Question 1147211

{{{H + H = 2H = B}}}, so {{{H/B = 1/2}}} 

Then {{{D/J = 1/2}}}, so {{{J/D = 2}}}
 
Hence {{{K = 2}}}. 

We therefore also have a bunch of doubling pairs: 1-2, 2-4, 3-6, 4-8. But {{{K=2}}}, so {{{J}}} and {{{D}}} are either 3-6 or 4-8. 

Note that {{{J}}} &  {{{D}}} are one set, and {{{H}}} & {{{B}}} are the other distinct set. 

Further, {{{D}}} and {{{F}}} are a {{{tripled}}} pair. 

Note that {{{F = 3D}}}, so{{{ D}}} = 1,2,3. But from above, {{{D}}}=3,4, so {{{D = 3}}}. 

Then {{{J = 6}}} and {{{F = 9}}}. Further, {{{HB}}} must be the other double pair available, 4-8. Hence {{{H = 4}}} and {{{B = 8}}}. 

Now, {{{F * D = 9 * 3 = 27 = KG}}}, so {{{G = 7}}}. 

Since {{{J = 6}}} and {{{J + A = CC}}}, {{{J+A < 16}}} so {{{CC = 11}}} and {{{C = 1}}}. 

Further, {{{A = 5}}}. Then {{{A* H = KE}}} => {{{5 * 4 = 2E = 20}}}, so {{{E = 0}}}.


solution: {{{A - K - B - J - C - H - D - G - E - F}}} is {{{5-2-8-6-1-4-3-7-0-9}}}