Question 778871
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The product of two consecutive integers is 71 more than the square of the preceding smaller integer. Find the numbers.
Ans:
Let the integers be n and n+1. Then the preceding smaller integer is (n-1).
It is given that
{{{n*(n+1) = (n-1)^2 + 71}}}
{{{n^2 + n = n^2 - 2*n + 1 + 71 = n^2 - 2*n + 72}}}
Cancelling out n^2 and simplifying
{{{3*n = 72}}}
{{{n = 24}}}
The numbers are 24 and 25.
Hope you got it :)
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