<s>[INST] (CHMMC #8 Individual, 2016) Find the smallest $n$ such that $n^2 \,\% \, 5 < n^2 \,\%\, 7 < n^2 \,\%\, 11 < n^2 \,\%\, 13?$ 
Define $n\,\%\,d$ as the remainder when n is divided by d. What is the smallest
  positive integer n, not divisible by 5, 7, 11, or 13, for which $n^2 \,\%
\, 5 < n^2 \,\%\, 7 < n^2 \,\%\, 11 < n^2 \,\%\, 13?$

So basically I tried using quadratic residues, but as each mod generates 3,4,6, or 7 quadratic residues, I would have to bash out all $3\cdot 4\cdot6\cdot7$ possibilities to find an answer. Also, through a computer program, I found that $n=19$. Is there an easier method or a method to solve this without a computer?
 [/INST] I found an elementary solution only using congruences and trial-and-error. Given $x\in\Bbb{N}$ such that $5\not\mid x$ and $7\not\mid x$, then $x^2\equiv 1,4\pmod 5$ and $x^2\equiv 1,2,4\pmod 7$. So if $n^2\equiv 4\pmod 5$ it's impossible to have $n^2 \,\%
\, 5 < n^2 \,\%\, 7$. Now, $n^2\equiv 4\pmod 5$ iff $n\equiv2, 3\pmod 5$, therefore if $n$ holds the required condition, $n$ must satisfy $n\equiv \pm1\pmod 5$. 
It's obvious that if $n^2\le 13$ it's impossible to have the required condition, so $n^2>13$, which give us $n\ge 4$. We now check case by case:
If $n=4$, then $16\,\% \, 11=5$, but $16\,\% \, 13=3$.
If $n=5$, then $n$ is divisible by $5$.
If $n=6$, then $36\,\% \, 5=36\,\% \, 7=1$.
If $n=7$, then $n$ is divisible by $7$.
If $n=8$, then $8\not\equiv \pm1\pmod 5$.
If $n=9$, then $81\,\% \, 7=81\,\% \, 11=4$.
If $n=10$, then $n$ is divisible by $5$.
If $n=11$, then $n$ is divisible by $11$.
If $n=12$, then $12\not\equiv \pm1\pmod 5$.
If $n=13$, then $n$ is divisible by $13$.
If $n=14$, then $n$ is divisible by $7$.
If $n=15$, then $n$ is divisible by $5$.
If $n=16$, then $256\,\% \, 7=4$, but $256\,\% \, 11=3$.
If $n=17$, then $17\not\equiv \pm1\pmod 5$.
If $n=18$, then $18\not\equiv \pm1\pmod 5$.
If $n=19$, then $361\,\% \, 5=1$, $361\,\% \, 7=4$, $361\,\% \, 11=9$ and $361\,\% \, 13=10$.
Hence, the smallest number which satisfies the required condition is $n=19$.
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