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What are the products of two consecutive positive integers?
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) ).
The product of four consective integers is one less than a perfect square?
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.
Product of three concecutive integer is 210 then find the sum of smallest two integers?
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.
Why is an odd number times an odd number an odd number?
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.
What is the integer if the product of two consecutive odd integers is one less than three times their sum?
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.
