<p>It's on pg 714, #8. (New blue book btw)</p>
<p>If a and b are positive integers and (1/a^2*1/b^3)^6=432, what is the value of ab?
a) 6
b)12
c)18
d)24
e)36</p>
<p>thanks in advance!</p>
<p>I could be wrong, but I believe you wrote down the problem incorrectly. I will go with (a^(1/2) x b^(1/3))^6 = 432, as is written in my copy. Just to clarify: a^1/2 is not necessarily equal to 1/a^2 (one situation in where it is equal is for 1. 1^(1/2) = 1/1^2). Getting back to the problem, the idea is to get rid of the 6. This is done by multiplying all exponents by 6. It follows from the rule that (ab)^n = a^n x b^n. Hence, a^(1/2 x 6) x b(1/3 x 6) = a^3 x b^2. That still equals 432. The question says they are positive integers, so that means we need to find a factorization of 432. We want one number to be a perfect cube and one to be a perfect square. We find that 27 x 16 works (to achieve this, find the prime factorization 2^4 x 3^3, call 3^3 the perfect cube, and 2^4 the perfect square). Call 27 = a^3, which means a = 3. Call 2^4 (which is 16) = b^2. b must then equal 4. Hence, the answer should be (B), or 12. Hope that helped.</p>
<p>^Thanks! Yeah you’re right, I actually read what was said in the book as what I had written above. Guess I need to take a more careful look next time!</p>
<p>432 = 2^4 * 3^3
a^3 * b^2 = 2^4 * 3^3 =
3^3 * 4^2
a=3, b=4, ab=12.</p>
<p>Possibly a quicker way to get the answer of “12”, is to first rewrite the term:</p>
<p>(a^1/2 x b^1/3) ^ 6 = a x (ab)^2</p>
<p>So you want one positive integer times another positive integer squared to equal 432. The problem asks for that second positive integer – the ab. You can go quickly through the list of choices (6, 12, 18, 24, and 36). ab can’t be 24 or 36 – the square is greater than 432. ab can’t be 18 – the square is 324, and so a would be 432/324 – certainly not an integer. ab can’t be 6, since that would mean that a is 12 (which is bigger than ab!). That leaves 12 squared (144), and observe that 144*3 is 432.</p>