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I wish to solve following problem $$x+11y+11z=n(n\in N^{+})$$ has $16653$ triples $(x,y,z)$ of postive integers.

Find $n_{\min}$

Of course, I can't solve it by Now, so there any solution?

Problem 2: let $a_{n}$ be numbers of triples $(x,y,z)$ of postive integers such that $$x+11y+11z=n$$ Find a closed form for the $a_{n}$?

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  • $\begingroup$ 1) What have you tried ? 2) What do you mean by "find closed $a_n$ "? A closed form for the $a_n$ ? $\endgroup$ – Jean Marie Mar 7 '16 at 9:49
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The problem can be separated in two operations:

(op. 1) find all $x,k \in \mathbb{N^*}$ such that $x+11k=n \ \ (1) \ $, then

(op. 2) find all $y,z \in \mathbb{N^*}$ such that $y+z=k \ \ (2)$.

Let $m=\lfloor n/11 \rfloor $ (integer part of $n/11$).

For any $k=2\cdots m$, there are $k-1$ solutions to (2), thus $k-1$ sol. to (1) as well.

As $k$ cannot exceed $m$ (as can be seen from (2)), the total number of solutions in $\mathbb{N^*}$ is:

* from here, changes have been made *

$$\begin{cases}1+2+\cdots+(m-1)=m(m-1)/2 & \text{if n is not a multiple of 11}\\ 1+2+\cdots+(m-2)=(m-2)(m-1)/2 & \text{otherwise} \end{cases}$$

This separation in two cases, signaled by @geromty (I had not seen it), comes from the fact that if $n$ is a multiple of $11$, one must not give value $0$ to $x$.

Thus:

  • either $n$ is not a multiple of $11$ and we solve $m(m-1)/2=16653$, with integer solution $m=183$, then find the smallest $n$ such that $\lfloor n/11 \rfloor =183$, with $n$ non multiple of $11$. This value is $n=11 \times 183+1=2014$.

  • or we have to solve $(m-2)(m-1)/2=16653$ giving $m=184$ and thus $n = 11 \times 184 = 2024$, which is higher than $2014$.

So the answer to the problem is $n=2014$.

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  • $\begingroup$ Hello,why $\dfrac{m(m+1)}{2}=16653?$ not $\dfrac{m(m-1)}{2}=16653?$$ $\endgroup$ – geromty Mar 7 '16 at 13:32
  • $\begingroup$ you are right, it's a typo. I correct it. $\endgroup$ – Jean Marie Mar 7 '16 at 13:39
  • $\begingroup$ In fact, I think I have a shift of one unit. I have to correct it. $\endgroup$ – Jean Marie Mar 7 '16 at 13:48
  • $\begingroup$ Thanks,ButI found have some wrong,because $[n/11]=183$,I think $n=11\times 183+1=2014?$ $\endgroup$ – geromty Mar 7 '16 at 13:59
  • $\begingroup$ No, you have not to add 1. $\endgroup$ – Jean Marie Mar 7 '16 at 14:15

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