Question 268563
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Your question doesn't make sense as posed.  What does "71 more than the integers" mean?


If your problem were:


<i>Find the product of two consecutive integers with the sum being 71.</i>


I can do that.


Let *[tex \Large x] represent the smaller integer.  Then the next consecutive integer must be *[tex \Large x\,+\,1]


So, if the sum is 71:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ x\ +\ 1\ =\ 71]


Solve for *[tex \Large x]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 1\ =\ 71]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ 70]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 35]


So the integers are 35 and 36.  Multiply 35 times 36 to get the product.


WRONG!  Got an update on the actual problem.  It reads:


The product of two consecutive integers is 71 more than their sum.  That means that <u>unlockmath</u> (see below) has the correct solution.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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