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

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]

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

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. 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 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? 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]

If a is an even integer and b is an odd integer then prove a ? b is an odd integer Let a be our even integer Let b be our odd integer We can express a = 2x (Standard form for even numbers) for some integer x We can express b = 2y + 1 (Standard form for odd numbers) for some integer y a - b = 2x - (2y + 1) a - b = 2x - 2y - 1 Factor our a 2 from the first two terms: a - b = 2(x - y) - 1 Since x - y is an integer, 2(x- y) is always even. Subtracting 1 makes this an odd number. [MEDIA=youtube]GDVuQ7bGHx8[/MEDIA]

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

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

Free Product of Consecutive Numbers Calculator - Finds the product of (n) consecutive integers, even or odd as well. Examples include:
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Let us take an integer x which is both even [I]and[/I] odd. [LIST] [*]As an even integer, we write x in the form 2m for some integer m [*]As an odd integer, we write x in the form 2n + 1 for some integer n [/LIST] Since both the even and odd integers are the same number, we set them equal to each other 2m = 2n + 1 Subtract 2n from each side: 2m - 2n = 1 Factor out a 2 on the left side: 2(m - n) = 1 By definition of divisibility, this means that 2 divides 1. But we know that the only two numbers which divide 1 are 1 and -1. Therefore, our original assumption that x was both even and odd must be false. [MEDIA=youtube]SMM9ubEVcLE[/MEDIA]

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

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

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 + 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 [B]{11, 13, 15, 17, 19, 21, 23, 25, 27, 29}[/B]

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]0T2PDuQIIwI[/MEDIA]

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 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]