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Difference between revisions of "2007 iTest Problems/Problem 11"

(Solution to Problem 11)
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Since <math>2^{2^{2^{\cdot^{\cdot^{\cdot^{2}}}}}}</math> with <math>2006</math> twos is a multiple of four, the units digit is <math>\boxed{\textbf{(G) }6}</math>.
 
Since <math>2^{2^{2^{\cdot^{\cdot^{\cdot^{2}}}}}}</math> with <math>2006</math> twos is a multiple of four, the units digit is <math>\boxed{\textbf{(G) }6}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=10|num-a=12}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 03:31, 10 June 2018

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 $\boxed{\textbf{(G) }6}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 10 		Followed by:
Problem 12
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