Quarter Wit, Quarter Wisdom: An Interesting Property of Exponents

Quarter Wit, Quarter WisdomToday, let’s take a look at an interesting number property. Once we discuss it, you might think, “I always knew that!” and “Really, what’s new here?” So let me give you a question beforehand:

For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?

Think about it for a few seconds – could you come up with the answer in the blink of an eye? If yes, great! Close this window and wait for the next week’s post. If no, then read on. There is much to learn today and it is an eye-opener!

Let’s start by jotting down some powers of numbers:

Power of 2: 1, 2, 4, 8, 16, 32 …

Power of 3: 1, 3, 9, 27, 81, 243 …

Power of 4: 1, 4, 16, 64, 256, 1024 …

Power of 5: 1, 5, 25, 125, 625, 3125 …

and so on.

Obviously, for every power of 2, when you multiply the previous power by 2, you get the next power (4*2 = 8).

For every power of 3, when you multiply the previous power by 3, you get the next power (27*3 = 81), and so on.

Also, let’s recall that multiplication is basically repeated addition, so 4*2 is basically 4 + 4.

This leads us to the following conclusion using the power of 2:

4 * 2 = 8

4 + 4 = 8

2^2 + 2^2 = 2^3

(2 times 2^2 gives 2^3)

Similarly, for the power of 3:

27 * 3 = 81

27 + 27 + 27 = 81

3^3 + 3^3 + 3^3 = 3^4

(3 times 3^3 gives 3^4)

And for the power of 4:

4 * 4 = 16

4 + 4 + 4 + 4 = 16

4^1 + 4^1 + 4^1 + 4^1 = 4^2

(4 times 4^1 gives 4^2)

Finally, for the power of 5:

125 * 5 = 625

125 + 125 + 125 + 125 + 125 = 625

5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 5^4

(5 times 5^3 gives 5^4)

Quite natural and intuitive, isn’t it? Take a look at the previous question again now.

For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?

A) 18

(B) 32

(C) 35

(D) 64

(E) 70

Which two powers when added will give 2^(36)?

From our discussion above, we know they are 2^(35) and 2^(35).

2^(35) + 2^(35) = 2^(36)

So x = 35 and y = 35 will satisfy this equation.

x + y = 35 + 35 = 70

Therefore, our answer is E.

One question arises here: Is this the only possible sum of x and y? Can x and y take some other integer values such that the sum of 2^x and 2^y will be 2^(36)?

Well, we know that no matter which integer values x and y take, 2^x and 2^y will always be positive, which means both x and y must be less than 36. Now note that no matter which two powers of 2 you add, their sum will always be less than 2^(36). For example:

2^(35) + 2^(34) < 2^(35) + 2^(35)

2^(2) + 2^(35) < 2^(35) + 2^(35)

etc.

So if x and y are both integers, the only possible values that they can take are 35 and 35.

How about something like this: 2^x + 2^y + 2^z = 2^36? What integer values can x, y and z take here?

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Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!