Introduction

Divisibility rules are a collection of specific criteria that apply to a number to determine whether or not it is divisible by a given number. There are some well-known divisibility tests for numbers 2 to 20. It enables us to find factors and multiples of numbers without dividing them by a large number.

Using divisibility rules, a person can determine whether a number is divisible by another integer in their head. Let’s look at divisibility tests in more detail.

What are Divisibility Rules?

A divisibility rule is a type of shortcut that allows us to determine whether a given integer is divisible by a divisor by looking at its digits rather than going through the entire division procedure. When many divisibility rules are applied to the same integer, the prime factorization can be determined quickly. A number’s divisor is an integer that divides the number fully without leaving any remainder.

Martin Gardner, a prominent math and science writer, discussed divisibility principles for 2–12 in a 1962 Scientific American article, explaining that the rules were well-known during the Renaissance and were used to reduce fractions with high numbers to their simplest terms. 

Because no number is divided by any other integer, a leftover other than zero may be left. Some rules can help us figure out what a number’s true divisor is merely by looking at its digits. 

Divisibility Rules Chart and Examples

Let’s look at some instances to help us comprehend how to prove divisibility checks.

• Is the number 280 divisible by two? Yes, because the unit’s place digit is 0, 280 is divisible by two.
• Is the number 345 divisible by three? Yes, 345 is divisible by three since the sum of the digits is 3+4+5, which equals 12, which is divisible by three. As a result, 345 is divisible by three.
• Is the number 450 divisible by four? No, because the number generated by the last two digits starting from the right, i.e. 50, is not divisible by four.
• Is the number 3900 divisible by 5? Yes, 3900 is divisible by 5 since the digit at the unit’s position is 0, which meets the 5 divisibility condition.

• Is the number 350 divisible by 6? The sum of 350’s digits is 8, so it isn’t divisible by three. As a result, it is not divisible by 6, because to be a multiple of 6, a number must be a common multiple of both 2 and 3.
• 357 is divisible by 7 because we get 35-14=21 when we subtract twice of the one’s place digit, 7 2 = 14, from the remaining digits 35, which is divisible by 7. As a result, 357 can be divided by 7.
• Because the number generated by the last three digits 238 is not divisible by 8, 79238 is not divisible by 8.
• Because the sum of all the numbers, 8+7+5= 20, is not divisible by 9, 875 is not divisible by 9.

Inequalities are frequently proved by mathematical induction. There are several families of assertions in which there is an inequality for each natural number. Often, such claims appear to be self-evidently correct, yet constructing proof can be difficult. If this is the case, PMI can help prove disability rules. 

One hint: the inductive phase in a PMI demonstration of inequality usually consists of reasoning that isn’t particularly sharp.

Properties of Mathematical Induction

The above-mentioned property a) is only a factual statement. In these scenarios, when the proposition is true in all cases (assuming n=5), step a) should begin with n=5 and we should test the outcome for n=5, i.e., P. (5).

A conditional property is shown in property b) (conditional property is the occurrence of an event B in relationship to an event A given that event A has already occurred.) since it does not prove that the supplied statement is correct for n=k, but if it is correct for n=k, it should also be correct for n=k+1.

PMI Solution and Proof

P(n), where n denotes the natural number, is an example of a statement. Then apply the following approach to determine the validity of P(n) for each n:

Step 1: Verify that the given statement is correct when n = 1.

Step 2: Assume that the above assertion P(n) holds for n = k, with k being any positive integer.

Step 3: For any positive integer k, prove that the result is true for P(k+1).

If the preceding requirements are met, it can be argued that P(n) is true for all n natural integers.

A Solution to the Problem:

Let’s understand how to state and prove disability rule;

Let’s define Statement P (n) as n 3 + 2n is divisible by 3.

Step 1: The Fundamentals

We start by proving that p (1) is correct. Assume n = 1 and write n 3 + 2n.

2(1) + 1 (3) = 3

Because 3 is divisible by 3, p (1) is true.

Divisibility Rules Tips and Tricks:

• When testing prime numbers, dividing rules are quite important.
• They’re useful for solving word issues.
• They come in handy for rapid computations.
• Every even number can be divided by two.
• Every leap year has a divisor of four.

Conclusion 

All of the above information relates to the applications of PMI in proving divisibility rules!

Hopefully, these notes will assist you in understanding the major themes and remembering the crucial elements for the exam.