It is n! or n factorial.
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The product of two consecutive positive integers can be found by multiplying the smaller integer by the larger integer. If the smaller integer is represented as ( n ), then the larger integer would be ( n + 1 ). Therefore, the product of two consecutive positive integers is ( n \times (n + 1) ).
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.