Question

)Consider the non-negative integer solutions to x1 + x2+ x3 + x4 + x5 = 2020.

(A) How many solutions does Equation (1) have satisfying 0 ≤ x1 ≤ 100? Explain.

(B) Remember to explain your work. How many solutions does Equation (1) have satisfying 0 ≤x1 ≤ 100, 1 ≤x2 ≤ 150, 10 ≤x3 ≤ 220?

Answer 1

ANSWER'. We know that If we have 2,742743....?n=r 7,30,3930.-, zenzo then Number of non-hegative is here Salution cor How Here given that X, +42+ 3+ X4+ *5 = 2020 Tötel Number of sal non-negative sala .X120., X2=0,X320, 420, X5 20} o n=5, r=2020 5 +2020-1 Czozo Czozo oh-regative solu = 2024 A In this part only change is that X, €100. Now need to remove all the possible Sul" which is x, 2101 & X-10820 X,+ x2 + x3+xn* X5 = 2020-101 X + X₂ + *3 + xy + x3 = 1919 Sulution's n=s, r=1919 Ś+1919-1 +18 Cigig = 1923 (1919 Remove 1923 (1g1g from the Totel Salh & we get result - 2024 1923. C2020 47919 - Cagig Are

B). x= 100 EX 25150, LOC83E22o Om X2101, X231504 X₂ 31, X3210, X32221 no 1923 5+(2020-101) – 6-101)-1 cigig C2020-101 for X₂ = 150 4X2 S+(2020-150)-) C2020-150 s+(2020-7)+) C2020-1 20-150 5+ (2020-14) 5+(2020-150)-, C2020-7 -2020-150 = 2023 (2015 - 1874 (1970 for Xz210, 83=221 4 5+(2020-221)-1 St12020-10)-) C2020-22) 2020-10 - 5+(20.20-10)- 5+(2020-22)-1 2020-10 tol = 2014–2010 - 1002 Cina 2020-22 | Hence Remove all this from titel Number of Sulni. 2024 1923 2023 1874 2014 1803 ė - C1919 - ċ toc - C2015 C2010 C1799 1870 1903