=1, a_{2}=0, m_{1}=2,m_{2}=4.$

For example: $2,6,10,14,18,...$.

I am sorry I say about x, satisfy all congruences. xeroxia wrote: Let $m_{1},m_{2},\ldots,m_{k}$ be integers with $2\leq m_{1}$ and $2m_{1}\leq m_{i+1}$ for all $i$. Show that for any integers $a_{1},a_{2},\ldots,a_{k}$ there are infinitely many integers $x$ which do not satisfy any of the congruences \[x\equiv a_{i}\ (\bmod \ m_{i}),\ i=1,2,\ldots k.\] In this case problem is trivial. Let $X_{i}=\{x|x\not =a_{i}(mod \ m_{i})\}$. Because $m_{i}\ge 2$ set $X_{i}$ have infinetely many elements. All $x\in X_{i}$ for one i don't satisfyed any congruence.

rust i can't understand what are u talking about the problem asks you to prove your solution not say it again

i think it was a good problem. first we will find what is the number of elements from ${1,...,[m_{1},...,m_{k}]}$ that satisfies at least one of the congruences. say that number $T$. every congruence has at most $[m_{1},... ,m_{k}]\cdot\frac{1}{m_{i}}$ different solution. so we have $T=(\frac{1}{m_{1}}+...+\frac{1}{m_{k}})\cdot[m_{1},... ,m_{k}] \le(\frac{1}{m_{1}}+...+\frac{1}{m_{1}\cdot\ 2^{k-1}})\cdot[m_{1},...,m_{k}]\le\frac{2^{k}}{m_{1}\cdot\ 2^{k-1}}\cdot[m_{1},...,m_{k}]<[m_{1},...,m_{k}]$ hence we have at least one element that doesn't satisfy any of the congruences. moreover if $x$ is a solution $[m_{1},...,m_{k}]+x$ is a solution so there are infinitely many solutions.

It's Pen D 19