# Distributive Law for Indeterminate Problems

##### Ho Soo Thong

Copyright © June 2015 AceMath, Singapore.

First Revised Edition, July 2016

**Abstract**

This article extends the use of Arithmetic Distributive Law in the Bar Model Method to indeterminate distributive problems.

**Introduction**

Following the article *Distributive Law for Excess and Shortage problems* (see [1]), we will continue with the pictorial approach for solving some simple *indeterminate* problems at primary school level.

First, we recall the bar modelling approach with *Distributive Law* to a counting problem.

### Example 1

A basket of tomatoes arrived in bags of 8.

When the tomatoes were repacked into bags of 5, there were 4 more bags and 1 tomato leftover.

How many tomatoes were there in the basket?

### Solution

First, we construct bar models for both scenarios. With the *condition* that there are 4 more bags of 5, we express the *difference *4×5 + 1 = 7×3, as 7 times of 3 (= 8 − 5 )as shown in Figure 1. By identity of *Distributive Law*, there were 7 bags of 8 tomatoes.

Notes :

(a) In the comparison bar models

number of 8 = number of 5 = number of 3 (=8 − 5)

(b) In an algebraic approach, we assume that there are n bags of 8 and m bags of 5 and solve the simultaneous equations

8n = 5m + 1

m = n + 4 ( condition )

## Indeterminate Problems

Next, we proceed further with a similar problem without the condition.

### Example 2

A basket of tomatoes arrived in bags of 8. When the tomatoes were repacked into bags of 5, there were 2 tomato left.

What is the least number of tomatoes in the basket?

### Solution

In Figure 2, we construct bar models for both scenarios and look for the sum 5 + 5 + 2 = 12 least multiple of 3 (= 8 − 5 )then obtain the difference = 4× 3.

Applying the identity of Distributive Law, the corresponding least number of bags of 8 is 4 as shown.

Therefore the least number of tomatoes is 4×8 = 32.

Note : In **Example 2**, we solve the equation 8n = 5m + 2 without any condition.

The equation 8n = 5m + 2 without any condition is known as an *indeterminate equation*. It is also known as a *Linear Diophantine Equation* (named after Diophantus of Alexandria (Ancient Greek).

There are multiple solutions to this indeterminate problem.

Next we will continue to find other possible values of the difference, as a multiple of 3, and the corresponding possible number of tomatoes:

Figure 3 shows a scenario where the difference as a larger multiple of 3 is 3×5 + 12 = 27 = 9×3 and the corresponding number of bags of 8 is 9. Therefore, there are 9×8 = 72 tomatoes.

Similarly, we have another scenario where the next larger number of tomatoes is 14×8 = 112 as shown in Figure 4.

There are infinitely many possible differences that are multiples of 3 and so there are infinitely many possible solutions.

Figure 5 shows a general solution.

The number of tomatoes is (5k + 4) ×8 = 40k + 32 where k = 0, 1, 2, 3,….

This above example illustrates an intuitive view of the general scenario of an indeterminate equation.

An algebraic approach to the general solution to the Diophantine equation 8n = 5m + 2 is given in [2].

Finally, we will see how the bar modelling of Distributive Law can help with visualising and solving a challenging question similar to a past PSLE question (See [3]).

### Example 3

Some candies can be packed into boxes of 3 and boxes of 5 with no candies leftover. When the candies are packed into boxes of 8, there is 1 candy leftover.

What is the least possible number of candies?

### Solution

First, we construct bar models for the scenario where the candies can be packed into boxes of 3 and boxes of 5. So the candies can be put into boxes of 15 with no candies leftover as shown in Figure 6.

Next we construct bar models for the resulting scenario and the other scenario in Figure 7. To apply the Distributive Law, we obtain the differences 9, 17, 25, 33, 41, 49 ( = 7 × 7 ) as the least multiple of 7 ( = 15 − 8 ).

Therefore, the least number is 7 × 15 = 105.

Note that the bar models with bars of 3 and the bars of 5 gives the bar model with bars of 15 ( = 3×5), the numbers 3 and 5 are said to be *relatively prime* and this nature of numbers was mentioned in Euclid Element II. This nature of numbers is extremely useful in advanced counting problems. An example is the well known Chinese Remainder Problem posed by Sun Tsu Suan-Ching (4th century AD) (See [2].)

There is an unknown number x.

When x is divided by 3, the remainder is 2.

When x is divided by 5, the remainder is 3.

When x is divided by 7, the remainder is 2.

What is x?

Figure 8 shows the bar modelling of the three scenarios.

In Figure 8, the 1st and the 3rd bar models have the same remainder 2. Since the numbers 3 and 7 are relatively prime, the resulting bar model consists of a bar of 2 followed by bars of 3×7= 21 as shown in Figure 9.

In Figure 9, the least number for x is 23 and the numbers 5 and 21 (= 3×7) are relatively prime numbers. The resulting bar model consists of a bar of 23 and followed by bars of 5×21 = 105.

There are infinitely possible numbers of x

23, 23 + 105, 23 + 2×105, … 23 + k×105 where k = 0,1,2,…

and it is also an indeterminate problem.

*Remarks*

The pictorial bar modelling approach enables intuitive problem solving process and highlights the use of basic properties of numbers and mathematical laws. It provides and encourages intuitive learning process in mathematics at school level.

**References**

[1] Ho Soo Thong, *Distributive Law for Excess and Shortage Problems in PSLE Math AceMath*, Singapore 2015

[2] Ho Soo Thong, *Bar Model Approach to Linear Diophantine Equations*, AceMath, Singapore 2013

[3] *PSLE Examination Questions 2005 – 2009*, Educational Publishing House Pte Ltd