<p>For 14, I honestly just went ahead and used a guess and check method and got the right answer fairly quickly. Is there a formal way of doing this problem?</p>
<p>For 16, I found the largest "tri-factorable" integer:</p>
<p>990 (9x10x11) and then I made a list starting with 1</p>
<p>1x2x3
2x3x4
3x4x5
4x5x6
5x6x7
6x7x8
7x8x9
8x9x10
9x10x11</p>
<p>Is there a quicker way to do these?</p>
<ol>
<li><p>f(x)=[3x-17]. either f(a)=3a-17 or -f(a)=3a-17 (the latter assumes 3x-17 is negative, the first assumes that 3x-17 is positive). For both cases, set a=f(a). So, for the first case, a=3a-17 and a=17/2=8.5. for the second case, -a=3a-17 so 4a=17 and a=17/4=4.25. Thus, by simple reasoning you can figure that the answer is between, but not including, 4.25 and 8.5 (generally, to make sure you’d either carry the inequalities through- which is risky, or put the two values on a line chart and test values less than 4.25, between 4.25 and 8.5, and greater than 8.5; 0, 6, and 10 for example). Guessing and checking is easier, especially since they just ask for one value, but you did ask if there was a formal way. </p></li>
<li><p>You can save a lot of time by not writing out anything until 9x10x11. Since that equals 990, you can be certain that the next three integers will not be less than 1000. Then, you realize that there is an integer group with factors staring with: 1, 2, 3, 4, 5, 6, 7, 8, 9 which is 9 total integers.</p></li>
</ol>
<p>Ah yeah, I didn’t really recognize that pattern on #16, in which every consecutive factor group started with the next consecutive integer :P. And thanks for the first explanation. I didn’t think about finding the boundaries like that.</p>
