Is x < 45^{45}? 1) x < 45! 2) x <67!/22!

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Is \(x < 45^{45}\)?

1) \(x < 45! \)

Clearly,
45! < \(45^{45}\)
45*44*43.....2*1 < 45*45*45...45(45 times)

Sufficient.

2) \(x <\frac{67!}{22!}\)
\(x <\frac{67*66*65....23*22!}{22!}\)
\(x <67*66*65....23\)
67 - 23 = 44 + 1 = 45 numbers

So we have pairs as
45 + 22 and 45 - 22
45 + 21 and 45 - 21
and 45 the individual number itself i.e. 22 numbers and one 45, totalling 23 numbers.

Notice that none of the multiples(22 pairs) are greater than \(45^2\) which has 22 pairs and one 45 itself.

Hence (45 + 22)*(45 - 22)*(45 + 21)*(45 - 21)....(45 + 1)*(45 - 1)*45 < \((45*45)^{22}*45\)

Sufficient.

Answer D.

If we take a simple set of numbers, let's say, 2 x 2 x 2 = 8 Now if we reduce one term by 1 unit and increase the other by the same we get 1 x 2 x 3 = 6. I notice that by increasing and decreasing two numbers by the same unit our overall product decreases BUT when we have a set of numbers, let's say, 2 + 2 + 2 = 6 Now if we follow the same process we get 1 + 2 + 3 = 6 the overall sum remains that sum.

Why is that while MULTIPLYING our overall value decreases (despite increasing and decreasing the values by the same unit) while ADDING our overall value remains the same? IanStewart KarishmaB I am curious to understand what's going on "behind the scenes"

If you start with the list 2, 2, 2, and then you subtract 1 from one number, and add 1 to another to get the list 1, 2, 3, because you've added 1 and subtracted 1, the sum won't change.

The multiplication case is more interesting, and there are (at least) two ways to see why we'll get a smaller product if we add 1 to one number and subtract 1 from another, if we start from the list 2, 2, 2. The first way is more algebraic, and the second more conceptual. If we take the list 1, 2, 3, and rewrite each number using its distance from 2, we get this list:

2 - 1, 2, 2 + 1

and now if you multiply this out, when you multiply the first and last terms, you get 2^2 - 1, which is less than the 2^2 you get when you do the same with the original list 2, 2, 2. So you'll get a smaller product overall. Assuming you're dealing with positive numbers, that will always happen when you compare the product of a list like b, b, b with the product of a list like b - k, b, b + k.

But there's a more conceptual way to see what's going on -- instead of thinking in terms of adding and subtracting, we can think in terms of multiplication. Again starting from 2, 2, 2, if you write each term in the new list 1, 2, 3 as a multiple of 2, notice that to get the '1' we're reducing 2 by 50%, and to get the '3' we're increasing 2 by 50%. So our new list is

(0.5)(2), 2, (1.5)(2)

and when you multiply this out, you get 2^3 * (1.5)(0.5). So we're really just getting our old product, 2^3, then increasing it by 50% (multiplying it by 1.5), and then decreasing it by 50% (multiplying it by 0.5). And if you increase a number by 50%, then decrease the result by 50%, you always end up with something smaller than what you started with, because the 50% decrease is being taken from a larger overall value than the 50% increase was taken from.

The rest of this post goes well beyond the scope of the GMAT, so anyone who only cares about studying for the test won't benefit from reading further. But the question you're asking about is really a special case of one of the most important theorems in mathematics, the "harmonic mean-geometric mean-arithmetic mean inequality" (sometimes extended to include the "quadratic mean"), often abbreviated as the HM-GM-AM inequality, which just says that for any set of positive numbers that are not all equal, HM < GM < AM. Here, the AM is the arithmetic mean, or the familiar average, so for the set 1, 2, 3 it is just equal to 2. The GM is the geometric mean, which is sort of the "product average" -- for a set of n numbers, it's the nth root of the product. So for the set 1, 2, 3, the geometric mean is the 3rd root, or cube root, of the product, which is 1*2*3 = 6, so the GM here is roughly 1.82. If you think about how this relates to your question, if you have a set of three numbers, the GM-AM inequality says that the cube root of the product is always less than the traditional average, or in other words, the product is less than the average cubed. So you could be sure just looking at a set like 1, 2, 3 that the product is less than 2^3, and you can be sure just looking at a set like 981, 990, 999, that the product is less than 990^3, because we can identify the average instantly with these kinds of sets. Or if you had a set like 81, 89, 100, say, doing a quick calculation, the normal average is 90, so from the GM < AM inequality, the product 81*89*100 must be less than 90^3 (the set doesn't need to be equally spaced).

The last kind of 'mean' in the HM-GM-AM inequality, the HM or harmonic mean, is actually one every GMAT test taker has used many times. The harmonic mean is the reciprocal of the average of the reciprocals, which probably sounds confusing at first. For the set 1, 2, 3, the harmonic mean would be the reciprocal of the average of 1/1, 1/2 and 1/3, so it is the reciprocal of (1/1 + 1/2 + 1/3) / 3 = 11/18, so it would be 18/11, which is roughly 1.64. So as the HM-GM-AM inequality says must happen, the HM is less than the GM is less than the AM for the set 1, 2, 3. If you know the 'rates formula' that some people use to solve standard rates problems, that formula says 1/t = 1/a + 1/b, and if you solve for t, you get

\(\\
t = \frac{1}{\frac{1}{a} + \frac{1}{b}}\\
\)

Notice on the right side, you have the reciprocal of the sum of the reciprocals of a and b, so the right side is exactly half of the harmonic mean of a and b. So the answer to a typical combined two-worker problem is just half the harmonic mean of a and b, so the harmonic mean is something GMAT test takers see all the time. We know HM < AM is always true (if our numbers aren't equal) from the famous theorem I mentioned above, and you can actually use this to get good estimates in rates problems. If you have a question like "one worker takes 10 hours to do a job, the other takes 15 hours to do a job, how long does it take them together to do the job?", from the above, we know the answer is half of the harmonic mean of 10 and 15. But the normal average (AM) of 10 and 15 is 12.5, and we know HM < AM, so HM/2 < AM/2 = 6.25. Since HM/2 is the answer to the question, the answer must be less than 6.25 (and when the two numbers aren't extremely far apart, it won't be much less, so you could guess the answer would be very close to 6 here and that turns out to be exactly right). There are better ways to estimate answers in most GMAT-level rates problems than this, which is why I wouldn't suggest learning it, though I have used this principle exactly one time to get a quick solution to an otherwise very awkward official question. I've seen many thousand official questions though, so it's essentially never important to know. So this is all for interest only (unless you want to get a head start on MBA level math). It's also a difficult theorem to prove in general (for special types of sets like equally spaced sets at least the GM-AM inequality is easy to prove, and I haven't given any thought to the HM-GM inequality in that situation, but when the sets aren't equally spaced, any proof gets very complicated), so don't ask me to do that.