Detailed Solution and Discussion: Australian Mathematics Competition 2021 – Intermediate, Problem 27

Let's discuss a problem from the AMC 2021 Intermediate: Problem 27 which revolves around basic counting.

Problem

What is the smallest natural number \(n\) such that the number

\(N=100000 \times 100002 \times 100006 \times 100008+ n\)

is it a perfect square? (AMC 2021 Intermediate: Problem 27)

Solution

Let's $100000 = x$

Thus, $100002 = x+2$, $100006 = x + 6$, $100008 = x + 8$

Thus the above expression becomes:

\(N=100000 \times 100002 \times 100006 \times 100008+n\) = \(x \times (x+2) \times (x + 6) \times (x+8) + n\)

If we calculate: \(x \times (x+2) \times (x + 6) \times (x+8) + n\) = \((x^2 + 2x) \times (x+6) \times (x+8) + n\)

= ( (x^3 + 2x^2 + 6x^2 + 12x) \times (x+8) + n\)

= \(x^4 + 2x^3 + 6x^3 + 12x^2 + 8x^3 + 16x^2 + 48x^2 + 96x + n\)

= \(x^4 + 16x^3 + 76x^2 + 96x + n\) -----------------------(a)

Let's compare it with a general formula:

\((x^2 + ax + b)^2 = (x^2 + ax + b) \times (x^2 + ax + b) \)

Let's expand this general expression:

\(x^4 + 2ax^3 + a^2x^2 + 2x^2b + 2axb + b^2\) -----------------------------------(b)

Comparing the equation (a) and (b) we get:

\(2a = 16, a = 8\) ----------------(c)

\( (a^2 + 2b = 76\)

\(2ab = 96 \Rightarrow ab = 48\)

Thus, \(b = 48 \div 8 = 6\)

Again, \(b^2 = n = 36\)

Thus \(N = (x^2 + 8x + 6)^2\)

So, 36 is the smallest value of n to make the expression a perfect square number.

What is AMC (Australian Mathematics Competition)?

The Australian Mathematics Competition (AMC) is one of Australia's largest and oldest annual mathematics competitions, aimed at fostering interest and excellence in mathematics among students.