What is the tens digit of 6^17?

Question

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

Bunuel · Accepted Answer

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.

shrouded1 · Answer

Just like cyclicity of the last digit, we can observe the cyclicity of the last 2 digits in this case : (which is possible because there is no cyclicity in the unit's digit, it is always 6)
6^2 = 36
6^3 = 16
6^4 = 96
6^5 = 76
6^6 = 56
6^7 = 36
.. and then it repeats

So for 6^17, it will have the same tens digit as 6^12, 6^7, 6^2 ... or 3

Answer is (b)

prab · Answer

shrouded can you please explain it a bit more, didnt get why 6^17, it will have the same tens digit as 6^12, 6^7, 6^2 .. thank you in advance

shrouded1 · Answer

So the logic here is simple. Consider the number 6^x, lets say that you know the tens digit of this number, can you find out the tens digit of 6^(x+1) ?

What we know is that the last digit of 6^x will always be 6 (which is easy enough to see). Now the fact of the matter is that the ten's digit of 6^(x+1) is only dependent on the tens digit of 6^x.

Because Ten's digit of 6^(x+1) = 6*(Ten's digit of 6^x) + 3 (carried over during the multiplication of the units digits 6 with the new 6).

Like 6^3 = 216
So 6^4, units digit is 6 and ten's digit is 6*1+3 = 9

Hence, as soon as the ten's digit of 6^x becomes the same as the ten's digit of 6^y, the pattern of tens digit will start to repeat itself

6^2 = 36
6^3 = 16
6^4 = 96
6^5 = 76
6^6 = 56
6^7 = 36

The pattern here is 3,1,9,7,5,3,1,9,7,5,3,1,9,7,5,3,1,9,7,5,3.....
The cyclicity of the pattern is five, so every 5th element in this series will be the same hence 2nd,7th,12th,17th have to be the same

nonameee · Answer

Bunuel,can you calculate it with modulo as below:

1) periodicity of the ten's digit is 5
2) 17 mod 5 = 2
3) 6^17 will have the same digit as 6^2

rajeevrks27 · Answer

well, this question demands calculation to see a pattern of tens digits
keep calculating till it's confirmed that u have hit a pattern.
6^1 = 6
6^2 = 36
6^3 = 216
now don't multiply 216 by 6, rather we are interested in only first two digits to know the outcome so
6^4 = 96 ( 16 x 6)
6^5 = 576 ( 96 x 6)
6^ 6 = 456 ( 76 x 6)
7^ 6 =336 ( 56 x 6)
so now we have the pattern in tens digit i.e. 
 3 in (6^2),
 1 in (6^3),
 9 in (6^4),
 7 in (6^5),
 5 in (6^6),
3 in (6^7),

so the tens digit is 3 for the 2,7,12 and 17 times..
IMO B

Bunuel · Answer

Yes, that's correct.

mbaiseasy · Answer

6^2 = 36 
 6^3 = 36 * 6 = n16 
 6^5 = 6^2 * 6^3 = 36 * n16 = n76

Note that when you multiply, you don't have to finish it all the way, knowing the tens digit should suffice... .
 Also, using the table we have we can calculate  6^{10}  and  6^{17} . We work with what we already have above/

6^10 = 6^5 * 6^5 = n76 * n76 = n76 
 6^7 = 6^5 * 6^2 = n36 * n76 = n36 
 6^{17} = 6^{10} * 6^{7} = n76 * n36   (We already know what happens to n76 * n36 as calculated above...)   =n36

Answer: B

chiccufrazer1 · Answer

bunuel would you please post me a link on the topic of exponents and powers from gmat math book if it has been finished..i want to learn and master way u hav solved the problem

Posted from my mobile device

Bunuel · Answer

For more on number theory and exponents check: https://gmatclub.com/forum/math-number-theory-88376.html DS questions on exponents: https://gmatclub.com/forum/search.php?se ... &tag_id=39 PS questions on exponents: https://gmatclub.com/forum/search.php?se ... &tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: https://gmatclub.com/forum/tough-and-tri ... 25967.html Tough and tricky PS exponents and roots questions with detailed solutions: https://gmatclub.com/forum/tough-and-tri ... 25956.html Hope it helps.

chiccufrazer1 · Answer

i have noticed that every number has 6 as the unit digit..is it the same for other numbers that they repeat each of the unit's digit throughout when it is being raised to powers of consecutive integers

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Bunuel · Answer

No. You could test that very easily yourself. Is the units digit of 2^2 equal 2? No, its 4. • Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base. • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4. For more check here: https://gmatclub.com/forum/math-number-theory-88376.html Hope it helps.

AbhimanyuDhar · Answer

If Question like this appears in GMAT. and we are asked to find the last two digits this could be used

Rule Express Even numbers in the form ( 2^10) ^even which will have last two digits as 76 or ( 2^10) ^odd where last two digits is 24

For Odd numbers Express them 3^4k, 7 ^ 4k, 9 ^2k

Question was 6^17

So (2^ 17 ) ( 3^17)

= {(2^10)^1 * 2^7} { (3^4)^3 * 3^5}

So last two digits (2^10)^1= 24
 So last two digits 2^7= 28

Last two digits of this number (3^4)^3 = (81)^3= last two digits are 41 (1 will be the last digit and second last digit will be 4 that i got by multiplying 8 *3 
last two digits of this number 3^5 = 43

So (24 * 28) * (41* 43)

Don't do complete multiplication just do it till you get two digits

72 * 63= 36

So digit in tenth place is 36

KarishmaB · Answer

Here is a post discussing the use of pattern recognition for last two digits: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/1 ... ns-part-i/

Anoop Singh · Answer

6^17 = (2×3)^17 = 2^17 × 3^17
Now, 2^17 = 2^10 × 2^7 = 24 × 28 = 72
And 3^17 = 3^16 × 3 = 81^4 ×3 = 21 × 3 = 63 
Lasr two digit = 72 × 63 = 36 
Tens digit = 3

Posted from my mobile device

KanishkM · Answer

IF you are able to identify the pattern you should be good, I find this approach i fool proof one

6
 36 
216
1296
7776
46656
 xxxx36

Now after that point you need not solve, as tens digit will be same after  this , After every 5th term the series will start again

Remove the first term 6 from the series, you will be left with 16 terms, making, 3 as the correct answer

B

gmatapprentice · Answer

Bunuel  Does the same rule of cyclicity apply of 5 apply for the tens digit of other numbers as well? This is the first time I have seen a question on cyclicity of the tens digits.

Bunuel · Answer

Check the links below:
 Cyclicity and the remainders on the GMAT 
 Units digits, exponents, remainders problems 
 The Units Digits of Big Powers

Hope it helps.

allan89 · Answer

Second last digit is nothing but remainder of 6^17 divided by 100

remainder of 6^17 / 100 = 36 so 2nd last digit is 3

What is the tens digit of 6^17?

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What is the tens digit of 6^17?

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