CAT Rules to determine the last two digits of the number. MCQ

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Rules to determine the last two digits of the number.

Introduction

Determining the last two digits of a number is an important concept in quantitative aptitude. It helps in simplifying calculations and making them more efficient. This concept is frequently tested in management entrance exams like CAT, MAT, XAT, SNAP, etc. In this lesson, we will learn the rules

While the cyclicity concept helps to find the unit digit in a power series, the process to determine the last two digits of an exponent involves a more complex concept involving modular arithmetic and Euler's theorem. Modular arithmetic is the arithmetic of congruences, while Euler's theorem is a generalisation of Fermat's little theorem that relates powers of integers to their inverses modulo a prime number.

Euler's Totient Function (φ)

Euler's Totient function φ(n) is an important function when we discuss number theory. The function φ(n) represents the count of numbers that are less than n and relatively prime to n (gcd(x, n) = 1).

Euler's Theorem

Euler's theorem states that if 'a' and 'n' are relatively prime $\mathrm{(i.e., gcd(a,n) = 1), then \; a^{\varphi (1)} \equiv 1 (mod \; n)}$ . This theorem can be used to simplify large exponents, like finding the last two digits of a power.

Modular Arithmetic

Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" after reaching a certain value, known as the modulus. When we are concerned with the last two digits of a number, we operate under mod 100, as the last two digits are the remainder when a number is divided by 100.

Applying Cyclicity, Euler's Theorem, and Modular Arithmetic

When asked to find the last two digits of a large exponent, the steps are as follows:

1. Determine the value of $\mathrm{\varphi (100) = \varphi (2^2 \times 5^2) = 100(1-\frac{1}{2})(1-\frac{1}{5}) = 40}$ . Here we use the property of the Euler's Totient function which states that $\mathrm{\varphi (p1^k1 * p2^k2 *...* pn^kn) = p1^k1 * (1-1/p1) * p2^k2 * (1-1/p2) *...* pn^kn * (1-1/pn) where \; p1, p2,..., pn}$ are prime factors.

2. If the number a is coprime with 100, the last two digits of a^n would be the same as the last two digits of $\mathrm{a^{(n\; mod 40)}}$ (according to Euler's Theorem).

3. If the number a is not coprime with 100, we will have to solve this in a different way. Separate out the part which is not coprime and the part which is coprime with 100, solve them separately, and multiply the results.

Example: Find the last two digits of $2^{567}$ .

Solution: To find the last two digits of $2^{567}$ , we can use Euler's Totient Function and Euler's Theorem.

Euler's Theorem states that if a and n are coprime (i.e., gcd(a, n) = 1), then:

$\mathrm{[ a^{\phi(n)} \equiv 1 \mod n]}$

where $\mathrm{$\phi(n)$}$ is the Euler's Totient Function of n.

The Euler's Totient Function, $\mathrm{$\phi(n)$}$ , gives the number of integers less than (n) that are coprime to n. For a prime number $\mathrm{(p), (\phi(p) = p-1\)}$ .

For our problem, we want to find the last two digits of $2^{567}$ , so we are interested in the modulus 100.

The prime factorization of 100 is $(2^2 \times 5^2)$ . Using properties of the Euler's Totient Function, we have:

$\phi(100) = \phi(2^2) \times \phi(5^2)$
$[ \phi(2^2) = 2^2 - 2^1 = 2 ]$

$[ \phi(5^2) = 5^2 - 5^1 = 20 ]\\$

$[ \phi(100) = 2 \times 20 = 40 ]$

Now, using Euler's Theorem:

$2^{40} \equiv 1 \mod 100$

$(2^{40})^{14} \equiv 1^{14} \mod 100$

$2^{560} \equiv 1 \mod 100$ $2^{560} \equiv 1 \mod 100$

Now, multiplying both sides by $$2^7$:$

$2^{567} \equiv 2^7 \mod 100$

Now, we can compute $2^7 \mod 100$ to get the last two digits of $2^{567}$ .

Let's calculate $2^{7}$ mod 100.

We know:

$2^7 = 128$

Now, to find $(2^7 \mod 100)$ , we simply take the remainder when 128 is divided by 100.

$[2^7 \mod 100 = 128 \mod 100 = 28]$

Thus, the last two digits of $(2^{567})$ are 28.

Key Takeaways

- Cyclicity for the last two digits involves advanced number theory concepts including modular arithmetic and Euler's theorem.

- Euler's theorem helps simplify calculations involving large exponents.

- The last two digits of an exponent can be found using the above-mentioned process. For numbers that are coprime with 100, Euler's theorem can be applied directly. For numbers that are not coprime with 100, they must be broken down and solved separately.