3 digit number math puzzle

Submitted by Atanu Chaudhuri on Thu, 17/12/2020 - 23:30

$3 digit number math puzzle$

Use basic concepts and find patterns that can be used in reasoning out the solution

The puzzle

If one-fifth of a 3 digit positive integer number equals product of its digits, find the number.

No trial and error please.

Recommended time: 20 minutes.

Comments

Essentially, this is not a hard puzzle, but it is interesting and intriguing enough to engage attention of common folks like us with only the basic number system concept base.

The puzzle can be solved easily by trial and error or a combination of reasoning based on basic number system concepts and trial and error.

When trial and error is to be completely avoided, the problem attains it best.

This puzzle is a good example of problem solving in math and is rich in potential for learning use of number system concepts among others.

Solution to the math puzzle - one-fifth of a 3 digit integer equals product of its digits

Let's repeat the description of the essentials of the puzzle for easy reference.

Puzzle: If one-fifth of a 3 digit positive integer number equals product of its digits, find the number.

First stage of problem solving: Precise problem definition

In any word problem in math the first step is to express the description in terms of math relations. Only then the analysis, reasoning and manipulation of the terms using maths become possible.

Expressing the puzzle in more mathematical form,

Puzzle: If (abc)/5=abc, find (abc), where (abc) is the 3 digit positive integer (the expression is not fully mathematical yet, but at the start this is what is needed).

It is expected that basic number system concept is to be applied to the equation to transform it for easier analysis.

But before any further action, the second step of drawing out all information hidden in the description is to be taken.

What more information can we draw from the puzzle description?

As $a$, $b$ and $c$ are the digits of the three digit number each of these three can have values only in the range of 1 to 9.

This completes the precise problem definition.

None of the digits can be 0, because then the RHS will become 0, and the puzzle won't have a unique answer.

Second stage of problem solving: Analysis and pattern identification: Identifying and applying most appropriate number system concept

What is the basic concept that is most appropriate in this case?

Guiding criterion for choosing the Place value mechanism for solving number system problems or puzzles is satisfied by this problem description,

Whenever a number system problem or puzzle has both the number and its digits related in an expression (description), the place value mechanism concept is to be used.

This is analysis of the math relation and identifying the key pattern at this point of solution.

Using place values for the digits of the number,

$\displaystyle\frac{100a+10b+c}{5}=a\times{b}\times{c}$.

Now the equation has become easier to simplify to,

$20a+2b+\displaystyle\frac{c}{5}= a\times{b}\times{c}$.

This is the time to apply basic number system concept to form the reasoning,

The product on the RHS being a whole number, $20a+2b+\displaystyle\frac{c}{5}$ must be an integer and so, $c$ must be a multiple of 5. As it cannot be more than 9, $c=5$.

This is the first major breakthrough.

Substitute value of $c$ to simplify the equation,

$20a+2b+1=5a\times{b}$.

Third stage of problem analysis: Identifying and applying most appropriate number system concept: Advanced stage

What important conclusion can we draw now?

But before thinking any further, as a must-do step to be taken at any intermediate stage, let us simplify the LHS taking 2 as a factor of first two terms,

$2(10a+b)+1=5a\times{b}$.

It is easier to apply the mathematical reasoning on this form,

As RHS has the factor 5, the unit's digit of the LHS has to be 5 to satisfy the equation (because of the third term of 1, the LHS can only be 5 and not 0).

By divisibility concept for 5, any number to be divisible by 5 its unit's digit must be 0 or 5.

It implies then,

For unit's digit of $2(10a+b)+1$ to be 5, the unit's digit of $2(10a+b)$ must be 4. And so, $(10a+b)$ can have its unit's digit only 7 or 2 (so that multiplying with 2, the unit's digit of $10a+b$ becomes 4).

This is going deeper into unit's digit analysis, a very important tool in solving number system problems.

What more can we discover at this point?

The next conclusion is staring at our face,

For $10a+b$ to have the unit's digit 7 or 2, effectively value of $b$ must be 7 or 2, that simple.

Fourth stage of problem analysis: Identifying and applying most appropriate number system concept: Advanced stage

The title of this and the earlier stage are nearly same. It is so because in abstraction (that's what is being done in both stages), only the context (or situation) and outcomes (or results) are different.

As was done before trying to discover a new concept to apply, at this last stage also, the equation is further transformed to a suitable form,

$20a+2b+1=5ab$.

As target is now to get an additional binding condition on the value of $b$, the reasoning is simply,

With both $a$ and $b$ non-zero positive integers, for the equation $20a+2b+1=5ab$ to be true,

$5ab > 20a$,

Or, $b > 4$.

This is the powerful concept and technique of inequality algebra (or inequality analysis).

Finally, as $b$ can only be 2 or 7 and $b > 4$,

$b=7$.

And value of $a$ given by,

$20a+2b+1=5ab$,

Or, $20a+15=35a$,

Or, $a=1$.

The number is 175.

This is systematic problem solving in maths by identifying key patterns and applying the concepts using the patterns.

Use of a number of basic number system concepts along with mathematical reasoning for solving one problem enriches overall basic maths concepts.

Practice of this approach strengthens ability to solve larger and more complex problems (of various types, not only mathematical, because essentially the approach is subject independent).