Change money math puzzle: Find three money amounts all made up of same three digits and sum of two amounts equal to the third amount. Time to solve 15 mins.

### Change money math puzzle

That was quite some time ago when things were cheap. Dipu found that the amount he had just paid for the roses he bought was a rearrangement of the digits of the amount of money he had in his wallet before the purchase.

When he checked his wallet he was surprised again to find that amount left after the purchase was also nothing but a rearrangement of the same three digits like the other two amounts.

What are the three amounts of money?

**Information:** 100 paise make 1 rupee.

**Recommended time to solve:** 15 minutes.

### Comment

15 minutes time may not be enough if you try out all of the possible combinations of 3 out of 10 digits by trial an error.

**Hint:** There is a faster way to solve the puzzle. And if you take that path you will reach the solution easily in 15 minutes of time.

Try your hand in solving the puzzle first before going through the step by step reasoning based solution.

### Solution to the change money math puzzle: Solving a more convenient essentially same puzzle

We know for certain that the three amounts are in the form of **x.yz** where **x** represents rupees and **0.yz** represents paise. In the three amounts that we are hunting, the three digits **x**, **y** and **z** are in rearranged form.

Each of the three digits may take one of the 10 values 0 to 9 and one amount equals sum of two other amounts.

We recognize that,

Key pattern 1:When we add two three digit money amounts to get a third 3 digit amount all of same 3 digits, the problem is essentially same as adding two three digit numbers to get a third three digit number with all three numbers made up of same three digits.

Decision 1:So we decide to solve this more convenient second problem of finding three numbers of three digit numbers instead of the three money amounts in the original puzzle.

**Reason is:** Both the additions follow the same rules of arithmetic number addition because 100 paise create a carry-over of 1 rupee same as the possible carry-over for the hundred's place in adding the three numbers.

### Solution to the change money math puzzle: Finding three numbers with same three digits

Let us assume the sum number to be "**abc**".

A possible situation is then,

**abc = bac + cab**, **bac** and **cab** being the two components of **abc**.

As none of the three digits can exceed 9, there may only be **two possible relations between the two most significant hundred's place leftmost digits of the two component numbers** - *these two digits may either be different or the same.*

Let's explore first the possibility 1.

### Solution to the change money math puzzle: Hundred's place digits of two components are different

This may be represented by,

**abc = bca + cba**, so that **a = b + c**.

It follows,

**bc = ca + ba**

$\Rightarrow$ **10b + c = 10(b + c) + 2a =12a.**

As **a > both b and c,** the RHS is larger than the LHS and the **above equation is invalid.**

Here we have assumed the largest of the three digits **a** in unit's place for both the component numbers **ensuring the RHS to have the minimum value possible.**

So our firm conclusion is,

Key pattern 2:Thehundred's place digits of the two component numbers must be sameand hundred's place digit of the sum number is twice the hundred's place digit of a component number plus 1 of carry-over.

In the form of an equation then,

**a = 2b + 1** for **abc = bca + bac**, and,

**a = 2c + 1** for **abc = cab + cba**.

### Solution to the change money math puzzle: Hundred's place digits of two components are same

There can then be two possibilities,

**abc = bca + bac**, with **a = 2b + 1**

**abc = cab + cba**, with **a = 2c + 1**.

Let's analyze the first situation **abc = bca + bac**, where **a = 2b + 1,**

$\Rightarrow$ **100(a - 2b) + bc = ca + ac**

$\Rightarrow$ **100 = 10c + a + 10a + c - 10b - c = 11a - 10b +10c = 11(b + c) - 10b + 10c = b + 21c.**

This again is invalid as with its highest possible value 4 of **c**, **21c** reaches 84. Reaching 100 by adding single digit **b** is impossible.

It follows,

Key pattern 3:The unit's digit of the original amount is the same as hundred's place digits of the two components, that is,abc = cab + cba.

### Solution to the change money math puzzle: Hundred's place digits of two components are same as unit's place digit of sum number

We'll analyze **abc = cab + cba** with certainty of validity of the equation.

**abc = cab + cba**, with **a - 2c = 1**,

$\Rightarrow$ **100 + 10b + c = 10a + b + 10b + a**

$\Rightarrow$ **100 = 11a + b - c.**

As all three digits a, b and c are of values 0 to 9, **b - c** **has value less than 9.**

With **a = 8** the equation would be invalid.

The only possible values of the three are then,

**a = 9, b - c = 1.**

But as **a = 9 = 2c + 1**,

**c = 4** and** b = 5**.

Sum and components are,

**954 = 459 + 495**.

In money terms answer is,

**Rs. 9.54 = R. 4.59 + Rs. 4.95.**

We can't say which of the two components is the change in the wallet and which is the amount paid for the roses.

### End note

Only the simple **concept of place values** is used for the solution.

Though it may seem to be long, when you actually solve the puzzle this way without the responsibility of explaining in details, the solution won't take much time.

The main thing that you need is to have the confidence that the simple looking not so simple puzzle can be solved systematically.

This approach is what we call **systematic problem solving.**

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