Reverse cheque puzzle solution

Many ways to the solution through exploration and critical reasoning

reverse cheque puzzle solution

We will repeat the problem for ease of explanation.


The problem

This is an incident that happened many years ago.

One day Bhabababu, a fairly wealthy man, decided to test the world-worthiness of his young grandson and asked him, “Khokon, can you get this cheque cashed in my bank? You know where it is. Last month I took you with me.”

Khokon was a smart young lad full of enthusiasm. He said proudly, “Yes Dadu, of course.”

So he took the cheque from his Dadu, went to bank, and confidently cashed the cheque.

At home when he handed over the money to his Dadu, Bhabababu counted the money twice. He couldn’t believe it. With deep surprise mixed with a tinge of anxiety he asked, “Khokon, what did you do with the money?”

Khokon replied apprehensively, “Why Dadu, I have spent only 20 paise from the money I got from the Bank as tram fare. Did I do anything wrong?” His Dadu replied, “No Khokon, you didn’t. The cashier did.”

Being keen on mathematical puzzles, he understood the mystery.

Continuing, he instructed Khokon, “The money that I got is exactly double of what I wrote in the cheque. By mistake the cashier had given you the rupees written as paise and paise written as rupees in my cheque. Here, take this extra money. I have added 40 paise for your two-way tram fare. Go quick Khokon. The cashier might be in trouble!”

Two questions:

  1. What was the cheque amount, and how did you get the answer?
  2. Can you deduce it mathematically, meaning, can you produce a deduction that is full of equations and mathematical procedures?

Information: 100 paise made 1 rupee.


The mathematical solution

Solution 1: Mathematical and deductive

Problem definition

Let us define the problem in specific terms.

There are four event stages.

Stage 1: Cheque written: Let us assume, $x$ be the cheque rupees and $y$ be the cheque paise.

Stage 2: Cheque cashed in bank. Making a grave mistake cashier gave $y$ rupees and $x$ paise to Khokon.

Stage 3: Khokon spent 20 paise from the amount he had received.

Stage 4: Bhabababu received from Khokon double the amount in cheque.

Domain condition: In real world transactions, the paise written in cheque or received from bank must be less than 100. If any paise figure equals or exceeds 100, 1 is added to Rupee figure and the rest paise forms the Paise figure.

Problem solving

Deduced condition: In stage 1 and stage 2, as both $x$ and $y$ have to represent paise, these must each be less than 100.

Given condition: The final received amount is double the cheque amount. So,

$A_{final}=2A_{cheque}$, where the symbol $A$ represents amount

Or, $A_{bank given}-\text{Paise}(20)=A_{final}=2A_{cheque}$, as after expending 20 paise from bank given amount the final amount is formed, final mount is 20 paise less than the bank given amount

Or, $100y+x- 20=2(100x+y)$, converting the rupees to paise to get the total amount as well as express all variables in unit of Paise, where,

$A_{cheque}=\text{Rs.}(x) +\text{Paise}(y)$, and because of reversing,

$A_{bank given}=\text{Rs.}(y) +\text{Paise}(x)$, and

$\text{Rs.}(1)=\text{Paise}(100)$.

We will call this as Doubling equation.

As rupee amount figures have factor 100, the doubling equation to be valid, the rupee figures $y$ on the LHS and $x$ on the RHS play the most important roles, and mathematically,

$y \geq 2x$.

Scenario 1: $y=2x$

Substituting in doubling equation,

$200x+x-20=200x+4x$,

Or, $3x=-20$.

As $x$ can't be negative, $y$ can't be equal to $2x$.

So valid Scenario 2 is,

$y \gt 2x$.

Let us assume,

$y=2x +c$, where $c \gt 0$,

As $y \lt 100$, $x \lt 50$.

Substituting $y=2x+c$ in the doubling equation $100y+x-20=2(100x+y)$ we get,

$200x + 100c + x - 20=200x + 4x +2c$,

Or, $98c-20=3x$.

The RHS being less than 150 (as $x \lt 50$), value of $c$ cannot exceed 1. If $c=2$ the LHS becomes 176 which is larger than 150. With increasing value of $c$ the shortfall increases.

So, $c=1$, that is, $y=2x+1$.

Substituting this value of $y=2x+1$ finally in doubling equation $100y+x-20=2(100x+y)$ we get,

$200x +100+x-20=200x+4x+2$,

Or, $3x=78$,

Or, $x=26$,

And,

$y=2x+1=53$.

Original cheque amount was Rupees 26 and Paise 53.

The unique solution is possible because of the doubling relation, reversing event, value of both $x$ and $y$ serving the role of Paise being less than 100 and consequent inequality analysis.

Solution 2: Mathematical reasoning

The essence of the problem definition can be expressed in the first two steps followed by reasoning steps in series that lead finally to the solution.

Doubling: Bank given amount is double the cheque amount plus 20 paise, as after expending 20 paise from the bank given amount, the amount finally given to checque writer becomes double the cheque amount.

Reversing: Bank given rupee figure same as cheque paise figure and bank given paise figure same as cheque rupee figure. There are only two unique figures and as both played the role of paise figure in one amount or the other, both figures must be less than 100.

Reasoning: As bank given amount is more than double the cheque amount, bank given rupee figure, that is, the cheque paise figure must at least be double the cheque rupee figure, that is, the bank given paise figure. So, Bank given paise figure is less than the cheque paise figure.

Reasoning: bank given paise figure can be less than the cheque paise figure even after doubling it and adding 20 paise to the result, only if,

cheque paise figure is equal to or more than 50 (and less than 100) so that after doubling the figure, 100 paise out of it is converted to 1 rupee and added to double the cheque rupee figure, thus forming the bank given rupee figure, as well as, the rest of the paise (after doubling, subtracting 100 and adding 20 to it) becomes equal to cheque rupee figure which is less than the cheque paise figure.

This is the core internal mechanism of the problem. We have discovered by reasoning how the peculiar results could happen, but we are yet to find the actual values of two unique figures.

Observe how we have eliminated the need of considering the final amount figures, and could focus on the two sets of cheque figures and bank given figures that were reversed and intimately related.

Second stage of problem solving

Reasoning: As bank given rupee figure, that is, cheque paise figure is twice the cheque rupee figure plus 1, it must be odd and so it must at least be 51, not 50.

Reasoning: With minimum value of cheque paise figure being 51, the minimum value of cheque rupee figure must be 25. We are homing on to the feasible range of values of cheque rupee figure through more detailed look at the feasibilities.

Reasoning: As cheque rupee figure, that is, bank given paise figure is formed after doubling the cheque paise figure, subtracting 100 and adding 20, it must be even. So minimum feasible value of bank given paise figure, that is, cheque rupee figure is 26 and not 25.

If we are curious, with no clear further narrowing down avenues visible, we may straightway test out this minimum value to see whether it satisfies all the problem conditions and is the solution.

Testing cheque rupee figure of 26:

With cheque rupee figure 26, cheque paise figure as well as the bank given rupee figure each is $2\times{26}+1=53$.

Let us check the bank given paise figure now.

Bank given paise figure is,

$2\times{53}-100+20=26$.

We get this value of bank given paise figure exactly reversed from the cheque rupee figure and so, all problem conditions are satisfied. The solution to the reverse cheque problem thus is,

Cheque was written as,

Rupees 26 and Paise 53.

Though this path to the solution seems to be ok, we might not have reached the solution at the first test itself. So we have to consider this solution as not the assured and methodological solution.

Solution 3: Mathematical reasoning: Assured and methodological

Instead of testing out with minimum cheque rupee figure of 26 directly, we will apply Range division technique on the range of feasible values of cheque rupee figure as 26 to 48 to test all feasible values and speed up the process simultaneously.

Range division technique

This general technique is derived from our experience of searching a specific value out of a set of uniformly increasing (or decreasing) series of values, and is stated as,

Divide a uniformly increasing (or decreasing) set of values into two equal parts and select the middle (approximately) value for the operation intended on each of the values for getting the desired result. If the result of the operation produces the desired result, the selected value gives the solution. Otherwise depending on the result of the operation we repeat the range division, middle value selection and result analysis on the lower or upper half of the set of values.

The beauty of this simple technique is, on each range division process, we reduce the present feasible range of values by half and also move towards to solution inexorably.

Applying range division technique on the uniformly increasing series of feasible values of cheque rupee figure, 26, 28, 30,......,46, 48, we select the approximately middle value of 34 for our first test.

First test with cheque rupee figure 34

With cheque rupee figure as 34, cheque paise figure as well as bank given rupee figure each becomes, $2\times{34}+1=69$.

The bank given paise figure will then be, $2\times{69}-100+20=58$. It exceeds cheque rupee figure of 34 (to which it must be equal) by 24. This selection of test value of cheque rupee figure is not then the solution.

We further observe that if we increase the test value from 34, the excess amount increases uniformly. For the next higher test value of 36, the excess would be 30 an increase of excess of 24 by 6. So we need to test with cheque rupee figure smaller than 34.

At this stage, we can form a hypothesis with this result that with every decrease of 2 in cheque rupee figure the excess paise amount would reduce by 6.

At this point itself, following the hypothesis we may decide to reduce 34 by $4\times{2}=8$ to reduce the excess of $4\times{6}=24$ to zero. The selected value of cheque rupee figure would then be $34-8=26$ and on verification it would indeed confirm that our hypothesis was correct.

But again our hypothesis may not have been right which makes this solution not fully assured.

Fully assured solution 4 by mathematical reasoning using Short jump technique

In case forming a hypothesis on the basis of the test result is difficult or time consuming, recognizing that the test value to be selected should be less than 34, we won't select the next lower even value of 32. Instead we would take a short jump over this next value of 32 and land onto 30 to carry out the next test.

The logic behind this short jump technique is,

  • If the present test value satisfies problem conditions, the selected value will give us the solution. Otherwise,
  • If the nature of present test result is reversed from the nature of previous test result (excess becomes shortfall), the jumped over feasible value (of 32 in this case) should provide us the solution if at all a solution exists. Otherwise,
  • If the nature of present test result is same as the nature of previous test result (excess still remains an excess), take another short jump.

The advantage of using this technique is, we cut down the number of values to be tested again by half.

Following this technique and making no hypothesis, proceeding thus in a fully assured but at the same time fairly efficient path of solution, we would next test cheque rupee figure as 30.

Second test with cheque rupee figure as 30

With cheque rupee figure as 30, cheque paise figure as well as bank given rupee figure each becomes, $2\times{30}+1=61$.

The bank given paise figure will then be, $2\times{61}-100+20=42$. It exceeds cheque rupee figure of 30 (to which it must be equal) by 12 now. This selection of test value of cheque rupee figure is not then the solution.

As excess still persists, taking the next short jump we would select now cheque rupee figure as 26 and on verification would confirm it to be the solution.

Final remarks

While actually solving a puzzle, usually we, the commmon people, (excluding the gifted ones as well as people with random approach) adopt reasoning along with suitable methods and techniques to get to the solution with assurance as well as at reasonable speed. This is what we call problem solver's approach.

Assurance of the right solution is as important to us as cutting down the time to solve.

Precise and concise problem definition (the problem essence or problem core) and problem analysis to discover the anatomy of the problem and the main hurdles to the solution should contribute greatly in selecting the right methods and techniques to reach the solution in assured manner as well as efficiently.

Though we have shown the usefulness of forming a hypothesis from the results achieved midway to the solution only sketchily, in solving complex real life problems (including business problems of course), forming hypothesis in stages and testing out the hypothesis methodically play vital roles towards efficient and useful solution.

By our experience, mostly mathematical and conventional approaches to significant problems almost always prove to be inefficient as well as tend to hide the true nature of the problems.

Lastly, once we reach the solution to a problem, invariably we explore other avenues to the solution, if resources permit. This is application of the powerful many ways technique that,

  • enhances power of discovering new possibilities
  • reveals the geography around a problem as well as its anatomy through various solutions
  • gives an opportunity of creating and applying new methods and techniques, and
  • opens up the area of comparing various processes to a solution for assurance, efficiency and other desirables.

Recommendation: Try to improve the solutions and discover new solutions.