Question 773349
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Let the 3 integers be n-1, n and n+1.
Actually the smallest integer is not referred to in the problem at all, so the
problem can be solved for any 2 consecutive integers too. But that does not 
matter.
Break down the statement given step by step and derive the algebraic expression
for each.
1) Largest of the number = n+1
2) Half the largest = {{{(n+1)/2}}}
This is the first part.
3) Thrice the second integer = 3*n
4) 24 less than thrice the second integer = {{{3*n - 24}}}
5) 1/4th of, 24 less than thrice the second integer = {{{(3*n - 24)/4}}}
This is the second part.
Equating the 2, we get
{{{(n+1)/2 = (3*n - 24)/4}}}
Cross multiplying and simplifying
{{{2*(n+1) = 3*n - 24}}}
{{{2*n + 2 = 3*n - 24}}}
{{{n = 26}}}
So the 3 numbers are 25, 26, 27
:)
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