# Manhattan GMAT Challenge Problem of the Week – 10 Aug 2010

by on August 10th, 2010

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people enter our challenge, the better the prizes.

As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

## Question

The operation x&n for all positive integers n greater than 1 is defined in the following manner:

x&n = x raised to the power of x&(n–1).

If x&1 = x, which of the following expressions has the greatest value?

(A) (3&2)&2
(B) 3&(1&3)
(C) (2&3)&2
(D) 2&(2&3)
(E) (2&2)&3

First, we need to figure out what this strange operation means for a few small integers n. Let’s build upward from 1:

x&1 = x

x&2 = x raised to the power of x&1 (which is just x), so x&2 = xx = x^x (we’ll use the caret symbol ^ to represent exponentiation, since as we’ll see, we’re going to do it a lot!)

x&3 = x raised to the power of x&2, so x&3 = x^(x^x)

x&4 = x^(x^(x^x))

So the number after the & sign tells you how many x’s are in the exponential expression. Now we can translate the answer choices. As always, do the operation inside the parentheses first.

(A) (3&2)&2

3&2 = 3^3 = 27

27&2 = 27^27 = (3^3)^27 = 3^81

(B) 3&(1&3)

1&3 = 1^(1^1) = 1^1 = 1

3&1 = 3

(C) (2&3)&2

2&3 = 2^(2^2) = 2^4 = 16

16&2 = 16^16 = (2^4)^16 = 2^64

Because both the base and the exponent of this answer choice are smaller, we can tell that choice A is still the winner at this point.

(D) 2&(2&3)

2&3 = 2^(2^2) = 2^4 = 16

2&16 = 2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^2)))))))))))))))

There are sixteen 2’s in this “tower of powers”! This number is incredibly large, far larger than 3^81. Let’s start to collapse the layers to see why.

2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^2)))))))))))))))

= 2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^4))))))))))))))

= 2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^(2^16)))))))))))))

2^16 = 65,536. You aren’t expected to know that, of course, but now imagine 2 raised to that power. This number has thousands of digits.

Now imagine 2 raised to THAT power.

Then 2 raised to THAT power.

And so on, over 10 more times!

This number is the winner by far among the first four answer choices.

(E) (2&2)&3

2&2 = 2^2 = 4

4&3 = 4^(4^4) = 4^(256) = 2^512

While enormous, this number is still far smaller than answer choice (D).

By the way, the operation represented by the & sign in this problem is sometimes called “tetration.” The reason is that just as multiplication is repeated addition, and exponentiation is repeated multiplication, so-called “tetration” is repeated exponentiation. (“Tetra-“ means “four,” and this operation is fourth in line: addition, multiplication, exponentiation, tetration.) Tetration is also called superexponentiation, ultraexponentiation, hyper-4, and power tower.