The tens digit of 6^17 is

Question

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

KarishmaB · Accepted Answer

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)

82vkgmat · Answer

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.

PareshGmat · Answer

Answer = B = 3

6^1 = 6

6^2  =  3 6

6^3  = 2 1 6

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

AliciaSierra · Answer

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

Bunuel · Answer

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 	imes 6^1 = 36  (tens digit = 3)
 
 6^3 = 6 	imes 6^2 = ..16  (tens digit = 1)
 
 6^4 = 6 	imes 6^3 = ..96  (tens digit = 9)
 
 6^5 = 6 	imes 6^4 = ..76  (tens digit = 7)
 
 6^6 = 6 	imes 6^5 = ..56  (tens digit = 5)
 
 6^7 = 6 	imes 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.

DIII · Answer

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.

nick1816 · Answer

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}

BrushMyQuant · Answer

We need to find the tens digit of 6^17

Last Two digits of 6 follow following pattern

Last Two Digits of 6^1 = 06 
 Last Two Digits of 6^2 = 36 
 Last Two Digits of 6^3 = 36*6 = 16 
 Last Two Digits of 6^4 = 16*6 = 96 
 Last Two Digits of 6^5 = 96*6 = 76 
 Last Two Digits of 6^6 = 76*6 = 56 
 Last Two Digits of 6^7 = 56*6 = 36   
 Last Two Digits of 6^8 = 36*6 = 16

=> So, from 6^2 cycle of tens' digit repeats as 3, 1, 9, 7 , 5, 3, 1 9, 7 , 5
=> We have a cycle of 5 from the 2nd term 
 => Tens' digit of 6^17 = Now 17 = 1(6^1 not part of cycle) + 15 (cycle of 5*3) + 1 (as cycle starts from 2nd term) = First term of the cycle = 3

So, Answer will be B 
Hope it helps!

Link to Theory for Last Two digits of exponents here    .

Link to Theory for Units' digit of exponents here    .

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The tens digit of 6^17 is

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