The sum of three consecutive positive integers is a perfect . . . Considering that the middle number ends with at least 5 zeros, it implies that n+1 ≥ 100000 Meanwhile, the sum of these three numbers is also a perfect cube Hence the equation n + (n+1) + (n+2) = k³ is established, where k is a positive integer By simplifying, we get 3n + 3 = k³
Art of Problem Solving The sum of consecutive positive integers is a perfect square The smallest possible value of this sum is Solution Solution 1 Let be the consecutive positive integers Their sum, , is a perfect square Since is a perfect square, it follows that is a perfect square The smallest possible such perfect square is when , and the sum is Solution 2
Sum of consecutive integers is a perfect square, perfect cube Because $p,q,r,s,t$ are consecutive positive integers, we know immediately that $q=r-1,p=r-2,s=r+1,t=r+2$ This means that $q+r+s=r-1+r+r+1=3r$ and that $p+q+r+s+t=5r$ If $3r$ is a perfect square, then this means that the prime factorization of $r$ must have $3^{2n+1}$ for some $n$; similarly, for $5r$ to be a perfect cube, we must have that
What is the smallest positive perfect cube that can be . . . Step-by-step explanation: A perfect cube is the result of multiplying a number three times by itself We can also say that perfect cubes are the numbers that have exact cube roots 1, 8, 27, 64, 125, 216, 343, 512, 729, 1,000, 1,331, 1,728, 2,197, 2,744, 3,375 Find Math textbook solutions? 1 Pancreas 2 Liver ️ ️ ️ ️ ️ ️ 3 Brain 4
Three consecutive positive integers are such that the sum of . . . To solve the problem of finding three consecutive positive integers such that the sum of the square of the first integer and the product of the other two is 46, we can follow these steps: Step 1: Define the integers Let the three consecutive integers be: - First integer: \( x \) - Second integer: \( x + 1 \) - Third integer: \( x + 2 \) Step 2