Counting solutions - part II

Given the equation 3 x + 4 y + 12 z + 24 v = 192 3x+4y+12z+24v=192 , find the number of all different solutions ( x , y , z , v ) (x,y,z,v) , where x x , y y , z z , and v v are whole numbers.


The answer is 525.

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