Suppose the smallest of the integers is n. Then the product of the four consecutive integers is n*(n+1)*(n+2)*(n+3) =(n2+3n)(n2+3n+2) = n4+6n3+11n2+6n So product +1 = n4+6n3+11n2+6n+1 which can be factorised as follows: n4+3n3+n2 +3n3+9n2+3n + n2+3n+1 =[n2+3n+1]2 Thus, one more that the product of four consecutive integers is a perfect square.
Let's assume that the three consecutive integers are n-1, n, and n+1. We know that their product is 210, so we can set up the equation (n-1)(n)(n+1) = 210. By solving this equation, we can find that the three integers are 6, 7, and 8. Therefore, the sum of the smallest two integers is 6 + 7 = 13.
Suppose m and n are integers. Then 2m + 1 and 2n +1 are odd integers.(2m + 1)*(2n + 1) = 4mn + 2m + 2n + 1 = 2*(2mn + m + n) + 1 Since m and n are integers, the closure of the set of integers under multiplication and addition implies that 2mn + m + n is an integer - say k. Then the product is 2k + 1 where k is an integer. That is, the product is an odd number.
Numbers are n and n + 2, so product = n2 + 2n and sum = 2n + 2. ie n2 + 2n + 1 = 6n + 6 or n2 - 4n - 5 = 0 Factors are (n + 1)(n - 5) making n = 5 and the integers 5 & 7. Check: 5 x 7 = 35; 5 + 7 = 36. OK There is also an answer of n = -1, making the integers -1 and 1, product being -1 and sum zero.
The product of four consecutive integers is always one less than a perfect square. The product of four consecutive integers starting with n will be one less than the square of n2 + 3n + 1
You can solve this in two ways.1) Trial and error. That is, try multiplying two consecutive integers; if the product is too large, try smaller integers; if the product is too small, try larger consecutive integers. 2) Call the two consecutive integers "n" and "n+1", and solve the equation: n(n+1)=210
Call the two consecutive integers n and n+1. Their product is n(n+1) or n2 +n. For example if the integers are 1 and 2, then n would be 1 and n+1 is 2. Their product is 1x2=2 of course which is 12 +1=2 Try 2 and 3, their product is 6. With the formula we have 4+2=6. The point of the last two examples was it is always good to check your answer with numbers that are simple to use. That does not prove you are correct, but if it does not work you are wrong for sure!
Suppose the smallest of the integers is n. Then the product of the four consecutive integers is n*(n+1)*(n+2)*(n+3) =(n2+3n)(n2+3n+2) = n4+6n3+11n2+6n So product +1 = n4+6n3+11n2+6n+1 which can be factorised as follows: n4+3n3+n2 +3n3+9n2+3n + n2+3n+1 =[n2+3n+1]2 Thus, one more that the product of four consecutive integers is a perfect square.
Let's assume that the three consecutive integers are n-1, n, and n+1. We know that their product is 210, so we can set up the equation (n-1)(n)(n+1) = 210. By solving this equation, we can find that the three integers are 6, 7, and 8. Therefore, the sum of the smallest two integers is 6 + 7 = 13.
Suppose m and n are integers. Then 2m + 1 and 2n +1 are odd integers.(2m + 1)*(2n + 1) = 4mn + 2m + 2n + 1 = 2*(2mn + m + n) + 1 Since m and n are integers, the closure of the set of integers under multiplication and addition implies that 2mn + m + n is an integer - say k. Then the product is 2k + 1 where k is an integer. That is, the product is an odd number.
It is n factorial, written as n!
For integers greater than 1 the product down to 1 is called factorial, indicated mathematically as N! wher N is the highest integer For example 5! = 5 factorial = 5x4x3x2x1 = 120
Numbers are n and n + 2, so product = n2 + 2n and sum = 2n + 2. ie n2 + 2n + 1 = 6n + 6 or n2 - 4n - 5 = 0 Factors are (n + 1)(n - 5) making n = 5 and the integers 5 & 7. Check: 5 x 7 = 35; 5 + 7 = 36. OK There is also an answer of n = -1, making the integers -1 and 1, product being -1 and sum zero.
The product of four consecutive integers is always one less than a perfect square. The product of four consecutive integers starting with n will be one less than the square of n2 + 3n + 1
One way to find out is write a formula. Let N and N+1 be the two integers, then N(N+1) = 182 N^2 + N - 182 = 0 This is a quadratic equation. If the factors are not obvious, (N -13)(N + 14) , then use the quadratic formula to find N. The factors tell you there are two possible solutions for N; 13 and -14. Now add 1 to these to get the two consecutive integers. 13 & 14 will work and -14 & -13 will work.
The "integers of 15" appear to be '1' and '5'. Their product is 5 .
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