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IMO u cannot get an answer here if we assume 2x3y5z mean 2*x*3*y*5*z
if f(m) = 9(v), if we assume that xyz for m and abc for v, xyz = 9 abc.
Now, x , y, z will individually have a relation with a b c which will be based on 3 factors of 9, i.e. 1 *9*1 and 1*3*3.
with each set, if v is abc we can have M as = [(a*1)(b*9)(c*1)] or (a*9)(b*1)(c*1) or (a*1)(b*1)(c*9)
this will give you m -v as 80, 800, or 8.
if we have the other set, M will be [(a*1)(b*3)(c*3)] or (a*3)(b*3)(c*1) or (a*3)(b*1)(c*3)
then m-v can be any value. as we have two possible digits if multiplied by 3 gives us another digit. so if u assume v to 222
then m is 266, 662, 626 / and if v is 333 then it m can be 399, 993, 939 so m-v can take different values.
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