2007 iTest Problems/Problem 11

Problem 11

Consider the "tower of power" $2^{2^{2^{\cdot^{\cdot^{\cdot^{2}}}}}}$ , where there are 2007 twos including the base. What is the last (units digit) of this number?

$\text{(A) }0\qquad \text{(B) }1\qquad \text{(C) }2\qquad \text{(D) }3\qquad \text{(E) }4\qquad \text{(F) }5\qquad \text{(G) }6\qquad \text{(H) }7\qquad \text{(I) }8\qquad \text{(J) }9\qquad \text{(K) }2007\qquad$

Solution

Note that $2^1 = 2$ , $2^2 = 4$ , $2^3 = 8$ , $2^4 = 16$ , and $2^5 = 32$ . The units digit of $2^n$ cycle every time $n$ is increased by $4$ .

Since $2^{2^{2^{\cdot^{\cdot^{\cdot^{2}}}}}}$ with $2006$ twos is a multiple of four, the units digit is $\boxed{\textbf{(G) }6}$ .

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2007 iTest Problems/Problem 11

Problem 11

Solution