BB2 Math question

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1

<p>I'm confused on this questions</p>

<h1>16 page 468:</h1>

<p>A positive integer is said to be "tri-factorable" if it is the product of three consecutive integers. How many positive integers less than 1000 are tri--factorable?</p>

<p>The answer is 9. I got this answer by just multiplying three consective integers until I got to 990, but I'm thinking there has got to be an easier way. Can anyone help me?</p>

2

<p>I think your way of solving the problem is pretty straight-forward. However, you might solve it slightly faster if you realize that because 10<em>10</em>10=1000, the biggest three integers you can have are 9<em>10</em>11. This way you wouldn’t have to actually multiply all of the 9 combinations; writing down and counting them would be enough.</p>

3

<p>Thanks. :)</p>