How many solutions are there to $$x+2y+3z=100$$ with $x,y,z\in\Bbb N$?
Can anyone tell me how to do it? I've got no idea how to start.
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shai horowitz
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ankit gupta
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hint choose z, then choose y such that $2y+3z<100$, then determine x from your equation – shai horowitz Oct 13 '16 at 11:24
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Does $0 \in \mathbb N$ ? – lhf Oct 13 '16 at 11:28
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2Compute the coefficient of $w^{100}$ in the Taylor series of $$\frac{1}{(1-w)(1-w^2)(1-w^3)}$$ around $w=0$, through partial fraction decomposition. You will get that the answer is $\color{red}{884}$ and that there are roughly $\frac{(N+3)^2}{12}$ natural solutions of $x+2y+3z=N$. – Jack D'Aurizio Oct 13 '16 at 12:32
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Related: http://math.stackexchange.com/questions/1529121/the-number-pn-of-triplets-x-y-z-x2y3z-n – Jack D'Aurizio Oct 13 '16 at 12:38
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@JackD'Aurizio - hmm, I get that 884 nicely with a one-liner in Pari/GP using your recipe. However, if I follow Shai Horowitz' brute force recipe I cannot produce that 848 by whatever seeming meningful. However, for the number of $z,y$-combinations only I get 748 and for the overall sum $C$ of possible combinations $(x,y,z)$ I get $C=25736$. Something that I'm missing? – Gottfried Helms Oct 13 '16 at 15:15
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@GottfriedHelms: me and Shai are making different assumptions about $0\in\mathbb{N}$, the discrepancy lies in that. – Jack D'Aurizio Oct 13 '16 at 15:30
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@Jack : ah, I see. I've included that variant in my answer. – Gottfried Helms Oct 13 '16 at 16:46
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Does this answer your question? Counting bounded integer solutions to $\sum_ia_ix_i\leqq n$ – Parcly Taxel Oct 04 '22 at 13:20
2 Answers
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Hint:
If $z\geq 33$, there is no solution.
If $z=32$, then you can set $y=1$ and you must then set $x=2$.
If $z=31$, you need $x+2y=7$, so you get $y=1$ or $y=2$ or $y=3$.
If $z=30$, you can have $y=1,2,3,4$
If $z=29$, you can have $y=1,2,3,4,5,6$
Can you write a pattern? How many options are there for odd values of $z$? How many for even values of $z$?
Remark:
If you consider $0\in\mathbb N$, the solution changes, but the basic idea remains the same.
5xum
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Let's denote the limiting value in your question as $V=100$ and let's assume $x,y,z>0$ (so $ \small 0 \not \in \Bbb N$).
Checking using brute force some smaller $V$, all divisible by $\small 6$, give a polynomially interpolatable sequence of values, producing the interpolating polynomial: $$ P(x)= (2 x^3 -15 x^2+30 x)/72 $$
For values $V$, which are not divisible by $6$ we get a fractional residue , actually $$\small \{P(6k ) \} = 0 \qquad \qquad \{P(6k+1 )\} = 17/72 \qquad \{P(6k+2 ) \} = 16/72 \\ \small \{P(6k+3 ) \} = 9/72 \qquad \{P(6k+4 ) \} = 8/72 \qquad \{P(6k+5 ) \} = 25/72 $$
The function $$ C(x) = \lfloor P(x) \rfloor = \left \lfloor {2 x^3 - 15 x^2+ 30 x\over 72} \right \rfloor $$ gives correct values for all x, for instance $V=100$ we get $C(V)= 25736$ , and for $V=12$ we get $C(V)=23$ (which can be checked by hand).
If we assume $x,y,z \ge 0$ (thus allow $ 0 \in \Bbb N$ ) we get the slightly changed function $$ C_0(x)= \left \lfloor { 2 x^3 + 21 x^2 + 66 x + 72 \over 72} \right \rfloor $$
and get $C_0(100) = 30787$
Gottfried Helms
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