Quite Strange Commons, aren't they ?

Probability Level 3

How many non-negative integer solutions of the equation $a+2s+3d+4f+5g+6h=99$ are also the solutions of $q+w+o+r+t+y=99$

Details and assumptions :-

$\bullet \quad$ Solutions here means the oredered 6-tuples $(a,s,d,f,g,h)$ and $(q,w,o,r,t,y)$

Are also solutions of means the 6-tuple (a,s,d,f,g,h) which satisfies 1st equation, must also satisfy the 2nd equation (If $(a_0,s_0,d_0,f_0,g_0,h_0)$ is some solution tuple of 1st equation, then $q=a_0 , w=s_0,e=d_0,r=f_0,t=g_0,y=h_0$ must satisfy second equation.

$\bullet \quad \quad a,s,d,f,g,h,q,w,o,r,t,y \in \mathbb{N} \cup$ {0}. (All non-negative integers).

The answer is 1.

1 solution

Aditya Raut
Aug 8, 2014

See that if the solutions are common at all, then you get $a=q,s=w,d=o,f=r,g=t,h=y$

Thus what we obtain is actually

$a+2s+3d+4f+5g+6h=99$ .....................(i)

and also

$a+s+d+f+g+h=99$ .....................(ii)

After a subtraction of second equation from the first, we remain with

$s+2d+3f+4g+5h=0$

Because we have non-negative solutions, we can conclude from this step that $s=d=f=g=h=0$

Thus there will be only one common 6-tuple and it will be $(99,0,0,0,0,0)$

Answer is $\boxed{1}$

Looks tough.But it's too easy.

akhilesh agrawal - 6 years, 10 months ago

Quite Strange Commons, aren't they ?

The answer is 1.

1 solution

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