Solving Mathematical Puzzles: A Diophantine Equation Approach

Introduction

Mathematical puzzles like those involving distribution and remainders can be quite intriguing and challenging. This article explores a specific puzzle, providing a step-by-step solution and explanation. We will delve into the methods used to solve such problems using Diophantine equations, which are equations requiring integer solutions.

The Puzzle

The puzzle in question involves distributing sweets among children and determining the total number of sweets. The puzzle states that when sweets are distributed among 25 children, 8 sweets remain. When distributed among 28 children, 22 sweets remain. We need to find the total number of sweets.

Mathematical Formulation

Let ( N ) be the total number of sweets. According to the problem, we can write:

When distributed among 25 children: [ N 25k 8 ] When distributed among 28 children: [ N 28m 22 ]

Setting these two expressions equal to each other gives:

[ 25k 8 28m 22 ]

Rearranging the terms, we obtain the Diophantine equation:

[ 25k - 28m 14 ]

Solving the Diophantine Equation

First, we rearrange the equation for clarity:

[ 25k 28m 14 ]

We need to find integer solutions for ( k ) and ( m ). This can be simplified by expressing ( k ) in terms of ( m ):

[ k frac{28m 14}{25} ]

For ( k ) to be an integer, ( 28m 14 ) must be divisible by 25. We can simplify this by reducing the equation modulo 25:

[ 28m 14 equiv 3m 14 pmod{25} ]

Setting this congruence to 0, we get:

[ 3m 14 equiv 0 pmod{25} ]

Subtracting 14 from both sides, we obtain:

[ 3m equiv -14 pmod{25} ]

Since (-14) is equivalent to (11) modulo 25 (because (-14 25 11)), we have:

[ 3m equiv 11 pmod{25} ]

Next, we find the multiplicative inverse of 3 modulo 25. Using the Extended Euclidean Algorithm, we have:

[ 25 8 cdot 3 1 ]

This shows that the inverse of 3 modulo 25 is 17, since multiplying both sides by 17 gives us:

[ 3 cdot 17 equiv 1 pmod{25} ]

Multiplying both sides of the congruence (3m equiv 11 pmod{25}) by 17 gives:

[ m equiv 11 cdot 17 pmod{25} ] [ m equiv 187 pmod{25} ] [ m equiv 12 pmod{25} ]

This means ( m 12 25t ) for any integer ( t ). Substituting ( m 12 ) back into the expression for ( N ):

[ N 28 cdot 12 22 ] [ N 336 22 ] [ N 358 ]

Verification

Let's verify the solution:

For 25 children, distributing 358 sweets, we get: [ 358 div 25 14 text{ remainder } 8 ] For 28 children, distributing 358 sweets, we get: [ 358 div 28 12 text{ remainder } 22 ]

The solution is correct.

Conclusion

The total number of sweets is ( boxed{358} ).