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S97-15

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S97-15 [#permalink]

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New post 16 Sep 2014, 01:51
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

50% (02:08) correct 50% (01:26) wrong based on 48 sessions

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Re S97-15 [#permalink]

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New post 16 Sep 2014, 01:51
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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Re: S97-15 [#permalink]

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New post 17 Mar 2015, 07:27
It can also be done fast by breaking 6 into: 2 to the power 17 and 3 to the power 17.
and then check the cyclicity....

:) :)
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Re: S97-15 [#permalink]

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New post 07 Dec 2017, 10:11
Can you please explain how?
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Re: S97-15 [#permalink]

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New post 07 Dec 2017, 10:14
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What is the tens digit of 6^17?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

There are several ways to deal with this problems some easier some harder, but almost all of them are based on the pattern recognition.

The tens digit of 6 in integer power starting from 2 (6^1 has no tens digit) repeats in pattern of 5: {3, 1, 9, 7, 5}:
The tens digit of 6^2=36 is 3;
The tens digit of 6^3=216 is 1;
The tens digit of 6^4=...96 is 9 (how to calculate: multiply 16 by 6 to get ...96 as the last two digits);
The tens digit of 6^5=...76 is 7 (how to calculate: multiply 96 by 6 to get ...76 as the last two digit);
The tens digit of 6^6=...56 is 5 (how to calculate: multiply 76 by 6 to get ...56 as the last two digits);
The tens digit of 6^7=...36 is 3 again (how to calculate: multiply 56 by 6 to get ...36 as the last two digits).

Hence, 6^2, 6^7, 6^12, 6^17, 6^22, ... will have the same tens digit of 3.

Answer: B.
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Re: S97-15   [#permalink] 07 Dec 2017, 10:14
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