## Divisibility by 4

This is the second post in the Divisibility Rules Series. In the last post, we discussed about divisibility by 2. In this post, we discuss divisibility by 4.

Now, how do we know if a number is divisible by $4$?

Four divides $100$ because $4 \times 25 = 100$. It is also clear that four divides $200, 300, 400$ and all multiples of $100$. Therefore, four divides multiples of $1000$, $10 000$, and $100 000$. In general, $4$ divides $10^n$, where $n$ is an integer greater than $1$.

Now, how do we know if a number that is not a power of $10$ is divisible by $4$. Let us try a few examples.

Example 1: Is $148$ divisible by $4$? $148$ is equal to $100 + 48$ and $100$ is divisible by $4$. Since $48$ is also divisible by $4$, therefore, $148$ id divisible by $4$.

Example 2: Is $362$ divisible by $4$? $362$ is equal to $300 + 62$. Now, $300$ is divisible by $4$. Since $62$ is not divisible by $4$, therefore, $362$ is not divisible by $4$.

Example 3: Is $3426$ divisible by $4$? $3426 = 3400 + 26$. Now, $3400$ is divisible by $4$ (it’s a multiple of 100), and $26$ is not divisible by $4$. Therefore, $3426$ is not divisible by $4$.

By now, you would have realized that we just test the last 2 digits of the numbers if we want to find out if it is divisible by 4: 148, 362, and 3426. » Read more

## Divisibility by 2

How do we know if a number is divisible by a certain number?

In this post, the first post in the Divisibility Rules Series, we examine why a number is divisible by 2 if it is even.  In this post, since we are talking about divisibility rules, when we use the word number, we mean integer.

Since multiplication and division are inverses of each other, we can examine what happens if a number is multiplied by 2. Let’s try a few examples:

0 x 2 = 0
1 x 2 = 2
2 x 2 = 4
3 x 2 = 6
4 x 2 = 8
5 x 2 = 10
6 x 2 = 12
7 x 2 = 14
8 x 2 = 16
9 x 2 = 18

From the list above, we make the following observations: (1) the ones digit of numbers  multiplied by 2 is either 0, 2, 4, 6, or 8; and (2) if the numbers are consecutive the pattern repeats.

Since we have exhausted all 1-digit numbers in the list above, it is clear that the ones digit of a number multiplied by 2 cannot be 1, 3, 5, 7 or 9.  Therefore, we can conclude that a number is divisible 2 if its ones digit is even.

## Problem Set 2

PROBLEMS

1.) Find a linear function $f(x)$ such that $f(1) = 42$ and $f(2) = 47$.

2.) Solve for $x$: $4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 170.$

3.) Prove that the product of $3$ consecutive numbers is always divisible by $6$.

4.) Prove that if $p$ is prime, $a$ and $b$ are integers, and $a \equiv b\mod p$, then $a^p \equiv b^p \mod p$.

SOLUTIONS AND PROOFS

Post Date: October 20, 2009

1. Solution: This is just the same as saying, find the equation of the line passing through $(1,42)$ and $(2,1337)$. So, by point slope formula, we have, $y - y_1 = \displaystyle\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\Rightarrow y - 42 = \displaystyle\frac{1337 - 42}{2 - 1}(x - 1). \Rightarrow y = f(x) = -18x + 92.$

2.) Solution:$4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 4^x(1 + 4 +4^2 +4^3) = 170 \Rightarrow 4^x(85) = 170\Rightarrow 4^x = 2 \Rightarrow 2^{2x}=2^1 \Rightarrow x = \displaystyle\frac{1}{2}.$

3.) Proof: A number is divisible by $6$ if it is divisible by $2$ and $3$. A product of $3$ consecutive numbers is divisible by $2$ because at least one of them is even, so it remains to show it is divisible by $3$.

If a number is divided by $3$, its possible remainders are $0, 1,$ and $2$.  Assume $n, n +1$ and $n+2$ be the three consecutive numbers, and $r$ be the remainder if $n$ is divided by $3$.

Case 1: If $r=0$, we are done.

Case 2: If $r = 1$, then $n + 2 \Rightarrow r=0$

Case 3: If $r = 2$, then $n + 1 \Rightarrow r = 0$.

Since the product of the three consecutive numbers is even, and for each case of $r$, one of the consecutive numbers is divisible by $3$, the product of three consecutive numbers is divisible by $6. \blacksquare$

4.) Proof: From definition, $a^p \equiv b^p \mod p \Leftrightarrow b = a + kp$ for some $k \in \mathbb{Z}.$

Raising both sides of the equation to $p$, we have $b^p = (a + kp)^p.$ By the binomial theorem,  $b^p = (a + kp)^p = a^p + \displaystyle {p \choose 1}a^{p-1}kp + \displaystyle{p \choose 2}a^{p-2}k^2p^2 + \ldots + k^pp^p$.

Notice that every term aside from $a^p$ is divisible by $p^2$. (Why?). Therefore,  $a^p \equiv 0 \mod p^2 .$

Hence, then $a^p \equiv b^p \mod p.$ $\blacksquare$