document.write( "Question 373247: Using the units 0,1,2,3,4,5,6,7,8,9 place the number in the correct position to make the following true equations. Each number can be used only once. (and for GH G= a unit number and and H= a unit number so for example 15 would be G=1 H=5) \r
\n" ); document.write( "\n" ); document.write( "\"2%2BA=B\"
\n" ); document.write( "\"C%2FD=2\"
\n" ); document.write( "\"E%2AF=GH\"
\n" ); document.write( "\"I-1=J\"\r
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\n" ); document.write( "my deductions are as follows
\n" ); document.write( "I Get that \"C%2FD=2\" has only 3 possibilities as \"8%2F4\" \"6%2F3\" and \"2%2F1\"
\n" ); document.write( "I Get that \"2%2BA=B\" B cant = \"1\" or \"0\" , A cant = \"8\" or \"9\"
\n" ); document.write( "I Get that \"E%2AF=GH\" E and F cant be \"1\" or \"0\" and cant be \"2%2A3\" or \"2%2A4\"
\n" ); document.write( "I Get that \"I-1=J\" I cant be \"0\"
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\n" ); document.write( "any and all help is greatly appreciated
\n" ); document.write( "Ive been working on this puzzle for 32hrs and seeking help from all you other bright minds. \n" ); document.write( "


Algebra.Com's Answer #265764 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
\r
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\n" ); document.write( "\n" ); document.write( "I had to hammer-hand this one -- basically eliminating everything that was impossible and doing trial and error on what was left.\r
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\n" ); document.write( "\n" ); document.write( "First there is one possibility and one exclusion that you missed. C = 4, D = 2 is possible. G can be no larger than 7 because the largest E * F can be is 72. (Although in the final analysis neither of these facts had any bearing on the results)\r
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\n" ); document.write( "\n" ); document.write( "I started by saying \"Let A = 0, then B is 2. Then C can't be 2 or 4 (if 4, D = 2), so if C is 6, D is 3. Then I examined all of the possibilities of E*F given the constraints of neither E nor F nor the resulting GH containing 0, 2, 3, or 6 and E or F not equal to 1. For the very few possibilities I got, I listed A through F against a list of the 10 digits to see if the two remaining unused digits were adjacent (so that I and J could differ by 1).\r
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\n" ); document.write( "\n" ); document.write( "Then I tried A=0, B=2, C=8, D=4. Then A=1, B=3, C=4, D=2 (C=2, D=1 repeats 1), and so on.\r
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\n" ); document.write( "\n" ); document.write( "In point of fact, unless you missed including some criteria that wasn't considered, there are 4 answers.\r
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\n" ); document.write( "\n" ); document.write( "A = 7
\n" ); document.write( "B = 9
\n" ); document.write( "C = 6
\n" ); document.write( "D = 3
\n" ); document.write( "E = 5 or 8
\n" ); document.write( "F = 8 or 5
\n" ); document.write( "G = 4
\n" ); document.write( "H = 0
\n" ); document.write( "I = 2
\n" ); document.write( "J = 1\r
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\n" ); document.write( "\n" ); document.write( "A = 7
\n" ); document.write( "B = 9
\n" ); document.write( "C = 8
\n" ); document.write( "D = 4
\n" ); document.write( "E = 5 or 6
\n" ); document.write( "F = 6 or 5
\n" ); document.write( "G = 3
\n" ); document.write( "H = 0
\n" ); document.write( "I = 2
\n" ); document.write( "J = 1\r
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\n" ); document.write( "\n" ); document.write( "There is probably a better way to go about this, and I'm going to chew on it for a couple of days. Write back early next week and I'll let you know if I discovered anything new. Thanks for the interesting challenge.\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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\"The

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