Question 940253

the first digit is {{{odd}}}:  could be {{{1}}},{{{3}}},{{{5}}},{{{7}}},{{{9}}}    
 the first digit is {{{larger}}} than the {{{third}}}  digit ...could be {{{0}}},{{{1}}},{{{2}}},{{{3}}}   so, first digit is {{{3}}}  

since the first digit is  larger than the third  digit, then the third  digit   is   {{{2}}} (looking at {{{0}}},{{{1}}},{{{2}}} it is {{{third}}})

the fourth digit is three times the first...so, it could be only{{{ 9}}} because {{{3*3=9}}}


 the final digit is the {{{product}}} of the second and third digit ....   remaining digits that could be the second  digit are  {{{4}}},{{{5}}},{{{6}}},{{{7}}} and {{{8}}}, and only product that gives us {{{one}}} digit number using third digit {{{ 2 }}} is {{{2*4=8}}} because {{{2*5=10}}} and the rest is same

so,  the second  digit is {{{4}}} and the final digit is {{{8}}}

then we have all digits to place in empty boxes:
the first digit is  {{{3}}}
 the second  digit is {{{4}}}
the third digit is {{{ 2 }}}
the fourth digit is {{{9}}}
the final digit is {{{8}}}