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! If no, then read on.
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?
