Let $n$ be a positive integer such that $12n^2+12n+11$ is a $4$-digit number with all $4$ digits equal. Determine the value of $n$.
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wcao9311
03.06.2014 15:56
SMOJ wrote: Let $n$ be a positive integer such that $12n^2+12n+11$ is a $4$-digit number with all $4$ digits equal. Determine the value of $n$. By this there would exist an $m$ such that $1111m=12n^2+12n+11$ m must be equal to $1$ mod $4$,because that way when subtracted by 11 it is still a multiple of 4. So there are 3 cases to check,$1,5,9$. We further use the property that the sum of the digits mod 3 = 2, so when subtracted by 11, which mod 3 =2, it becomes divisible by 3(factors of 12). then the equation becomes $5544=12n^2+12n$ which becomes $n^2+n-462=0$ and then we find the solutions of this equation, only one is positive. This factorizes into $(n-21)(n+22)$ so the POSITIVE integer solution would be 21. QED.
DIffCALCFTW
23.06.2016 00:59
We know that the 4-digit number has to be odd, therefore let's say we have a number of the form $xxxx$ where $x=1,3,5,7$ or $9$. If we take $\mod 3$ then that means that $12n^2+12n+111 \equiv 2 \mod 3$. Therefore our four-digit number $xxxx \equiv 2 \mod 3$ and $4x \equiv 2 \mod 3$. This problem is now trivial by CRT, the only congruence that fits the description above is when $x=5$ so now our equation is $$12n^2+12n+11=5555$$which simplifies to $$n^2+n-462$$. You can factorize this into $$(n-21)(n+22)$$where the only positive integer solution is $21$. $$Q.E.D$$
Math_CYCR
23.06.2016 06:39
Since any four digit number with all four digits equal is divisible by $11$, we get
$12n^2+12n+11 \equiv n^2+n=n(n+1) \equiv 0 \pmod{11}$
Therefore $11 \mid n$ or $11 \mid n+1$
Check the cases $n=10, 11, 21, 22$. If $n$ it's greater than that, then $12n^2+12n+11$ would be greater than $10000$.
After checking, we get the only solution is $n=21$
talkinaway
23.06.2016 08:17
$12n^2 + 12 + 11$
$12n^2 + 12n + 12 - 1$
$12(n^2 + n + 1) - 1$
$12(n^2 + n + 1)$ is one of $1112, 2223, 3334, 4445, 5556, 6667, 7778, 8889, 10000$.
The only one that is a multiple of $12$ is $5556$.
Setting that equal gives $n^2 + n + 1 = 463$. Since $21 < \sqrt{463} < 22$, we're looking for $n = \boxed{21}$, which works.
Side note: Sometimes you can do too much work. I "simplified" $n^2 + n + 1$ into $\frac{n^3 - 1}{n - 1}$. But that doesn't really prove to be necessary, and actually hurts the solution.