Question 935268

{{{BAT=90}}}   -> factors of  {{{90}}}  are {{{2*5*3*3}}};  since we have three letters, we use factors {{{2*5*9}}}
{{{LET=168}}} -> factors of  {{{168}}}->three of them that make {{{168}}} are {{{ 8*3*7}}}
{{{BET=105}}} ->three of them that make {{{105}}} are {{{3*5*7}}}


{{{BAT=90}}} => factors  are->  {{{2*5*9}}}
{{{BET=105}}} =>factors are {{{5*3*7}}}

since we have letter {{{B}}} in {{{BAT}}} and {{{BET}}} -> common factor in these words is {{{5}}}, so {{{highlight(B=5)}}}


a letter {{{T}}} is in all three  {{{BAT}}} ,{{{LET}}} and {{{BET}}} 
{{{BAT=90}}} => factors  are->  {{{2*5*9}}}
{{{LET=168}}}-> factors are->{{{8*3*7}}}
{{{BET=105}}} =>factors are {{{5*3*7}}}

-> since {{{9=3*3}}} greatest divisor of {{{90}}} is {{{9}}}, so,  {{{highlight(T=9)}}}

since we have letter {{{E}}} in {{{LET}}} and {{{BET}}} 

{{{LET=168}}}-> factors are->{{{8*3*7}}}
{{{BET=105}}} =>factors are {{{5*3*7}}}-> common factor in these words is {{{7}}}, so {{{highlight(E=7)}}}


a letter {{{A}}} is {{{BAT}}},  since   {{{B=5}}} and {{{T=9}}}, letter {{{A}}} must be {{{2}}};so, {{{highlight(A=2)}}}


a letter {{{L}}} is {{{LET}}}, {{{LET=168}}}-> factors are->{{{8*3*7}}}

 since   {{{E=7}}} and {{{T=9}}}, letter {{{L}}} must be  {{{8}}};so, {{{highlight(L=8)}}}


so, the value of {{{TABLE=9*2*5*8*7=highlight(5040)}}}