Star equation

Write digits instead of stars so that the sum of the written digits is odd and is true equality:

42 · ∗8 = 2 ∗∗∗

Result

b1 =  48
b2 =  68
c1 =  2016
c2 =  2856

Solution:

a=42 s1=a⋅ 18=42⋅ 18=756 s2=a⋅ 28=42⋅ 28=1176 s3=a⋅ 38=42⋅ 38=1596 s4=a⋅ 48=42⋅ 48=2016 s5=a⋅ 58=42⋅ 58=2436 s6=a⋅ 68=42⋅ 68=2856 s7=a⋅ 78=42⋅ 78=3276 2000<=s<=2999 S={2016,2436,2856} x1=4+0+1+6=11 x2=5+4+3+6=18 x3=6+8+5+6=25 x1=nepar b1=48a=42 \ \\ s_{1}=a \cdot \ 18=42 \cdot \ 18=756 \ \\ s_{2}=a \cdot \ 28=42 \cdot \ 28=1176 \ \\ s_{3}=a \cdot \ 38=42 \cdot \ 38=1596 \ \\ s_{4}=a \cdot \ 48=42 \cdot \ 48=2016 \ \\ s_{5}=a \cdot \ 58=42 \cdot \ 58=2436 \ \\ s_{6}=a \cdot \ 68=42 \cdot \ 68=2856 \ \\ s_{7}=a \cdot \ 78=42 \cdot \ 78=3276 \ \\ 2000 <=s <=2999 \ \\ S=\{ 2016, 2436, 2856 \} \ \\ x_{1}=4+0+1+6=11 \ \\ x_{2}=5+4+3+6=18 \ \\ x_{3}=6+8+5+6=25 \ \\ x_{1}=nepar \ \\ b_{1}=48
b2=68b_{2}=68
c1=a⋅ b1=42⋅ 48=2016c_{1}=a \cdot \ b_{1}=42 \cdot \ 48=2016
c2=a⋅ b2=42⋅ 68=2856c_{2}=a \cdot \ b_{2}=42 \cdot \ 68=2856

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