# The number of non-negative integer solutions of $a + 2b + 3c + 4d + 5e + 6f = 10000000$ [closed]

Find how many solutions are there to the equation $a + 2b + 3c + 4d + 5e + 6f = 10000000$ where $a, b, c, d, e, f$ are all non-negative integers?

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why should we find it? More seriously, what did you try? Do you have any idea where to start? Where did that question come from? –  mau Oct 2 '13 at 10:01
i have found a similar question on stackexchange The solutions of x+2y+3z=n,(x,y,z∈N) –  Agarwal Shubham Oct 2 '13 at 10:11
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## closed as off-topic by Arthur, azimut, Daniel Rust, Davide Giraudo, SeiriosOct 2 '13 at 10:57

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## 1 Answer

Just over $1.1574 \times 10^{30}$

See OEIS A001402 for approximate and exact formulae.

This can be solved by recursion: my Java applet would try to calculate it with "Partitions of $10000000$" with "any number of terms" and "each part no more than $6$", but seems to get vey slow on my machine for partitions of more than $4000000$.

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Thanks, I got a formula on OEIS and finally got the answer. –  Agarwal Shubham Oct 2 '13 at 10:41
how we solve if i use 5a + 6b + 7c + 8d + 9e + 10f instead of a + 2b + 3c + 4d + 5e + 6f. –  Agarwal Shubham Oct 2 '13 at 10:43
That is the number of partitions of $n$ where each part is between 5 and 10. There will be a simple generating function to provide an answer. –  Henry Oct 2 '13 at 22:47
u mean 1/((1 - x^5)*(1 - x^6)*(1 - x^7)*(1 - x^8)*(1 - x^9)*(1 - x^10)) but the problem statement ask to solve 11a+12b+13c+14d+15e+16f=1234567891011121314 here is the link: [The number of solutions(2) ](mathalon.in/Mathalon/?page=show_problem.php&pid=570) –  Agarwal Shubham Oct 3 '13 at 11:39
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