odd numbers - Numbers not divisible by 2

1 Die Roll

Free 1 Die Roll Calculator - Calculates the probability for the following events in the roll of one fair dice (1 dice roll calculator or 1 die roll calculator):

* Probability of any total from (1-6)

* Probability of the total being less than, less than or equal to, greater than, or greater than or equal to (1-6)

* The total being even

* The total being odd

* The total being a prime number

* The total being a non-prime number

* Rolling a list of numbers i.e. (2,5,6)

* Simulate (n) Monte Carlo die simulations.

1 die calculator

* Probability of any total from (1-6)

* Probability of the total being less than, less than or equal to, greater than, or greater than or equal to (1-6)

* The total being even

* The total being odd

* The total being a prime number

* The total being a non-prime number

* Rolling a list of numbers i.e. (2,5,6)

* Simulate (n) Monte Carlo die simulations.

1 die calculator

1, 9, 25, 49, .......... What is next

1, 9, 25, 49, .......... What is next
1^2 = 1
3^2 = 9
5^2 = 25
7^2 = 49
So this pattern takes odd numbers and squares them. Our next odd number is 9:
9^2 = [B]81[/B]

2 dice roll

Free 2 dice roll Calculator - Calculates the probability for the following events in a pair of fair dice rolls:

* Probability of any sum from (2-12)

* Probability of the sum being less than, less than or equal to, greater than, or greater than or equal to (2-12)

* The sum being even

* The sum being odd

* The sum being a prime number

* The sum being a non-prime number

* Rolling a list of numbers i.e. (2,5,6,12)

* Simulate (n) Monte Carlo two die simulations. 2 dice calculator

* Probability of any sum from (2-12)

* Probability of the sum being less than, less than or equal to, greater than, or greater than or equal to (2-12)

* The sum being even

* The sum being odd

* The sum being a prime number

* The sum being a non-prime number

* Rolling a list of numbers i.e. (2,5,6,12)

* Simulate (n) Monte Carlo two die simulations. 2 dice calculator

A is the set of odd integers between 4 and 12

A is the set of odd integers between 4 and 12
Let A be the set of odd numbers between 4 and 12:
[B]A = {5, 7, 9, 11}[/B]

Find four consecutive odd numbers which add to 64

Find four consecutive odd numbers which add to 64.
Let the first number be x. The next three numbers are:
x + 2
x + 4
x + 6
Add them together to get 64:
x + (x + 2) + (x + 4) + (x + 6) = 64
Group like terms:
4x + 12 = 64
Using our [URL='http://www.mathcelebrity.com/1unk.php?num=4x%2B12%3D64&pl=Solve']equation calculator[/URL], we get:
[B]x = 13[/B]
The next 3 odd numbers are:
x + 2 = 13 + 2 = 15
x + 4 = 13 + 4 = 17
x + 6 = 13 + 6 = 19
So the 4 consecutive odd numbers which add to 64 are:
[B](13, 15, 17, 19)[/B]

Find the odd number less than 100 that is divisible by 9, and when divided by 10 has a remainder of

Find the odd number less than 100 that is divisible by 9, and when divided by 10 has a remainder of 7.
From our [URL='http://www.mathcelebrity.com/divisibility.php?num=120&pl=Divisibility']divisibility calculator[/URL], we see a number is divisible by 9 if the sum of its digits is divisible by 9.
Starting from 1 to 99, we find all numbers with a digit sum of 9.
This would be digits with 0 and 9, 1 and 8, 2 and 7, 3 and 6, and 4 and 5.
9
18
27
36
45
54
63
72
81
90
Now remove even numbers since the problem asks for odd numbers
9
27
45
63
81
Now, divide each number by 10, and find the remainder
9/10 = 0
[URL='http://www.mathcelebrity.com/modulus.php?num=27mod10&pl=Calculate+Modulus']27/10[/URL] = 2 R 7
We stop here. [B]27[/B] is an odd number, less than 100, with a remainder of 7 when divided by 10.

Four cousins were born at two-year intervals. The sum of their ages is 36. What are their ages?

Four cousins were born at two-year intervals. The sum of their ages is 36. What are their ages?
So the last cousin is n years old. this means consecutive cousins are n + 2 years older than the next.
whether their ages are even or odd, we have the sum of 4 consecutive (odd|even) integers equal to 36. We [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof4consecutiveevenintegersis36&pl=Calculate']type this into our search engine[/URL] and we get the ages of:
[B]6, 8, 10, 12[/B]

Number Property

Free Number Property Calculator - This calculator determines if an integer you entered has any of the following properties:

* Even Numbers or Odd Numbers (Parity Function or even-odd numbers)

* Evil Numbers or Odious Numbers

* Perfect Numbers, Abundant Numbers, or Deficient Numbers

* Triangular Numbers

* Prime Numbers or Composite Numbers

* Automorphic (Curious)

* Undulating Numbers

* Square Numbers

* Cube Numbers

* Palindrome Numbers

* Repunit Numbers

* Apocalyptic Power

* Pentagonal

* Tetrahedral (Pyramidal)

* Narcissistic (Plus Perfect)

* Catalan

* Repunit

* Even Numbers or Odd Numbers (Parity Function or even-odd numbers)

* Evil Numbers or Odious Numbers

* Perfect Numbers, Abundant Numbers, or Deficient Numbers

* Triangular Numbers

* Prime Numbers or Composite Numbers

* Automorphic (Curious)

* Undulating Numbers

* Square Numbers

* Cube Numbers

* Palindrome Numbers

* Repunit Numbers

* Apocalyptic Power

* Pentagonal

* Tetrahedral (Pyramidal)

* Narcissistic (Plus Perfect)

* Catalan

* Repunit

Odd Numbers

Free Odd Numbers Calculator - Shows a set amount of odd numbers and cumulative sum

Product of Consecutive Numbers

Free Product of Consecutive Numbers Calculator - Finds the product of (n) consecutive integers, even or odd as well. Examples include:

product of 2 consecutive integers

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sum of 3 consecutive odd integers equals 1 hundred 17

sum of 3 consecutive odd integers equals 1 hundred 17
The sum of 3 consecutive odd numbers equals 117. What are the 3 odd numbers?
1) Set up an equation where our [I]odd numbers[/I] are n, n + 2, n + 4
2) We increment by 2 for each number since we have [I]odd numbers[/I].
3) We set this sum of consecutive [I]odd numbers[/I] equal to 117
n + (n + 2) + (n + 4) = 117
[SIZE=5][B]Simplify this equation by grouping variables and constants together:[/B][/SIZE]
(n + n + n) + 2 + 4 = 117
3n + 6 = 117
[SIZE=5][B]Subtract 6 from each side to isolate 3n:[/B][/SIZE]
3n + 6 - 6 = 117 - 6
[SIZE=5][B]Cancel the 6 on the left side and we get:[/B][/SIZE]
3n + [S]6[/S] - [S]6[/S] = 117 - 6
3n = 111
[SIZE=5][B]Divide each side of the equation by 3 to isolate n:[/B][/SIZE]
3n/3 = 111/3
[SIZE=5][B]Cancel the 3 on the left side:[/B][/SIZE]
[S]3[/S]n/[S]3 [/S]= 111/3
n = 37
Call this n1, so we find our other 2 numbers
n2 = n1 + 2
n2 = 37 + 2
n2 = 39
n3 = n2 + 2
n3 = 39 + 2
n3 = 41
[SIZE=5][B]List out the 3 consecutive odd numbers[/B][/SIZE]
([B]37, 39, 41[/B])
37 ? 1st number, or the Smallest, Minimum, Least Value
39 ? 2nd number
41 ? 3rd or the Largest, Maximum, Highest Value

Sum of Consecutive Numbers

Free Sum of Consecutive Numbers Calculator - Finds the sum of (n) consecutive integers, even or odd as well. Examples include:

sum of 2 consecutive integers

sum of 2 consecutive numbers

sum of 2 consecutive even integers

sum of 2 consecutive odd integers

sum of 2 consecutive even numbers

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Sum of the First (n) Numbers

Free Sum of the First (n) Numbers Calculator - Determines the sum of the first (n)

* Whole Numbers

* Natural Numbers

* Even Numbers

* Odd Numbers

* Square Numbers

* Cube Numbers

* Fourth Power Numbers

* Whole Numbers

* Natural Numbers

* Even Numbers

* Odd Numbers

* Square Numbers

* Cube Numbers

* Fourth Power Numbers

Sum of two consecutive numbers is always odd

Sum of two consecutive numbers is always odd
Definition:
[LIST]
[*]A number which can be written in the form of 2 m where m is an integer, is called an even integer.
[*]A number which can be written in the form of 2 m + 1 where m is an integer, is called an odd integer.
[/LIST]
Take two consecutive integers, one even, and one odd:
2n and 2n + 1
Now add them
2n + (2n+ 1) = 4n + 1 = 2(2 n) + 1
The sum is of the form 2n + 1 (2n is an integer because the product of two integers is an integer)
Therefore, the sum of two consecutive integers is an odd number.

Take a look at the following sums: 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 +

Take a look at the following sums:
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
a. Come up with a conjecture about the sum when you add the first *n* odd numbers. For example, when you added the first 5 odd numbers (1 + 3 + 5 + 7 + 9), what did you get? What if wanted to add the first 10 odd numbers? Or 100?
b. Can you think of a geometric interpretation of this pattern? If you start with one square and add on three more, what can you make? If you now have 4 squares and add on 5 more, what can you make?
c. Is there a similar pattern for adding the first n even numbers?
2 = 2
2 + 4 = 6
2 + 4 + 6 = 12
2 + 4 + 6 + 8 = 20
a. The formula is [B]n^2[/B].
The sum of the first 10 odd numbers is [B]100[/B] seen on our s[URL='http://www.mathcelebrity.com/sumofthefirst.php?num=10&pl=Odd+Numbers']um of the first calculator[/URL]
The sum of the first 100 odd numbers is [B]10,000[/B] seen on our [URL='http://www.mathcelebrity.com/sumofthefirst.php?num=100&pl=Odd+Numbers']sum of the first calculator[/URL]
b. Geometric is 1, 4, 9 which is our [B]n^2[/B]
c. The sum of the first n even numbers is denoted as [B]n(n + 1)[/B] seen here for the [URL='http://www.mathcelebrity.com/sumofthefirst.php?num=+10&pl=Even+Numbers']first 10 numbers[/URL]

The set of all odd numbers between 10 and 30

The set of all odd numbers between 10 and 30
[B]{11, 13, 15, 17, 19, 21, 23, 25, 27, 29}[/B]

The sum of 5 odd consecutive numbers is 145

The sum of 5 odd consecutive numbers is 145.
Let the first odd number be n. We have the other 4 odd numbers denoted as:
[LIST]
[*]n + 2
[*]n + 4
[*]n + 6
[*]n + 8
[/LIST]
Add them all together
n + (n + 2) + (n + 4) + (n + 6) + (n + 8)
The sum of the 5 odd consecutive numbers equals 145
n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 145
Combine like terms:
5n + 20 = 145
Using our [URL='http://www.mathcelebrity.com/1unk.php?num=5n%2B20%3D145&pl=Solve']equation solver[/URL], we get [B]n = 25[/B]. Using our other 4 consecutive odd numbers above, we get:
[LIST]
[*]27
[*]29
[*]31
[*]33
[/LIST]
Adding the sum up, we get: 25 + 27 + 29 + 31 + 33 = 145.
So our 5 odd consecutive number added to get 145 are [B]{25, 27, 29, 31, 33}[/B].
[MEDIA=youtube]3nN2ROooVlc[/MEDIA]

Three good friends are in the same algebra class, their scores on a recent test are three consecutiv

Three good friends are in the same algebra class, their scores on a recent test are three consecutive odd integers whose sum is 273. Find the score
In our search engine, we type in [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=3consecutiveintegerswhosesumis273&pl=Calculate']3 consecutive integers whose sum is 273[/URL] and we get:
[B]90, 91, 92[/B]

You are using a spinner with the numbers 1-10 on it. Find the probability that the pointer will sto

You are using a spinner with the numbers 1-10 on it. Find the probability that the pointer will stop on an odd number or a number less than 4.
We want P(odd number) or P(n<4).
[LIST]
[*]Odd numbers are {1, 3, 5, 7, 9}
[*]n < 4 is {1, 2, 3}
[/LIST]
We want the union of these 2 sets:
{1, 2, 3, 5, 7, 9}
We have 6 possible pointers in a set of 10.
[B]6/10 = 3/5 = 0.6 or 60%[/B]