SOLUTION: 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 a

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: 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 a      Log On

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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)
2%2BA=B
C%2FD=2
E%2AF=GH
I-1=J


my deductions are as follows
I Get that C%2FD=2 has only 3 possibilities as 8%2F4 6%2F3 and 2%2F1
I Get that 2%2BA=B B cant = 1 or 0 , A cant = 8 or 9
I Get that E%2AF=GH E and F cant be 1 or 0 and cant be 2%2A3 or 2%2A4
I Get that I-1=J I cant be 0


any and all help is greatly appreciated
Ive been working on this puzzle for 32hrs and seeking help from all you other bright minds.
Answer by solver91311(24713) About Me  (Show Source):

You can put this solution on YOUR website!


I had to hammer-hand this one -- basically eliminating everything that was impossible and doing trial and error on what was left.

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)

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).

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.

In point of fact, unless you missed including some criteria that wasn't considered, there are 4 answers.

A = 7
B = 9
C = 6
D = 3
E = 5 or 8
F = 8 or 5
G = 4
H = 0
I = 2
J = 1

A = 7
B = 9
C = 8
D = 4
E = 5 or 6
F = 6 or 5
G = 3
H = 0
I = 2
J = 1

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.

John

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