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

Related Posts Plugin for WordPress, Blogger...