Official Solution:The tens digit of \(6^{17}\) is

A. 1

B. 3

C. 5

D. 7

E. 9

We know that there must be a pattern, since we can’t be expected to expand \(6^{17}\) out to all its digits. In other words, we must be able to spot a repeating cycle of digits.

The only way forward is to compute tens digits for powers of 6, starting with \(6^1\), and see what we get. To go up, multiply the previous result by 6 and drop any higher digits than the tens, but we have to keep the units digit (which, as we’ll see, will be 6 every time).

\(6^1 = 6\) (no tens digit)

\(6^2 = 6 \times 6^1 = 36\) (tens digit = 3)

\(6^3 = 6 \times 6^2 = ..16\) (tens digit = 1)

\(6^4 = 6 \times 6^3 = ..96\) (tens digit = 9)

\(6^5 = 6 \times 6^4 = ..76\) (tens digit = 7)

\(6^6 = 6 \times 6^5 = ..56\) (tens digit = 5)

\(6^7 = 6 \times 6^6 = ..36\) (tens digit = 3)

Whew - the numbers finally started repeating! The cycle is 3, 1, 9, 7, 5 - which is 5 terms long. Every power will have the same tens digit as the 5th larger power, so \(6^2\), \(6^7\), \(6^{12}\), and most importantly \(6^{17}\) will all have 3 as their tens digit.

Notice that the pattern didn’t start until \(6^2\). \(6^1\) doesn’t have a tens digit (or has a tens digit of 0, but this digit is never repeated later in the cycle).

Answer: B

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