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# Linear Diophantine Equation #4

Linear Diophantine Equations
1. C + D = B + E
2. B + C = A + E
3. A - D = E - 2

## Hint

Sometimes, under the pressure of time, you need to make a trade-off between solving it analytically and solving it with trial and error. In this case, a combination of both will leads to the quickest solution. First, sum up all three equations, in order to eliminate 3 out 5 variables.

Sometimes, under the pressure of time, you need to make a trade-off between solving it analytically and solving it with trial and error. In this case, a combination of both will leads to the quickest solution. First, sum up all three equations, in order to eliminate 3 out 5 variables.

C+ D = B + E

B + C = A + E

A - D = E - 2

——————— +

2C + B + A = 3E + B + A - 2

Which we can simplify as:
• 2C = 3E - 2
Here, your quickest approach would be to try different values. Let's try out all numbers ranging from 1-5 for C.
• If C= 1, 3E should equal 4, which isn't possible with integers.
• If C= 2, 3E should equal 6, which isn't possible, because E = 2 would result in a non-unique solution.
• If C= 3, 3E should equal 8, which isn't possible with integers.
• If C= 4, 3E should equal 10, which isn't possible with integers.
• If C= 5, 3E should equal 12, which works! E = 4.
So now we have:
• C = 5
• E = 4
Now we can substitute these values in the first two equations:
• 5 + D = B + 4 (or D = B - 1)
• B + 5 = A + 4 (or A = B + 1)
From this, we know that D < B < A, so:
• D = 1
• B = 2
• A = 3
So the full solution is:
• A = 3
• B = 2
• C = 5
• D = 1
• E = 4