Find three consecutive natural numbers, whose product is 81204. (Select the equation.)
n x (n+1) x (n+2) = 81204 ?
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Find three consecutive natural numbers, whose product is 81204. (Select the equation.) n x (n+1) x (n+2) = 81204 ? |
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It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.
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You wish to solve $n(n+1)(n+2)=81204$. If there is a solution, then $n$ is approximately $\root 3\of {81204}\approx 43.3$. Now checking possible solutions: $\ \ \ 42\cdot43\cdot44=79464$ $\ \ \ 43\cdot44\cdot 45=85140$ None of these work, but these are the only possible solutions (other choices would give you something less than $79464$ or something greater than $85140$). So there is no solution to your problem. |
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Hint: if the first number is $n$, what are the next two? |
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27067 + 27068 + 27069 = 81204 81205 + 81206 - 81207 = 81204 But really I have no idea what your question really means. Edit: Okay you want the product. I recommend you try trial and error to find the value of n if it exists. |
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n*(n+1)*(n+2)=81204 expand this equation to get: n^3 + 3n^2 + 2n - 81204 = 0 use this 3rd degree Calculator, to get these answers: 42.31147753802359 -22.655738769011794 + 37.49550726280472 i -22.655738769011794 - 37.49550726280472 i the answers are not integers. (refer to your quest) or check this site to know the technique of solving a third degree eqn. |
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number1 = n number2 = n + 1 number3 = n + 2 sum = number1 + number2 + number3 81204 = n + (n + 1) + (n + 2) 81204 = 3n + 3 81201 = 3n n = 27067 solution: {27067, 27068, 27069} |
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