How many different pairs of integers (a, b) exist such that 2 <= a <=

MANHATTAN GMAT OFFICIAL SOLUTION:

A glance at the answer choices shows that you’re not going to be counting out all of the individual pairs, so there must be some kind of pattern. How can you find it?

A restriction is given on a, but not on b (except that it’s an integer), so it’s best to begin the reasoning with a.

Try a = 2. The equation becomes 2+b>2^b. How to solve that? Think theoretically just a little bit. In general, raising 2 to a power will result in a number that grows more quickly than merely adding 2 to something. What are a few circumstances in which the 2+b portion will actually be bigger? You would need values for b that would minimize 2^b.

b=0 would work: 2+0>2^0. So would b=1: 2+1>2^1. If you go any bigger than that though, then the 2b side will get too big (e.g. if b=2, then 2+2=2^2). What about the other direction: negative integers? If b=-1, then 2+(-1)<2^(-2). Remember that a negative exponent will not cause a positive integer base to become negative. Rather, the base will just become a positive fraction.) There are three possible b values then when a=2, so there are 3 pairs of integers here that fulfill the inequality.

Okay, so what about a=3? Using the same “testing cases” method shown above, 3+b>3^b is true for four values: b=-2,-1,0, 1.

Try a=4. This time 4+b>4^b, which is true for five values: b=-3,-2,-1,0, 1.

Hmm. So far:

When a=2, b has 3 values: -1,0,1.

When a=3, b has 4 values: -2,-1,0,1.

When a=2, b has 3 values: -3,-2,-1,0,1.

That seems like a pattern. Each time, b hasn’t been able to be larger than 1. Each time, b had added one more negative integer. Does it hold all the way to a=200?

If you raise 2 to an integer larger than 1, it’s going to start getting bigger very rapidly, much more rapidly than adding that same number to 2. It makes sense that b can never be larger than 1. Okay, so that side of the pattern holds.

What about the negative end? Especially, the a+b side needs to stay positive in order to be greater than the a^b side, because the a^b side can never be zero or negative. In that case, if b is negative, the magnitude (distance from zero) has to be smaller than a‘s magnitude. If a=3, then b can’t be -3,-4,-5, and so on. If a=4, then b can’t be -4 or smaller. So this side of the pattern also holds.

The pattern also shows that the number of pairs for any value of a is equal to a+1. The number of possible pairs, then, is 3+4+5+...+201. (The final possible value for a is 200.)

There are 201-3+1=199 items with an average value of (3+201)/2=102, so the sum is 199*102=20298.

The correct answer is (E).