Solving a Diophantine Equation: ( frac{100a10bc}{11} a^2b^2c^2 )

Diophantine equations are a fascinating area of number theory, involving finding integer solutions to polynomial equations. In this article, we will walk through a detailed solution to the Diophantine equation ( frac{100a10bc}{11} a^2b^2c^2 ).

Step 1: Eliminating the Fraction

To start, we simplify the equation by eliminating the fraction. Multiply both sides by 11:

100a10bc 11a^2b^2c^2

Reorganizing the equation, we get:

100a10bc - 11a^2 - 11b^2 - 11c^2 0

This can be further rewritten as:

11a^2 11b^2 11c^2 - 100a - 10b - c 0

Step 2: Analyzing the Variables

Given that (a), (b), and (c) are likely digits (i.e., integers from 0 to 9), we set their ranges as follows:

a, b, c in {0, 1, 2, ..., 9}

Step 3: Testing Possible Values

We will systematically test values for (a), (b), and (c) within their ranges. We calculate the left-hand side (LHS) and the right-hand side (RHS) for combinations of (a), (b), and (c).

Example Calculation

Let's start by testing some values:

Case (a 9): For (b 0), (c 0)

LHS 10091000 900

RHS 119^20^20^2 1181 891 ot 900

This is not a solution.

Case (a 8): For (b 0), (c 1)

LHS 10081001 801

RHS 118^20^21^2 116401 715 ot 801

This is also not a solution.

We will continue testing until we find a valid set of values for (a), (b), and (c).

Step 4: Finding a Valid Set

After testing various combinations, we find:

Case (a 6): For (b 5), (c 0)

LHS 10061050 650

RHS 116^25^20^2 1136250 1161 ot 650

This is still not a solution. Continue the process as needed, but we must check all combinations within the defined ranges.

Conclusion

After testing all combinations of (a), (b), and (c), we find that the solutions are limited. You can use a script or program to automate this testing if needed as it can be tedious to do manually. If you find a solution, it will be in the form of digits (a), (b), and (c) that satisfy the original equation. If no solutions exist within the range of digits, the equation may not yield integer solutions under the provided constraints.

Alternative Approach: Divisibility by 11

We can also approach the problem using the standard test of divisibility by 11. We begin with the equation:

9ab frac{a-bc}{11} frac{100a10bc}{11} a^2b^2c^2ldots 1

This simplifies to:

11 | a-bc

The standard test of divisibility by 11 applied to (n 100a 10b c) means:

Case i: (a - bc 0) Case ii: (a - bc 11)

Case i

If (a - bc 0), then (b ac). The equation reduces to:

10ac a^2b^2c^2 ldots 2

Here, (c) must be even. If (c eq 0), then (c geq 2), and the inequality:

2a^2 - 2a c - 1c^2 - 2 geq 2a^2 - 7a - 8 frac{1}{8}left(16a^2 - 56a - 64right) frac{1}{8}left(4a - 7right)^2 - 15 geq frac{15}{8}

This implies (c 0), and from equation (2), we get (b a 5). Thus:

n 550 11(5^2)(5^2)(0^2)

Case ii

If (a - bc 11), and if (b geq 1), then:

a^2b^2c^2 - left(9ab - frac{a-bc}{11}right) geq a^2b^212 - a^2 - 9ab1 geq 2a^2 - 33a 143 frac{1}{8}left(4a - 33right)^2 - 55 geq frac{55}{8}

This implies (b 0) and so (ac 11). The equation now gives:

9a 1 a^2 11 - a^2 2a^2 - 22a 121

This is a quadratic in (a), solving this we get (a 8) or (a frac{15}{2}). Neglecting the second solution, we get (a 8), so (c 3). Thus:

n 803 11(8^2)(0^2)(3^2)

The only two solutions are 550 and 803. blacksquare