What is the tens digit of 6^17?

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

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

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

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

Yes, that's correct.

\(6^2 = 36\)
\(6^3 = 36 * 6 = n16\)
\(6^5 = 6^2 * 6^3 = 36 * n16 = n76\)

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

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

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.

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

Posted from my mobile device

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.

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

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/

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

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

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.

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.

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