Letter Code

[Akansha]

Each stack is designated with a 1, 2, and 3 letter code where each letter is selected
from the 26 letters of the alphabet. If the letters may be repeated and if the same
letters used in a different order constitutes a different code, how many different
stocks is it possible to uniquely designate with these codes?
a. 2951
b. 8125
c. 15600
d. 16302
e. 18278

OA is E

[Frankenstein]

Hi,
Number of stacks with 1 letter code = 26
Number of stacks with 2 letter code = 26^2(repetition allowed)
Number of stacks with 2 letter code = 26^3(repetition allowed)
So, total number of diff. stacks is 26+26^2+26^3 = 18278

Hence, answer E

Cheers!

[Brian@VeritasPrep]

Great answer, Frankenstein - nice work!

I recognize this problem and have always loved it for this reason - even if you get to that point of 26 + 26^2 + 26^3, getting from there to 18278 is still a pretty tall task if you're doing the math by hand. But remember - even when they're not specifically testing Number Properties, you can use them to your advantage!! If you look at the answer choices, each ends with a different units digit (1, 5, 0, 2, 8), so if you can just predict the units digit of the correct answer you won't have to square or cube 26.

Well, 26 is just 26.

26^2 will end in a 6, because 6*6 yields a units digit of 6.

And 26^3 will also end in a 6. Any time you're multiplying a units digit of a 6 by another units digit of a 6, you'll yield a 6.

So our addition problem will be:

26 + XX6 + YYYY6

And adding up those units digits we'll end with a value of 8. E is the only plausible answer, so it must be correct.

Almost exactly this same problem exists in our books, and I joke with my students that the notes in my book from the first time I did this problem are pretty roundabout, but use the same principle. I did:

26 + 26^2 + 26^3 ---> Factor out a common 26 to get:
26 (1 + 26 + 26^2)
26 (a units digit of 3, since 1 + 6 + 6 is 13)
26 * 3 ---> a units digit of 8

So I was still right, but maybe took one or two unnecessary steps. But I like making that point - as long as you perform correct steps, you can't "break" math...you're just finding a different way of phrasing an expression and hoping that it's more useful to your task. So if you look at a problem that adds three different exponents of 26 and think that you should factor, you'll still get there even if the right "tool" to use was a number property. The key on the GMAT is knowing your toolkit for problems with large numbers so that you don't have to slog through too much work. Factoring, number properties, etc. are great tools to know you have at your disposal.