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The tens digit of 6^17 is

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The tens digit of 6^17 is  [#permalink]

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New post 17 Nov 2014, 12:42
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Re: The tens digit of 6^17 is  [#permalink]

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New post 17 Nov 2014, 22:40
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Bunuel wrote:

Tough and Tricky questions: Exponents/Powers.



The tens digit of 6^17 is

A. 1
B. 3
C. 5
D. 7
E. 9

Kudos for a correct solution.


For tens digit, pattern recognition is very effective.

6^2 = 36
6^3 = 216
6^4 = ...96 (Just multiply the last two digits since we only care about the tens digit)
6^5 = ...76
6^6 = ...56
6^7 = ...36
and hence starts the cycle again:

3, 1, 9, 7, 5,
3, 1, 9, 7, 5,
and so on

The new cycle with tens digit of 3 begins at the powers of 2, 7, 12, 17 etc
So 6^17 will have 3 as the tens digit.

Answer (B)
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Re: The tens digit of 6^17 is  [#permalink]

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New post 17 Nov 2014, 23:06
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Bunuel wrote:

Tough and Tricky questions: Exponents/Powers.



The tens digit of 6^17 is

A. 1
B. 3
C. 5
D. 7
E. 9

Kudos for a correct solution.



I got B as well. Started with 6, 36, 216, 1296, ....76, ...56, ...36. The pattern is 3, 1, 9, 7, 5, ...... So 6^2 has same tens digit as 6^17 which is 3.

Answer = B.
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Re: The tens digit of 6^17 is  [#permalink]

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New post 18 Nov 2014, 01:18
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Answer = B = 3

\(6^1 = 6\)

\(6^2\) = 36

\(6^3\) = 216

\(6^4\)>> Units place = 16*6 = 9

\(6^5\) >> Units place = 96*6 = 7

\(6^6\) >> Units place = 76*6 = 5

\(6^7\) >> Units place = 56*6 = 3

Cyclicity is 3, 1, 9, 7, 5

\(6^2, 6^7, 6^{12}, 6^{17}\) will have the same digit in units place = 3

Answer = 3
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Re: The tens digit of 6^17 is  [#permalink]

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New post 18 Nov 2014, 01:42
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In order to find tense place, we can check cyclicity of tense place of 6^1, 6^2, 6^3 etc.

Let us find out.

6^2 = 36 --> tense place is 3
6^3 = 216 --> tense place is 1
6^4= ..296 --> tense place is 9
6^5= ....76 --> tense place is 7
6^6=.....56 --> tense place is 5
6^7=.....36 --> tense place is 3 (Please note, here last two digits are 36, which is also same for 6^2)
6^8=.....16 --> tense place is 1 ( same as in 6^3)

so we can see that last two digits are repeating after 6^7
i.e. last two digit of 6^2= last two digit of 6^7
last two digit of 6^3= last two digit of 6^8
last two digit of 6^4= last two digit of 6^9
so on.

So cyclicity of tense place is 5.

So 6^2, 6^7, 6^12, 6^17 will have 3 in tense place.

Answer: B

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Re: The tens digit of 6^17 is  [#permalink]

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New post 18 Nov 2014, 08:02
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: The tens digit of 6^17 is  [#permalink]

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New post 24 Jun 2017, 08:48
I think multiplying till finding a pattern would not be a sure way to solve.
the method i used is congruency

6: 6mod100
6^2: 36mod100
6^4: 96mod100
6^8: 16mod100
6^16: 56mod100
6^17: 36mod100

so 36 would be the last two digits of the final number.
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Re: The tens digit of 6^17 is  [#permalink]

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New post 11 May 2019, 10:48
DIII wrote:
I think multiplying till finding a pattern would not be a sure way to solve.
the method i used is congruency

6: 6mod100
6^2: 36mod100
6^4: 96mod100
6^8: 16mod100
6^16: 56mod100
6^17: 36mod100

so 36 would be the last two digits of the final number.


Can you please elaborate your method?
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The tens digit of 6^17 is  [#permalink]

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New post 11 May 2019, 18:53
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1. When u divide any number by 100, remainder is last 2 digits of that number.
2. Remainder of X*Y, divided by 100= Remainder of X when divided by 100 * Remainder of Y when divided by 100
Dividend≡ (Remainder) mod divisor
6≡6mod100
6^2≡36mod100 {6*6=36}
6^4≡96mod100 {36*36=1296}
or 6^4≡(-4) mod 100

6^8≡16mod100 {-4*-4=16}
6^16≡56mod100 {16*16=256}
6^17≡36mod100 {56*6=336}





BarcaForLife wrote:
DIII wrote:
I think multiplying till finding a pattern would not be a sure way to solve.
the method i used is congruency

6: 6mod100
6^2: 36mod100
6^4: 96mod100
6^8: 16mod100
6^16: 56mod100
6^17: 36mod100

so 36 would be the last two digits of the final number.


Can you please elaborate your method?
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Re: The tens digit of 6^17 is  [#permalink]

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New post 15 May 2019, 10:00
Bunuel wrote:
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.


hi bunuel , a silly doubt
shouldn't we consider 6^1 =06 claiming tens digit as 0 ?
thanks a lot in advance
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Re: The tens digit of 6^17 is   [#permalink] 15 May 2019, 10:00
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