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

6 sided die probability to roll a odd number or a number less than 6 First, we'll find the set of rolling an odd number. [URL='https://www.mathcelebrity.com/1dice.php?gl=1&opdice=1&pl=Odds&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']From this dice calculator[/URL], we get: Odd = {1, 3, 5} Next, we'll find the set of rolling less than a 6. [URL='https://www.mathcelebrity.com/1dice.php?gl=4&pl=6&opdice=1&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']From this dice calculator[/URL], we get: Less than a 6 = {1, 2, 3, 4, 5} The question asks for [B]or[/B]. Which means a Union: {1, 3, 5} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5} This probability is [B]5/6[/B]

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]

2 rules. If either of them passes, then the number is divisible by 11: [LIST=1] [*]Sum of the odd digits - Sum of the even digits is divisible by 11 [*]Sum of the odd digits - Sum of the even digits = 0 (Ex. 121) [/LIST] [MEDIA=youtube]WpV87es0WAU[/MEDIA]

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]

if p=2x is even, then p^2 is also even p^2 = 2 * 2 * x^2 p^2 = 4x^2 This is [B]true [/B]because: [LIST] [*]If x is even, then x^2 is even since two evens multiplied by each other is even and 4x^2 is even [*]If x is odd, the x^2 is odd, but 4 times the odd number is always even since even times odd is even [/LIST]

Let n be an integer. If n^2 is odd, then n is odd Proof by contraposition: Suppose that n is even. Then we can write n = 2k n^2 = (2k)^2 = 4k^2 = 2(2k) so it is even [I]So an odd number can't be the square of an even number. So if an odd number is a square it must be the square of an odd number.[/I]

Let x be an integer. If x is odd, then x^2 is odd Proof: Let x be an odd number. This means that x = 2n + 1 where n is an integer. [U]Squaring x, we get:[/U] x^2 = (2n + 1)^2 = (2n + 1)(2n + 1) x^2 = 4n^2 + 4n + 1 x^2 = 2(2n^2 + 2n) + 1 2(2n^2 + 2n) is an even number since 2 multiplied by any integer is even So adding 1 is an odd number [MEDIA=youtube]GlzV80M33x0[/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

n^2+n = odd Factor n^2+n: n(n + 1) We have one of two scenarios: [LIST=1] [*]If n is odd, then n + 1 is even. The product of an odd and even number is an even number [*]If n is even, then n + 1 is odd. The product of an even and odd number is an even number [/LIST]

n^2-n = even Factor n^2-n: n(n - 1) We have one of two scenarios: [LIST=1] [*]If n is odd, then n - 1 is even. The product of an odd and even number is an even number [*]If n is even, then n - 1 is odd. The product of an even and odd number is an even number [/LIST]

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:
product of 2 consecutive integers
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Use proof by contradiction. Assume sqrt(2) is rational. This means that sqrt(2) = p/q for some integers p and q, with q <>0. We assume p and q are in lowest terms. Square both side and we get: 2 = p^2/q^2 p^2 = 2q^2 This means p^2 must be an even number which means p is also even since the square of an odd number is odd. So we have p = 2k for some integer k. From this, it follows that: 2q^2 = p^2 = (2k)^2 = 4k^2 2q^2 = 4k^2 q^2 = 2k^2 q^2 is also even, therefore q must be even. So both p and q are even. This contradicts are assumption that p and q were in lowest terms. So sqrt(2) [B]cannot be rational. [MEDIA=youtube]tXoo9-8Ewq8[/MEDIA][/B]

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]

Roberto has taken 17 photos photos are placed on each odd number page and the newspaper has 10 pages total. The pages with photographs will have 3 or 4 photos each. How many pages has 3 photos and how many pages have 4 photos? Odd pages are 1, 3, 5, 7, 9 17/5 = 3 with 2 remaining. So all 5 pages have 3 photos. Then with 2 left over, 2 pages get 4 photos. So 5 pages have [B]3 photos, and 2 pages have 2 photos[/B] 3(3) + 4(2) = 9 + 8 = 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

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]

There are 15 cards, numbered 1 through 15. If you pick a card, what is the probability that you choose an odd number or a two? We want the P(odd) or P(2). P(odd) = 1, 3, 5, 7, 9, 11, 13, 15 = 8/15 P(2) = 1/15 Add them both: 8/15 + 1/15 = 9/15 Simplified, we get [B]3/5[/B].

There is a stack of 10 cards, each given a different number from 1 to 10. Suppose we select a card randomly from the stack, replace it, and then randomly select another card. What is the probability that the first card is an odd number and the second card is greater than 7. First Event: P(1, 3, 5, 7, 9) = 5/10 = 1/2 or 0.5 Second Event: P(8, 9, 10) = 3/10 or 0.3 Probability of both events since each is independent is 1/2 * 3/10 = 3/20 = [B]0.15 or 15%[/B]

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]

You roll a red die and a green die. What is the size of the sample space of all possible outcomes of rolling these two dice, given that the red die shows an even number and the green die shows an odd number greater than 1? [LIST] [*]Red Die Sample Space {2, 4, 6} [*]Green Die Sample Space {3, 5} [*]Total Sample Space {(2, 3), (2, 5), (4, 3), (4, 5), (6, 3), (6, 5)} [*]The sie of this is 6 elements. [/LIST]