Question 818108
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I definitely concur with tutor ikleyn that this kind of problem is much better solved using a little logical reasoning, instead of formal mathematics.  Indeed, when you try to solve it using the traditional algebraic approach, you end up having to factor a quadratic by finding two numbers whose product is 1505 and whose difference is 8.  But THAT IS WHAT THE ORIGINAL PROBLEM asks you to do.<br>
So the formal algebra is a waste of time....<br>
And the "trick" tutor ikleyn uses is a very useful one.  If you know that
{{{x(x-8)=1505}}}
then you know that
{{{(x+4)(x-4) = 1505+4^2 = 1505+16 = 1521 = 39^2}}}
and so the answers are 39-4=35 and 39+4=43.<br>
It did not occur to me to use that trick on this problem.  I just started playing with numbers to find a pair with a product of 1505 and a difference of 8.  There weren't many possibilities....<br>
I of course first saw
{{{1505 = 5*301}}}<br>
Then I had to find a factorization of 301; when I found it, I had my answer.<br>
{{{301 = 7*43}}}, so {{{1505 = 5*7*43 = 35*43}}}.<br>
Answer: 35 rows of 43 seats each.