If *a* and *b *are positive integers, what is the greatest possible value of *a*?

(A) 2

(B) 5

(C) 10

(D) 11

(E) 13

## Explanation

Though you might be tempted to run for your calculator, this problem can be solved using a couple of logic-based approaches. Indeed, no intense calculation needed, just some application of exponents.

, which approximates to 500. So , or slightly less than 100. Therefore, the answer is most likely (D). However, you might argue, 500 isn’t quite 512, and that little bit of difference could make a whole lot of difference when taken to the eleventh power.

Notice the answers, though. There is no twelve, but only thirteen. So if it is a hard call, without a calculator, to know if that extra 500 will make the left side greater than the right, we don’t need to worry: 12 is not an answer. (E) 13, which would add two ‘500s’ to the left, or about would definitely go over . In other words, that slight difference between 500 and 512, taken to the eleventh power, would be much smaller than . Therefore, answer (D).

## Answer: D

hey i was able to solve this quickly by more or less the same method explained here.

i factored 500 as 2x5x2x5x5 and the approximated 5 as 2×2 so i got 500=2^8

to get the greatest possible value of a substitute b =1

2^8a < 2^100

8a<100

a<12.5

so it has to be 11.

I’m a freshman in undergrad and thought about the problem in the following way:

From previous math experience I know that log(5) is approximately 0.7 (really like 0.699) and log(2) is approximately 0.3 (really like 0.301).

If 500^ab < 2^100, then log(500^ab) < log(2^100).

By properties of logarithms, log(500^ab) becomes 2.7ab and log(2^100) becomes 100(0.3) and, hence,

2.7a < 30

Clearly, the largest integer value of a that satisfies the equation is 11.

I would assume that most have not memorized approximations of logarithms of numbers, but committing approximations of log(2), log(3) and log(5) [even though you really only need either log(2) or log(5) to know the other because of properties of logarithms] to memory has proven to be very useful in any problem requiring quick approximations of numbers with large exponents.

Hi Chris,

I am still an average Maths student. Here’s the method I tried for this problem :

Splitting : 500 = 512-12

Consider b=1

Therefore we can write : (500)^ab = (512-12)^ab

=2^9ab – 12^ab

For options

a) 2

2^18 – 12^2 Fine. Still not the greatest.

b) 5

2^45 – 12^5 Fine. Still not the greatest.

c) 10

2^90 – 12^10 Fine. Still not the greatest.

d) 11

2^99 – 12^11 May be!

e) 13

2^117 – 12^13 May be!

Using this approach, I am confused with option d) and e)

You can’t write (A-B)^C as (A^C – B^C).

Thanks Kunal. That was a big flaw.