High School Number Theory Archives - Page 7 of 11

Divisibility by 8

March 3, 2012 GB High School Number Theory

This is the seventh post in the Divisibility Rules Series. In this post, we will discuss divisibility by 8.

A number is divisible by $8$ if the last three digits is divisible by $8$ . For example, $25816$ is divisible by $8$ since $816$ is divisible by 8. On the other hand, $5780$ is not divisible by $8$ since $780$ is not divisible by $8$ . Why is this so?

Let us start with $25 816$ . First, we know that $1000$ is divisible by $8$ . Therefore, $2000$ , $3000$ , $4000$ , and all multiples of $1000$ are divisible by $8$ . Since $25 816 = 25000 + 816$ and $25 000$ is divisible by $8$ , we just have examine the last three digits. Notice that this is similar to $5780$ . Since $5780 = 5000 + 780$ , and $5000$ is divisible by $8$ , we are sure that it is not divisible by $8$ since the last three digits is not divisible by $8$ .

This observation can be generalized because all numbers greater than $1000$ can be expressed as multiple of 1000 + three-digit number (the hundreds, tens, and ones). Since all multiples of $1000$ are divisible by $8$ , we just have to examine the divisibility of the last three digit number.

Of course this observation is also similar with negative numbers. All negative numbers less than $-1000$ can can be expressed as multiple of -1000 + three-digit negative number.

2 comments divisibility by 8, divisibility rules, divisibility test

Divisibility by 7 and Its Proof

February 29, 2012 GB High School Mathematics, High School Number Theory

This is the 6th post in the Divisibility Rules Series. In this post, we discuss divisibility by 7.

Simple steps are needed to check if a number is divisible by 7. First, multiply the rightmost (unit) digit by 2, and then subtract the product from the remaining digits. If the difference is divisible by 7, then the number is divisible by 7.

Example 1: Is 623 divisible by 7?

3 x 2 = 6
62 – 6 = 56
56 is divisible by 7, so 623 is divisible by 7.

If after the process above, the number is still large, and it is difficult if to know if it is divisible by 7, the steps can be repeated. We take the difference as the new number, we multiply the rightmost digit by 2, and then subtract from the remaining digits.

Example 2: Is 3423 divisible by 7?

3 x 2 = 6
342 – 6 = 336

We repeat the process for 336. We multiply 6 by 2 and then subtract it from 33.

6 x 2 = 12
33 – 12 = 21
21 is divisible by 7, so 3423 is divisible by 7.

Note that if the number is still large, this process can be repeated over and over again, until it is possible to determine if the remaining digits is divisible by 7. » Read more

9 comments divisibility by 7, divisibility by 7 proof, divisibility rules, divisibility rules 1-12, divisibility test

Using Area to Prove the Arithmetic-Geometric Mean Inequality

February 16, 2012 GB High School Algebra, High School Mathematics, High School Number Theory

If we have real numbers $a$ and $b$ , we call $\frac{a + b}{2}$ the arithmetic mean (AM) and $\sqrt{ab}$ the geometric mean (GM) of $a$ and $b$ . In this post, we are going to examine the relationship of these two means.

To start off, let’s have a few examples. If $a = 3$ and $b = 12$ , then $GM = 6$ and $AM = 7.5$ ; if $a = 4$ and $b=16$ , then $GM = 8$ and $AM = 10$ ; if $a = 3$ and $b = 27$ , then $GM = 9$ and $AM = 15$ . What do you observe? Try a few example and see if your observations hold.

From a few examples above, and from your trials, you have probably observed that $GM \leq AM$ which means that $\sqrt{ab} \leq \frac{a + b}{2}$ for some positive real numbers $a$ and $b$ . This is actually true for all positive real numbers $a$ and $b$ . In the following discussion, we are going to use the concept of area to prove that the statement is true.

To begin the proof, we construct a square with side length $a + b$ made up of four rectangles and a square at the center (technically, a square is also a rectangle). Clearly, the area of each of the four rectangles is $ab$ , and the square at the center has are $(a - b)^2$ (Can you see why?). If we remove the square at the center, the remaining area is represented by the equation $(a + b)^2 - (a - b)^2=4ab$ . Note that $4ab$ is the total area of the four rectangles. » Read more

2 comments AM-GM inequality, arithmetic mean, arithmetic mean geometric mean inequality, geometric mean

« 1 … 5 6 7 8 9 … 11 »