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1 integer is 7 times another. If the product of the 2 integers is 448, then find the integers.
1 integer is 7 times another. If the product of the 2 integers is 448, then find the integers. Let the first integer be x and the second integer be y. We have the following two equations: [LIST=1] [*]x = 7y [*]xy = 448 [/LIST] Substitute (1) into (2), we have: (7y)y = 448 7y^2 = 448 Divide each side by 7 y^2 = 64 y = -8, 8 We use 8, since 8*7 = 56, and 56*8 =448. So the answer is [B](x, y) = (8, 56)[/B]

10 times the first of 2 consecutive even integers is 8 times the second. Find the integers
10 times the first of 2 consecutive even integers is 8 times the second. Find the integers. Let the first integer be x. Let the second integer be y. We're given: [LIST=1] [*]10x = 8y [*]We also know a consecutive even integer means we add 2 to x to get y. y = x + 2 [/LIST] Substitute (1) into (2): 10x = 8(x + 2) Multiply through: 10x = 8x + 16 To solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=10x%3D8x%2B16&pl=Solve']we type this equation into our search engine[/URL] and we get: [B]x = 8[/B] Since y = x + 2, we plug in x = 8 to get: y = 8 + 2 [B]y = 10 [/B] Now, let's check our work. Does x = 8 and y = 10 make equation 1 hold? 10(8) ? 8(10) 80 = 80 <-- Yes!

2 consecutive even integers that equal 118
Let x be the first even integer. That means the next consecutive even integer must be x + 2. Set up our equation: x + (x + 2) = 118 Group x terms 2x + 2 = 118 Subtract 2 from each side 2x = 116 Divide each side by 2 x = 58 Which means the next consecutive even integer is 58 + 2 = 60 So our two consecutive even integers are [B]58, 60[/B] Check our work: 58 + 60 = 118

2 consecutive odd integers such that their product is 15 more than 3 times their sum
2 consecutive odd integers such that their product is 15 more than 3 times their sum. Let the first integer be n. The next odd, consecutive integer is n + 2. We are given the product is 15 more than 3 times their sum: n(n + 2) = 3(n + n + 2) + 15 Simplify each side: n^2 + 2n = 6n + 6 + 15 n^2 + 2n = 6n + 21 Subtract 6n from each side: n^2 - 4n - 21 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-4n-21%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: n = (-3, 7) If we use -3, then the next consecutive odd integer is -3 + 2 = -1. So we have [B](-3, -1)[/B] If we use 7, then the next consecutive odd integer is 7 + 2 = 9. So we have [B](7, 9)[/B]

237 what is the place value of 3
237 what is the place value of 3 Place value for integers with no decimals from right to left is: 7 is the ones digit 3 is the [B]tens digit[/B]

2consecutiveevenintegerssuchthatthesmalleraddedto5timesthelargergivesasumof70
2 consecutive even integers such that the smaller added to 5 times the larger gives a sum of 70. Let the first, smaller integer be x. And the second larger integer be y. Since they are both even, we have: [LIST=1] [*]x = y - 2 <-- Since they're consecutive even integers [*]x + 5y = 70 <-- Smaller added to 5 times the larger gives a sum of 70 [/LIST] Substitute (1) into (2): (y - 2) + 5y = 70 Group like terms: (1 + 5)y - 2 = 70 6y - 2 = 70 [URL='https://www.mathcelebrity.com/1unk.php?num=6y-2%3D70&pl=Solve']Typing 6y - 2 = 70 into our search engine[/URL], we get: [B]y = 12 <-- Larger integer[/B] Plugging this into Equation (1) we get: x = 12 - 2 [B]x = 10 <-- Smaller Integer[/B] So (x, y) = (10, 12)

3 consecutive odd integers such that thrice the middle is 15 more than the sum of the other 2
3 consecutive odd integers such that thrice the middle is 15 more than the sum of the other 2. [LIST] [*]Let the first integer be n [*]The next odd one (middle) is n + 2. [*]The next odd one is n + 4 [/LIST] We are given 3(n + 2) = n + n + 4 + 15. Simplifying, we get: 3n + 6 = 2n + 19 [URL='http://www.mathcelebrity.com/1unk.php?num=3n%2B6%3D2n%2B19&pl=Solve']Plugging that problem[/URL] into our search engine, we get n = 13. So the next odd integer is 13 + 2 = 15 The next odd integer is 15 + 2 = 17

4 consecutive integers such that the sum of the first 3 integers is equal to the 4th
4 consecutive integers such that the sum of the first 3 integers is equal to the 4th Let n be our first consecutive integer. [LIST=1] [*]n [*]n + 1 [*]n + 2 [*]n + 3 [/LIST] The sum of the first 3 integers is equal to the 4th: n + n + 1 + n + 2 = n + 3 Simplify by grouping like terms: (n + n + n) + (1 + 2) = n + 3 3n + 3 = n + 3 3n = n n = 0 n = 0 n + 1 = 1 n + 2 = 2 n + 3 = 3 Check our work: 0 + 1 +2 ? 3 3 = 3 Our final answer is [B](0, 1, 2, 3}[/B]

A is the set of integers greater than or equal to -5 and less than or equal to -2
A is the set of integers greater than or equal to -5 and less than or equal to -2 [B]-5 <= A <= -2[/B]

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]

A set of 4 consecutive integers adds up to 314. What is the least of the 4 integers?
A set of 4 consecutive integers adds up to 314. What is the least of the 4 integers? First integer is x. The next 3 are x + 1, x + 2, and x + 3. Set up our equation: x + (x + 1) + (x + 2) + (x + 3) = 314 Group x terms and group constnats (x + x + x + x) + (1 + 2 + 3) = 314 Simplify and combine 4x + 6 = 314 [URL='http://www.mathcelebrity.com/1unk.php?num=4x%2B6%3D314&pl=Solve']Enter this in the equation solver[/URL] [B]x = 77[/B]

Ages are consecutive integers. The sum of ages are 111. What are the ages
Ages are consecutive integers. The sum of ages are 111. What are the ages In the search engine, we type [I][URL='http://www.mathcelebrity.com/consecintwp.php?num=111&pl=Sum']sum of 2 consecutive integers is 111[/URL][/I]. We get [B]55 and 56[/B].

are all integers whole numbers true or false
are all integers whole numbers true or false [B]False [/B] [LIST] [*]All whole numbers are integers but not all integers are whole numbers. [*]Whole numbers are positive integers. Which means negative integers are not whole numbers [*]-1 for instance is an integer, but not a whole number [/LIST]

Before Barry Bonds, Mark McGwire, and Sammy Sosa, Roger Maris held the record for the most home runs
Before Barry Bonds, Mark McGwire, and Sammy Sosa, Roger Maris held the record for the most home runs in one season. Just behind Maris was Babe Ruth. The numbers of home runs hit by these two athletes in their record-breaking seasons form consecutive integers. Combined, the two athletes hit 121 home runs. Determine the number of home runs hit by Maris and Ruth in their record-breaking seasons. We want [URL='https://www.mathcelebrity.com/consecintwp.php?num=121&pl=Sum']the sum of 2 consecutive integers equals 121[/URL]. [B]We get Maris at 61 and Ruth at 60[/B]

Consecutive Integer Word Problems
Calculates the word problem for what two consecutive integers, if summed up or multiplied together, equal a number entered.

Consecutive odd integers are odd integers that differ by ______ , such as ______ and 13. Consecutive
[INDENT]Consecutive odd integers are odd integers that differ by ______ , such as ______ and 13. Consecutive even integers are even integers that differ by ______ , such as 12 and ______ . Consecutive odd integers are odd integers that differ by ___2___ , such as ___11___ and 13. Consecutive even integers are even integers that differ by ___2___ , such as 12 and ___14___ .[/INDENT]

Find 3 consecutive integers such that the sum of twice the smallest and 3 times the largest is 126
Find 3 consecutive integers such that the sum of twice the smallest and 3 times the largest is 126. Let the first integer be n, the second integer be n + 1, and the third integer be n + 2. We have: Sum of the smallest and 3 times the largest is 126: n + 3(n + 2) = 126 Multiply through: n + 3n + 6 = 126 Group like terms: 4n + 6 = 126 [URL='https://www.mathcelebrity.com/1unk.php?num=4n%2B6%3D126&pl=Solve']Type 4n + 6 = 126 into our calculator[/URL], we get n = 30. Which means the next two integers are 31 and 32. [B]{30, 31, 32}[/B]

Find 3 Even Integers with a sum of 198
Find 3 Even Integers with a sum of 198 Let x be the first even integer. Then y is the next, and z is the third even integer. [LIST=1] [*]y = x + 2 [*]z = x + 4 [*]x + y + z = 198 [/LIST] Substituting y and z into (3): x + x + 2 + x + 4 = 198 Group x terms 3x + 6 = 198 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B6%3D198&pl=Solve']equation solver[/URL], we get: [B]x = 64[/B] y = 64 + 2 [B]y= 66[/B] z = 64 + 4 [B]z = 68[/B]

Find the largest of three consecutive even integers when six times the first integers is equal to fi
Find the largest of three consecutive even integers when six times the first integers is equal to five times the middle integer. Let the first of the 3 consecutive even integers be n. The second consecutive even integer is n + 2. The third (largest) consecutive even integer is n + 4. We are given 6n = 5(n + 2). Multiply through on the right side, and we get: 6n = 5n + 10 [URL='https://www.mathcelebrity.com/1unk.php?num=6n%3D5n%2B10&pl=Solve']Typing 6n = 5n + 10 into the search engine[/URL], we get n = 10. Remember, n was our smallest of 3 consecutive even integers. So the largest is: n + 4 10 + 4 [B]14[/B]

Find two consecutive integers if the sum of their squares is 1513
Find two consecutive integers if the sum of their squares is 1513 Let the first integer be n. The next consecutive integer is (n + 1). The sum of their squares is: n^2 + (n + 1)^2 = 1513 n^2 + n^2 + 2n + 1 = 1513 2n^2 + 2n + 1 = 1513 Subtract 1513 from each side: 2n^2 + 2n - 1512 = 0 We have a quadratic equation. We [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B2n-1512%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this into our search engine[/URL] and get: n = (-27, 28) Let's take the positive solution. The second integer is: n + 1 28 + 1 = 29

Find two consecutive odd integers such that the sum of their squares is 290
Find two consecutive odd integers such that the sum of their squares is 290. Let the first odd integer be n. The next odd integer is n + 2 Square them both: n^2 (n + 2)^2 = n^2 + 4n + 4 from our [URL='https://www.mathcelebrity.com/expand.php?term1=%28n%2B2%29%5E2&pl=Expand']expansion calculator[/URL] The sum of the squares equals 290 n^2 + n^2 + 4n + 4 = 290 Group like terms: 2n^2 + 4n + 4 = 290 [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B4n%2B4%3D290&pl=Solve+Quadratic+Equation&hintnum=+0']Enter this quadratic into our search engine[/URL], and we get: n = 11, n = -13 Which means the two consecutive odd integer are: 11 and 11 + 2 = 13. [B](11, 13)[/B] -13 and -13 + 2 = -11 [B](-13, -11)[/B]

Find two consecutive positive integers such that the difference of their square is 25
Find two consecutive positive integers such that the difference of their square is 25. Let the first integer be n. This means the next integer is (n + 1). Square n: n^2 Square the next consecutive integer: (n + 1)^2 = n^2 + 2n + 1 Now, we take the difference of their squares and set it equal to 25: (n^2 + 2n + 1) - n^2 = 25 Cancelling the n^2, we get: 2n + 1 = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B1%3D25&pl=Solve']Typing this equation into our search engine[/URL], we get: n = [B]12[/B]

Find two consecutive positive integers such that the sum of their squares is 25
Find two consecutive positive integers such that the sum of their squares is 25. Let the first integer be x. The next consecutive positive integer is x + 1. The sum of their squares equals 25. We write this as:: x^2 + (x + 1)^2 Expanding, we get: x^2 + x^2 + 2x + 1 = 25 Group like terms: 2x^2 + 2x + 1 = 25 Subtract 25 from each side: 2x^2 + 2x - 24 = 0 Simplify by dividing each side by 2: x^2 + x - 12 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3. This means, our next positive integer is 3 + 1 = 4. So we have [B](3, 4) [/B]as our answers. Let's check our work: 3^2 + 4^2 = 9 + 16 = 25

Four consecutive integers beginning with n
Four consecutive integers beginning with n consecutive meaning one after another. So we have: [LIST] [*][B]n[/B] [*][B]n + 1[/B] [*][B]n + 2[/B] [*][B]n + 3[/B] [/LIST]

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]

Help
Yes, $40 for both. This was my son's 8th grade problem. They are learning integers with special cases, so this makes sense, I hope. Thank you so much for responding. I did have this written, but I wasn't certain it was correct.

if a and b are odd then a + b is even
if a and b are odd then a + b is even Let a and b be positive odd integers of the form: [LIST] [*]a = 2n + 1 [*]b = 2m + 1 [/LIST] a + b = 2n + 1 + 2m + 1 a + b = 2n + 2m + 1 + 1 Combing like terms, we get: a + b = 2n + 2m + 2 a + b = 2(n + m) + 2 Let k = n + m a + b = 2k + 2 [B]Therefore a + b is even[/B]

If a, b, and c are positive integers such that a^b = x and c^b = y, then xy = ?
If a, b, and c are positive integers such that a^b = x and c^b = y, then xy = ? A) ac^b B) ac^2b C) (ac)^b D) (ac)^2b E) (ac)^b^2 xy = a^b * c^b We can use the Power of a Product Rule a^b * c^b = (ac)^b Therefore: xy = [B](ac)^b - Answer C[/B]

If first integer is 5y, then the next two consecutive integers are
If first integer is 5y, then the next two consecutive integers are integers increase by 1, so we have: [B]5y + 1[/B] 5y + 1 + 1 = [B]5y + 2[/B]

If x represents the first, or the smaller, of two consecutive odd integers, express the sum of the
If x represents the first, or the smaller, of two consecutive odd integers, express the sum of the two integers in terms of x If x is the first of two consecutive odd integers, then we find the next consecutive odd integer by adding 2 to x: x + 2 The sum of the two consecutive odd integers is expressed by x + (x + 2) Simplify by grouping like terms, we get: [B]2x + 2[/B]

Integers
This lesson walks you through what integers are, how to write integers, integer notation, and what's included in integers

Integers Between
This calculator determines all integers between two numbers (Decimals)

Let n be the middle number of three consecutive integers
Let n be the middle number of three consecutive integers This means: [LIST] [*]n is the second of three consecutive integers [*]The first consecutive integer is n - 1 [*]The third consecutive integer is n + 1 [/LIST] The sum is found by: n - 1 + n + n + 1 Simplifying, we get: (n + n + n) + 1 - 1 [B]3n[/B]

Let U be the set of all integers between ?3 and 3 (including ?3 and 3). Let A={?2,0,1,3}. Find Ac. G
Let U be the set of all integers between ?3 and 3 (including ?3 and 3). Let A={?2,0,1,3}. Find Ac. Give your answer in standard set notation Ac is anything not in A, but in U. So we have: Ac = [B]{-3, -1, 2}[/B]

M is the set of integers that are greater than or equal to -1 and less than or equal to 2
M is the set of integers that are greater than or equal to -1 and less than or equal to 2 We include -1 on the left, and include 2 on the right [B]M = {-1, 0, 1, 1, 2)[/B]

Modulus
Given 2 integers a and b, this modulo calculator determines a mod b or simplifies modular arithmetic such as 7 mod 3 + 5 mod 8 - 32 mod 5

natural numbers that are factors of 16
natural numbers that are factors of 16 Natural numbers are positive integers starting at 1. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Of these, [URL='https://www.mathcelebrity.com/factoriz.php?num=16&pl=Show+Factorization']the only factors of 16[/URL] are: {[B]1, 2, 4, 8, 16}[/B]

Need Help on this problem
What do you want to do with this number set? Express it as representation of integers?

Product of Consecutive Numbers
Finds the product of (n) consecutive integers, even or odd as well. Examples include:
product of 2 consecutive integers
product of 2 consecutive numbers
product of 2 consecutive even integers
product of 2 consecutive odd integers
product of 2 consecutive even numbers
product of 2 consecutive odd numbers
product of two consecutive integers
product of two consecutive odd integers
product of two consecutive even integers
product of two consecutive numbers
product of two consecutive odd numbers
product of two consecutive even numbers
product of 3 consecutive integers
product of 3 consecutive numbers
product of 3 consecutive even integers
product of 3 consecutive odd integers
product of 3 consecutive even numbers
product of 3 consecutive odd numbers
product of three consecutive integers
product of three consecutive odd integers
product of three consecutive even integers
product of three consecutive numbers
product of three consecutive odd numbers
product of three consecutive even numbers
product of 4 consecutive integers
product of 4 consecutive numbers
product of 4 consecutive even integers
product of 4 consecutive odd integers
product of 4 consecutive even numbers
product of 4 consecutive odd numbers
product of four consecutive integers
product of four consecutive odd integers
product of four consecutive even integers
product of four consecutive numbers
product of four consecutive odd numbers
product of four consecutive even numbers
product of 5 consecutive integers
product of 5 consecutive numbers
product of 5 consecutive even integers
product of 5 consecutive odd integers
product of 5 consecutive even numbers
product of 5 consecutive odd numbers
product of five consecutive integers
product of five consecutive odd integers
product of five consecutive even integers
product of five consecutive numbers
product of five consecutive odd numbers
product of five consecutive even numbers


Prove 0! = 1
Prove 0! = 1 Let n be a whole number, where n! represents the product of n and all integers below it through 1. The factorial formula for n is: n! = n · (n - 1) * (n - 2) * ... * 3 * 2 * 1 Written in partially expanded form, n! is: n! = n * (n - 1)! [U]Substitute n = 1 into this expression:[/U] n! = n * (n - 1)! 1! = 1 * (1 - 1)! 1! = 1 * (0)! For the expression to be true, 0! [U]must[/U] equal 1. Otherwise, 1! <> 1 which contradicts the equation above

Prove sqrt(2) is irrational
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]

Quotient-Remainder Theorem
Given 2 positive integers n and d, this displays the quotient remainder theorem.

Rational Exponents - Fractional Indices
This calculator evaluates and simplifies a rational exponent expression in the form ab/c where a is any integer or any variable [a-z] while b and c are integers. Also evaluates the product of rational exponents

Roster form of: A = {3x-2/x are integers between 0 and 8}
Roster form of: A = {3x-2/x are integers between 0 and 8} x = 0 = Undefined since we divide by 0 x = 1: 3*1 + 2/1 = 5 x = 2: 3*2 + 2/2 = 7 x = 3: 3*3 + 2/3 = 9.66666666666667 x = 4: 3*4 + 2/4 = 12.5 x = 5: 3*5 + 2/5 = 15.4 x = 6: 3*6 + 2/6 = 18.3333333333333 x = 7: 3*7 + 2/7 = 21.2857142857143 x = 8: 3*8 + 2/8 = 24.25 [B]A = {(0, undefined), (1, 5), (2, 7), (3, 9.6667), (4, 12.5), (5, 15.4), (6, 18.3333), (7, 21.2857142857143), (8, 24.25)}[/B]

Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which
Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which equation could be used to find Jeremy's age, j, if he is the younger man. Let Sam's age be s. Let' Jeremy's age be j. We're given: [LIST=1] [*]s = j + 2 <-- consecutive odd integers [*]sj = 783 [/LIST] Substitute (1) into (2): (j + 2)j = 783 j^2 + 2j = 783 Subtract 783 from each side: j^2 + 2j - 783 = 0 <-- This is the equation to find Jeremy's age. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=j%5E2%2B2j-783%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this quadratic equation into the search engine[/URL] and get: j = 27, j = -29. Since ages cannot be negative, we have: [B]j = 27[/B]

Storm Center 7 was tracking a cold front approaching Cedarburg. Before it rolled in, the temperature
Storm Center 7 was tracking a cold front approaching Cedarburg. Before it rolled in, the temperature was 7F. Then, the temperature decreased by 9F. What was the temperature after the cold front rolled in? Using signed integers, we start with 7 below or -7 -7 The temperature decreased by 9 which means we subtract: -7 - 9 or -7 + (-9) [B]-16F or 16 below 0 [MEDIA=youtube]oJjEhkdnTxA[/MEDIA][/B]

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
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
sum of 2 consecutive odd numbers
sum of two consecutive integers
sum of two consecutive odd integers
sum of two consecutive even integers
sum of two consecutive numbers
sum of two consecutive odd numbers
sum of two consecutive even numbers
sum of 3 consecutive integers
sum of 3 consecutive numbers
sum of 3 consecutive even integers
sum of 3 consecutive odd integers
sum of 3 consecutive even numbers
sum of 3 consecutive odd numbers
sum of three consecutive integers
sum of three consecutive odd integers
sum of three consecutive even integers
sum of three consecutive numbers
sum of three consecutive odd numbers
sum of three consecutive even numbers
sum of 4 consecutive integers
sum of 4 consecutive numbers
sum of 4 consecutive even integers
sum of 4 consecutive odd integers
sum of 4 consecutive even numbers
sum of 4 consecutive odd numbers
sum of four consecutive integers
sum of four consecutive odd integers
sum of four consecutive even integers
sum of four consecutive numbers
sum of four consecutive odd numbers
sum of four consecutive even numbers
sum of 5 consecutive integers
sum of 5 consecutive numbers
sum of 5 consecutive even integers
sum of 5 consecutive odd integers
sum of 5 consecutive even numbers
sum of 5 consecutive odd numbers
sum of five consecutive integers
sum of five consecutive odd integers
sum of five consecutive even integers
sum of five consecutive numbers
sum of five consecutive odd numbers
sum of five consecutive even numbers


Sum of Five Consecutive Integers
Finds five consecutive integers, if applicable, who have a sum equal to a number. Sum of 5 consecutive integers

Sum of Four Consecutive Integers
Finds four consecutive integers, if applicable, who have a sum equal to a number. Sum of 4 consecutive integers

Sum of Three Consecutive Integers
Finds three consecutive integers, if applicable, who have a sum equal to a number. Sum of 3 consecutive integers

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.

The ages of three siblings are all consecutive integers. The sum of of their ages is 39.
The ages of three siblings are all consecutive integers. The sum of of their ages is 39. Let the age of the youngest sibling be n. This means the second sibling is n + 1. This means the oldest/third sibling is n + 2. So what we want is the[URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof3consecutiveintegersequalto39&pl=Calculate'] sum of 3 consecutive integers equal to 39[/URL]. We type this command into our search engine. We get: n = 12. So the youngest sibling is [B]12[/B]. The next sibling is 12 + 1 = [B]13[/B] The oldest/third sibling is 12 + 2 = [B]14[/B]

The domain of a relation is all even negative integers greater than -9. The range y of the relation
The domain of a relation is all even negative integers greater than -9. The range y of the relation is the set formed by adding 4 to the numbers in the domain. Write the relation as a table of values and as an equation. The domain is even negative integers greater than -9: {-8, -6, -4, -2} Add 4 to each x for the range: {-8 + 4 = -4, -6 + 4 = -2. -4 + 4 = 0, -2 + 4 = 2} For ordered pairs, we have: (-8, -4) (-6, -2) (-4, 0) (-2, 2) The equation can be written: y = x + 4 on the domain (x | x is an even number where -8 <= x <= -2)

the left and right page numbers of an open book are two consecutive integers whose number is 235 fin
the left and right page numbers of an open book are two consecutive integers whose number is 235 find the page numbers Using our [URL='https://www.mathcelebrity.com/consecintwp.php?pl=Sum&num=+235']consecutive integer calculator[/URL], we get: [B]117, 118[/B]

The left and right page numbers of an open book are two consecutive integers whose sum is 403. Find
The left and right page numbers of an open book are two consecutive integers whose sum is 403. Find these page numbers. Page numbers left and right are consecutive integers. So we want to find a number n and n + 1 where: n + n + 1 = 403 Combining like terms, we get: 2n + 1 = 403 Typing that equation into our search engine, we get: [B]n = 201[/B] This is our left hand page. Our right hand page is: 201 + 1 = [B]202[/B]

The number -2.34 can be found between which two integers
The number -2.34 can be found between which two integers We want to take the integer of -2.34 which is -2 Since -2.34 is less than 0, we subtract 1: -2 -1 = -3 Therefore, -2.34 lies between [B]-2 and -3[/B] -3 < -2.34 < -2

The square of a positive integer minus twice its consecutive integer is equal to 22. find the intege
The square of a positive integer minus twice its consecutive integer is equal to 22. Find the integers. Let x = the original positive integer. We have: [LIST] [*]Consecutive integer is x + 1 [*]x^2 - 2(x + 1) = 22 [/LIST] Multiply through: x^2 - 2x - 2 = 22 Subtract 22 from each side: x^2 - 2x - 24 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2-2x-24%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: x = 6 and x = -4 Since the problem states [U]positive integers[/U], we use: x = 6 and x + 1 = 7 [B](6, 7)[/B]

The sum of 3 consecutive integers is greater than 30.
The sum of 3 consecutive integers is greater than 30. Let the first consecutive integer be n The second consecutive integer is n + 1 The third consecutive integer is n + 2 The sum is written as: n + n + 1 + n + 2 Combine like terms: (n + n + n) + (1 + 2) 3n + 3 The phrase [I]greater than[/I] means an inequality, which we write as: [B]3n + 3 > 30[/B]

The sum of the squares of two consecutive positive integers is 61. Find these two numbers.
The sum of the squares of two consecutive positive integers is 61. Find these two numbers. Let the 2 consecutive integers be x and x + 1. We have: x^2 + (x + 1)^2 = 61 Simplify: x^2 + x^2 + 2x + 1 = 61 2x^2 + 2x + 1 = 61 Subtract 61 from each side: 2x^2 + 2x - 60 = 0 Divide each side by 2 x^2 + x - 30 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL], we get: x = 5 and x = -6 The question asks for [I]positive integers[/I], so we use [B]x = 5. [/B]This means the other number is [B]6[/B].

The sum of three consecutive integers is 42
Let the 3 integers be x, y, and z. y = x + 1 z = y + 1, or x + 2. Set up our equation: x + (x + 1) + (x + 2) = 42 Group our variables and constants: (x + x + x) + (1 + 2) = 42 3x + 3 = 42 Subtract 3 from each side: 3x = 39 Divide each side of the equation by 3: [B]x = 13 So y = x + 1 = 14 z = x + 2 = 15 (x,y,z) = (13,14,15)[/B]

The sum of two consecutive integers if n is the first integer.
The sum of two consecutive integers if n is the first integer. consecutive means immediately after, so we have: n n + 1 [U]The sum is written as:[/U] n + n + 1 [U]Grouping like terms, we have:[/U] (n + n) + 1 [B]2n + 1[/B]

The sum of two consecutive integers plus 18 is 123
The sum of two consecutive integers plus 18 is 123. Let our first integer be n and our next integer be n + 1. We have: n + (n + 1) + 18 = 123 Group like terms to get our algebraic expression: 2n + 19 = 123 If we want to solve the algebraic expression using our [URL='http://www.mathcelebrity.com/1unk.php?num=2n%2B19%3D123&pl=Solve']equation solver[/URL], we get n = 52. This means the next integer is 52 + 1 = 53

There are 2 consecutive integers. Twice the first increased by the second yields 16. What are the nu
There are 2 consecutive integers. Twice the first increased by the second yields 16. What are the numbers? Let x be the first integer. y = x + 1 is the next integer. We have the following givens: [LIST=1] [*]2x + y = 16 [*]y = x + 1 [/LIST] Substitute (2) into (1) 2x + (x + 1) = 16 Combine x terms 3x + 1 = 16 Subtract 1 from each side 3x = 15 Divide each side by 3 [B]x = 5[/B] So the other integer is 5 + 1 = [B]6[/B]

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]

To make an international telephone call, you need the code for the country you are calling. The code
To make an international telephone call, you need the code for the country you are calling. The code for country A, country B, and C are three consecutive integers whose sum is 90. Find the code for each country. If they are three consecutive integers, then we have: [LIST=1] [*]B = A + 1 [*]C = B + 1, which means C = A + 2 [*]A + B + C = 90 [/LIST] Substitute (1) and (2) into (3) A + (A + 1) + (A + 2) = 90 Combine like terms 3A + 3 = 90 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3a%2B3%3D90&pl=Solve']equation calculator[/URL], we get: [B]A = 29[/B] Which means: [LIST] [*]B = A + 1 [*]B = 29 + 1 [*][B]B = 30[/B] [*]C = A + 2 [*]C = 29 + 2 [*][B]C = 31[/B] [/LIST] So we have [B](A, B, C) = (29, 30, 31)[/B]

Two consecutive even integers that equal 126
Two consecutive even integers that equal 126 Let the first integer equal x. So the next even integer must be x + 2. The sum which is equal to 126 is written as x + (x + 2) = 126 Simplify: 2x + 2 = 126 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2x%2B2%3D126&pl=Solve']equation calculator,[/URL] we get: x = 62 This means the next consecutive even integer is 62 = 2 = 64. So our two even consecutive integers with a sum of 126 are [B](62, 64)[/B]

What pair of consecutive integers gives the following: 7 times the smaller is less than 6 times the
What pair of consecutive integers gives the following: 7 times the smaller is less than 6 times the larger? Let x and y be consecutive integers, where y = x + 1 We have 7x < 6y as our inequality. Substituting x, y = x + 1, we have: 7x < 6(x + 1) 7x < 6x + 6 Subtracting x from each side, we have: x < 6, so y = 6 + 1 = 7 (x, y) = (6, 7)

When 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 intege
When 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 integers. What is the difference of the product minus the sum? Let the 3 consecutive positive integers be: [LIST=1] [*]x [*]x + 1 [*]x + 2 [/LIST] The product is: x(x + 1)(x + 2) The sum is: x + x + 1 + x + 2 = 3x + 3 We're told the product is equivalent to: x(x + 1)(x + 2) = 16(3x + 3) x(x + 1)(x + 2) = 16 * 3(x + 1) Divide each side by (x + 1) x(x + 2) = 48 x^2 + 2x = 48 x^2 + 2x - 48 = 0 Now subtract the sum from the product: x^2 + 2x - 48 - (3x + 3) [B]x^2 - x - 51[/B]

{x | x is an even integer between -3 and 5}
{x | x is an even integer between -3 and 5} We list even integers out in this range: [B]{-2, 0, 2, 4}[/B]