Divisibility Rules and its applications

Here is the link to the divisible rules

Here are more practices

Sum up the divisible rules:

For any power of 2:

If the number is 2ⁿ , check the last n digit number to see if it's divisible by 2ⁿ.

2¹= 2   Check the last digit to see if it's even.

For example: 6, 10, 38, 104 are divisible by 2.

2² = 4  Check the last two digit to see if it's divisible by 4.

For example, 312, 348, 997640 is divisible by 4 because the last two digit in each number, which

are "12", "48" and "40" is divisible by 4.

2³ = 8  Check the last three digit to see if it's divisible by 8.

For example, 312312, 348640, 5248 is divisible by 8 because the last three digit in each number, 

which is "312", "640" and "248" is divisible by 8.

etc,...

For any power of 3: 

Add up all the digit in that number and see if it's divisible by that multiple of 3.

3¹= 3, add up all the digits in that number and check if it's divisible by 3.

For example, 108, 3465, 6534 is divisible by 3 because the sum of the digit in each number, 

which is "9" (1+8), "18" (3 + 4 + 6 + 5) and again "18" is divisible.

3² = 9, add up all the digits in that number and check if it's divisible by 9.

For example, 225, 198765, 5634 is divisible by 9 because the sum of the digit in each number, 

which is "9" ( 2+ 2 + 5), " 36" (1 + 9 + 8 + 7 + 6 + 5) and  "18" (5 + 6 + 3 + 4) is divisible by 9.

An easier way to check is to cross out the multiple of 9 or the combination of digit sum that is a multiple

of 9.

3³ =  27  Check the last three digit to see if it's divisible by 27.
etc...

For divisible rule for 5, the unit (one's digit) digit has to end in either 5 or 0.

For divisible rule for 6, the number has to be divisible by both 2 and 3 because 6 = 2 x 3

For divisible rule for 10, the unit (or one's) digit has to end in 0.

For divisible rule for 11, the difference of the sum of alternative digit has to be multiple of 11,
such as 0, 11 or 22, etc...

For example, 908050. The sum of alternate digit is (9 + 8 + 5) and (0 + 0 + 0) and their difference
is 22 - 0 = 22 so 908050 is divisible by 11.

For divisible rule for 12, the number has to be divisible by 4 and 3 since 12 = 4 x 3

To check if a number is divisible by 2, 4 or 8, you can just check 8 since 2 and 4 are factors of 8 so 

any number that is divisible by 8 is divisible by 2 and 4 but not vice versa.

Questions to ponder: (Answer key and solutions below)

1. How many positive integers less than 200 are divisible by 2, 3 and 5?

2. How many positive three-digit integers are divisible by neither 3 nor 5?

3. The number A346A is divisible by 9. What number does the letter "A" stand for?

4. What is the smallest four-digit number that is divisible by 2, 3, 4, 5, 6, 7 and 8?

5. The number 9A8B is divisible by 11. How many pairs of distinct (A, B) meet the criteria?

6. What is the smallest 3 digit number that is divisible by 3 and 5 but not by 10?
7. What is the sum of all the positive integers less than 105 that is a multiple of 3 or 5? 

Answer:
#1: 6   The least common denominator of 2, 3, and 5 is 2 * 3 * 5 = 30 since the three numbers are relatively prime. = 6
#2:  480   There are 999-100 + 1 or 999 - 99 = 900 three digit numbers. Of those,  numbers are divided by 3 and  numbers are divided by 5 and   numbers are divided by both and thus are repeated (over-counted twice). Total there are 300 + 180 - 60 = 420  numbers that are divisible by 3 or 5. So there are 900 - 420 = 480 three digit numbers that are divisible by neither 3 nor 5.

#3: The divisible rule for 9 is to add up all the sum of the digits in that number and check if it's divisible by 9.
2A + 3 + 4 + 6 = 13 + 2A a multiple of 9 which means that it has to be an odd sum for the 2A to be integer. 9 * 3 = 27
13 + 2A = 27, 2A = 14 so A = 7

#4: The smallest number that is divisible by 2, 3, 4, 5, 6, 7 and 8 is 2 * 3 * 2 * 5 * 7 * 2 = 840. ( or 8 * 7 * 5 * 3 because both 2 and 4 are factors of 8 and for multiple of 6, you only need an extra 3)
The smallest four digit number is thus 840 * 2 = 1680

#5:  (9 + 8) - (A + B) = multiple of 11 so A + B = 6 or A + B = 17
(A, B) = (9, 8) ; (8, 9); (6, 0); (0, 6); (5, 1); (1, 5); (4, 2); (2, 4) and so 8 pairs

#6: To be divisible by 5, the unit digit has to be 5 or 0; however, since the question says it's not divisible by 10 so the unit digit is 5. To be divisible by 3, the sum of the digit is divisible by 3, so 105 is the smallest three digit number that works.
The next number would be 105 + 30, and continue adding multiple of 30. 135, 165, 195, etc...

#7: For multiples of 3 less than 105, you have 3, 6, 9, 12, 15 ...to 102.
For multiples of 5 less than 105, you have 5, 10, 15,...to 100.
The LCM of 3 and 5 is 15 and the multiples of 15 will be repeated twice for both number multiples so we
need to subtract one set to avoid duplicates.
3 (1 + 2 + 3 + ...34) + 5 (1 + 2 + 3 + ...20) - 15 (1 + 2 + ...6) = 3 *   + 5 *   - 15 *
(sum of the first n natural numbers) = 2520

Triangular Numbers & Word Problems: Chapter Level

Triangular Numbers :  From Math is Fun. 1, 3, 6, 10, 15, 21, 28, 36, 45...

Interesting Triangular Number Patterns: From  Nrich

Another pattern: The sum of two consecutive triangular numbers is a square number.

What are triangular numbers? Let's exam the first 4 triangular numbers:

The 1st number is  "1".
The 2nd number is "3" (1 + 2)
The 3rd number is "6" (1 + 2 + 3)
The 4th number is "10" (1 + 2 + 3 + 4)
.
.
The nth number is   It's the same as finding out the sum of the first "n" natural numbers.

Let's look at this question based on the song "On the Twelve Day of Christmas"
(You can listen to this on Youtube,)

On the Twelve Day of Christmas
On the first day of Christmas
my true love gave to me
a Partridge in a Pear Tree

On the second day of Christmas,
My true love gave to me,
Two Turtle Doves,
And a Partridge in a Pear Tree.

On the third day of Christmas,
My true love gave to me,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree

On the fourth day of Christmas,
My true love gave to me,
Four Calling Birds,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree.


The question is "How many gifts were given out on the Day of Christmas?"

Solution I"
1st Day:       1
2nd Day:      1 + 2
3rd Day       1 + 2 + 3
4th Day        1 + 2 + 3 + 4
.
.
12th Day      1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12

Altogether, you'll have 12 * 1 + 11 * 2 + 10 * 3 + 9 * 4 + 8 * 5 + 7 * 6 + 6 * 7 + 5 * 8 + 4 * 9 + 3 * 10 + 2 * 11 + 1 * 12 = 12 + 22 + 30 + 36 + 40 + 42 + 42 + 40 + 36 + 30 + 22 + 12 = 364

Solution II: The sum of the "n" triangular number is a tetrahedral number.
To get the sum, you use 
n = 12 and   

Here is a proof without words.

Another proof.

Applicable questions: (Answers and solutions below)

#1 Some numbers are both triangular as well as square numbers. What is the sum of the first three positive numbers that are both triangular numbers and square numbers?

#2 What is the 10th triangular number? What is the sum of the first 10 triangular numbers?

#3 What is the 20th triangular number?

#4 One chord can divide a circle into at most 2 regions, Two chords can divide a circle at most into 4 regions. Three chords can divide a circle into at most seven regions. What is the maximum number of regions that a circle can be divided into by 50 chords? 

#5: Following the pattern, how many triangles are there in the 15th image? 

Answers: 
#1 1262  The first 3 positive square triangular numbers are: 1, 36 (n = 8) and 1225 (n = 49).

#2 55; 220    a.  ;   b.

#3 210    

#4 1276 
1 chord :    2 regions or + 1
2 chords:   4 regions or + 1 + 2
3 chords:   7 regions or + 1 + 2 + 3
.
.
50 chords: + 1 + 2 + 3 + ...+ 50 = 1 + =1276

#5: 120
The first image has just one triangle, The second three triangles. The third 6 triangles total. 
It follows the triangular number pattern. The 15th triangular is 

For Young Mathletes 3

May 23th, 2012 homework for young Mathletes.

May 23rd, 2013 for Young Mathletes Download answer key here for easier self studying and feedback.

Please give me feedback on the comment.
I would like to use students' problems soon.

Thanks a lot. 

Prime Factorization : Part II

   Learn basic facts here.











Another fun trick is to find the total number of factors any number has. To do this, take the prime factorization of the number, add 1 to each of the exponents of the prime factor, and multiply these new numbers together.

Example:

#1: 12 = 2² x 3¹
(2+1) x (1+1) = 3 x 2 = 6
Therefore, 12 has 6 factors. {1, 2, 3, 4, 6, 12}

#2: 8= 2³ , (3+1) = 4 Therefore, 8 has 4 factors. {1, 2, 4, 8}

Problems to ponder: answers and solutions below.
1. What is the smallest number with 5 factors?
2. What is the smallest number with 6 factors?
3. What is the smallest number with 12 factors?
4. How many factors does any prime number have?
5. How many even factors does 36 have? (tricky question)
6. How many perfect square factors does 144 have? ( don’t just list them, think how you get the answers)









Answers:
1. 16  Since 5 itself is a prime, the smallest number that has 5 factors would look like this: n⁴  put in the smallest prime number and the answer is  2⁴ = 16 

2. 12  The number 6 can be factor as 6 x 1 or 3 x 2 so you can have either n⁵ or x² * y¹
so either 2⁵ or 2² * 3 so the answer is 12. 

3. 60  12 = 12 x 1 = 6 x 2 = 4 x 3 = 3 x 2 x 2  The last would give you the smallest number which is  
2² * 3¹ * 5¹ = 60

4. 2 Any prime number has two factors, which are 1 and itself. The smallest prime number is 2, which is the oddest prime -- the only even number that is a prime. (Why??)

5. 6  Any even number multiples another integer will give you another even number. To get only even factors, you  need to always leave the smallest even number, which is 2, with all the other factors. 
36 = 2² * 3²
= 2 (2¹ *3²)  There are(1 + 1) ( 2 + 1 ) = 6 even factors

6. 6  The number 144 = 2⁴ * 3²
To get only square factors, you need to keep the smallest square number of all the prime numbers, leave out the others that are not square and find how many factors that new arrangement has. 
( 2²)² * (3²)¹  There are (2 + 1) ( 1 + 1)= 6 square factors.

See what they are on the left.

Polygons: Part I: Interior and Exterior angles, Sum and Diagonals

Some helpful links: Looking at Polygons by Kadie Armstrong, a mathematician.

Year 7 Interactive Maths: Polygons

Notes below. It takes a few seconds to download the pages. Please be patient.
Send me feedback or notify me of errors. Answer key is below for self-study.Polygons, Interior Angles, Exterior Angles, Sum













Answer Key to Polygons, Interior Angles, Exterior Angles and Sum