<p>The question is from test 2, section 5, number 16 (grid-in):</p>
<p>A positive integer is said to be "trifactorable" if it is the product of three consecutive integers. How many positive integers less than 1000 are trifactorable?</p>
<p>The answer is 9. Please tell me how to derive this answers. I appreciate the help.</p>
<p>1000 = 10<em>10</em>10
1000 < 10<em>11</em>12
1000 > 9<em>10</em>11 = 990
1000 > 8… > 7… > 6… > … 1<em>2</em>3
0 does not count since it is not positive, although it is an integer</p>
<p>You basically list them all out.</p>
<p>Multiplying consecutively larger numbers makes the product get bigger really quickly. So, just start with 1<em>2</em>3 and keep listing until you get past 1000. It shouldn’t take too long.</p>
<p>1x2x3, 2x3x4, 3x4x5, etc…</p>
<p>Notice <last number=“” multiplied=“” -=“” 2=“”> yields the answer</last></p>
<p>We find that 9x10x11 = 990</p>
<p>So answer is 11 - 2= [9]</p>
<p>^Don’t do it that way, this way is MUCH faster. Takes around 15 seconds.</p>
<p>
</last></p>
<p>How would you know that without listing out or being a savant?</p>
<p>your just looking to count the number of terms right, so as you can 3x4x5 is the third term. 5-2 = 3 (3rd term). I didn’t think it was that difficult to see sorry. I probably should have elaborated. BTW, you can also just use the first term in the series. E.g. 5x6x7, 5th term.</p>
<p>Thanks a lot for the help</p>