**Number system puzzle problem 7:** Cube of a number exceeds its product with two other numbers by 2, the cube of the second number is less than...Read on.

## CAT Number System Puzzle Problem 7

Cube of a number exceeds its product with two other numbers by 2, the cube of the second number is less than their product by 3, and the cube of the third exceeds their product by 3. Find the lowest of the three numbers.

**Time to solve:** 90 seconds.

**Hint:** Solve using mathematical reasoning based on indices concepts and inequality analysis.

### Solution to the Number system puzzle problem 7: Mathematical reasoning using indices concepts and inequality analysis

Let $a$, $b$, and $c$ be the first, second and third number. By the problem description then,

$a^3-abc=2$...............(1)

$b^3+3=abc$..............(2)

$c^3-abc=3$...............(3)

Both $a^3$ and $c^3$ are greater than $abc$, but $b^3$ is less than $abc$. So,

**Conclusion 1:** $b^3$ is the lowest among the cubes of the three numbers. Or $b$ is the lowest among the three.

We have to find value of $b$.

Now we will analyze the relative values of $c$ and $a$. Subtracting equation 1 from equation 3,

$c^3-a^3=1$.

This is a key important derived relation.

It establishes that,

**Conclusion 2:** $c$ is greater than $a$.

**Conclusion 3:** From the difference of unity between cubes of two numbers it can be concluded that both $a$ and $c$ cannot be integers.

**Conclusion 4:** Either one or both of $a$ and $c$ must be a cube root of an integer, so that after having raised to the power of 3 their difference can become unity.

**Conclusion 5:** Both cannot have same base of cube roots to have the difference as unity. So even if both of these two are cube roots, their bases are different.

For example, it might be so that $c=7^{\frac{1}{3}}$ and $a=6^{\frac{1}{3}}$.

But as all three terms of all three equations must be integers (from the nature of equations),

**Conclusion 6:** $abc$ is an integer.

It follows from Conclusion 4 and Conclusion 6,

**Conclusion 7:** Both $a$ and $c$ cannot be cube roots, one of the two must be an integer. This is because, otherwise, the third factor $b$ in $abc$ couldn't have in any way provided the compensating shortfall in the power from an integer power as it is the lowest among the three.

For example, if $a=6^{\frac{1}{3}}$, with $c$ an integer, $b$ couldn't have been $b=6^{\frac{2}{3}}$.

**Firm Conclusion 8:** Only one of $c$ or $a$ is a cube root of an integer and the other an integer.

Knowing quite a bit about the nature of the three numbers, we are now in a position to answer the **critical question:**

Critical question:What can be the nature of the three numbers so that $abc$ be an integer?

Analyzing the conclusions above, it follows, the only way $a$ or $c$ be a cube root, $c$ or $a$ be an integer and also $abc$ be an integer is,

**Conclusion 9:** With one of $a$ or $c$ an integer, the other must be in the form, $x=p^{\frac{2}{3}}$ and $b=p^{\frac{1}{3}}$ so that,

$abc=yp$ an integer, where $x$ is the non-integer among $a$ and $c$ and $y$ is the other of the two, an integer.

It follows from the previous result,

**Conclusion 10:** $x=b^2$, a **crucial conclusion**.

Substitute these forms of $a$ and $c$ in equation 2,

$b^3+3=abc=bxy=yb^3$,

Or, $b^3(y-1)=3$.

Only way this relation to be true is,

$b^3=3$,

$y-1=1$.

We need not know the values of $x$ or $y$, that is, $c$ or $a$. The lowest among the three is $b$ and its value is,

$b=3^{\frac{1}{3}}$.

### End note

Though a rather boring series of reasoning and conclusions are shown to reach the end result, in actual exam environment we would reach the answer much faster and long before the end shown by simply trying out one or two small values for $a$ and $c$.

To go through **more such selected hard CAT Quant problems** with **step by step easy solutions,** click * here*.