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The tens digit of 6^17 is
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17 Nov 2014, 12:42
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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.
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Re: The tens digit of 6^17 is
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17 Nov 2014, 22:40
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
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17 Nov 2014, 23:06
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
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18 Nov 2014, 01:18
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
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18 Nov 2014, 01:42
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
Regards, Ammu



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Re: The tens digit of 6^17 is
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18 Nov 2014, 08:02
Official Solution:The tens digit of \(6^{17}\) isA. 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
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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
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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
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11 May 2019, 18:53
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
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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




Re: The tens digit of 6^17 is
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