Even stranger commons !!!

How many non-negative integer solutions of the equation \[a+4b+c+3d+3e+3f+6g+6h=20\]are also the solutions of \[j+2k+l+3m+n+x+p+6q=20\]

Details and assumptions:-

\(\bullet \quad \) Solutions here means the ordered 8-tuples \((a,b,c,d,e,f,g,h)\) for the first equation and \((j,k,l,m,n,x,p,q)\) for the second equation.

\(\bullet \quad \) Are also solutions of means the 8-tuple (a,b,c,d,e,f,g,h) which satisfies 1st equation, must also satisfy the 2nd equation.

(If \(a_0,b_0,c_0,d_0,e_0,f_0,g_0,h_0)\) is some solution tuple of 1st equation, then \(j=a_0 , k=b_0,l=c_0,m=d_0,n=e_0,x=f_0,p=g_0,q=h_0\) \(\color{Red}{\textbf{must}}\) satisfy second equation.

\(\bullet \quad \quad a,b,c,d,e,f,g,h,j,k,l,m,n,x,p,q \in \mathbb{N} \cup\){0}. (All non-negative integers).

Problem 1 of this type was a bit easier, and it gives hint for this one.


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