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Find the number of solutions to
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 \le 10
in nonnegative integers, where x_2 is even and x_5 is odd.

 May 9, 2024
 #1
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The pair (x2,x5) can be (0, 1), (0, 3), ..., (0, 9), (2, 1), (2, 3), ..., (2, 7), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (8, 1).

 

From here, 5 pairs add up to 9, 4 pairs add up to 7, 3 pairs add up to 5, 2 pairs add up to 3, and 1 pair add up to 1.

 

If x2+x5=k, then x1+x3+x4+x6+x7=10k. The number of nonnegative integer solutions of x1+x3+x4+x6+x7=N is (N+5151)=(N+44). Then, we simply calculate the required number to be 5(1+44)+4(3+44)+3(5+44)+2(7+44)+1(9+44)=1918.

 May 10, 2024
edited by MaxWong  May 10, 2024

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