Question

Formulate a conjecture about the final two decimal digits of the square of an integer. Prove your conjecture using a proof by cases.

Short answer

The final decimals of \({n^2}\) consists of.0,1,4,5,6,9.

Step-by-step solution

Introduction

An integer n can be expressed as \(\left( {10b + a} \right)\) here \(a\)is in units and \(b\)is in tens.

If \(n = 10b + a\)then \(b = \frac{{n - a}}{{10}}\).

Note that,

\(\begin{aligned}{c}{n^2} = {\left( {10b + a} \right)^2}\\ = 100{b^2} + 20ab + {a^2}\\ = 10b\left( {10b + 2a} \right) + {a^2}\end{aligned}\)

The final decimal digit of \({n^2}\)is the same as the final decimal digit of \({a^2}\).

Also the final decimal digit of \({a^2}\)is the same as the final decimal digit of:

\(\begin{aligned}{c}{\left( {10 - a} \right)^2} = 100 - 20a + {a^2}\\ = 10\left( {10 - 2a} \right) + {a^2}\end{aligned}\)

To prove the conjecture

If the final digits of n are 1 and 9:

As \({1^2} = 1\)or \({9^2} = 81\)then the final decimal digit of \({n^2}\)is the same as the final decimal digits which is 1.

If the final digits of n are 2 and 8:

As \({2^2} = 4\)or \({8^2} = 64\)then the final decimal digit of \({n^2}\)is the same as the final decimal digits which is 4.

If the final digits of n are 3 and 7:

As \({3^2} = 9\) or \({7^2} = 49\) then the final decimal digit of \({n^2}\)is the same as the final decimal digits which is 9.

More cases for final result

If the final digits of n are 4 and 6:

As \({4^2} = 16\)or \({6^2} = 36\)then the final decimal digit of \({n^2}\)is the same as the final decimal digits which is 6.

If the final digits of n are 5:

As \({5^2} = 25\)then the final decimal digit of \({n^2}\)is the same as the final decimal digits which is 5.

If the final digits of n are 0 and 10:

As \({0^2} = 0\)or \({10^2} = 100\)then the final decimal digit of \({n^2}\)is the same as the final decimal digits which is 0.

Thus, the final decimals of \({n^2}\)consists of 0, 1, 4, 5, 6, 9.