What is the tens digit of 11^{13}?

Question

What is the tens digit of  11^{13} ?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Kudos for a correct  solution .

noTh1ng · Accepted Answer

In this case ciclicity is helpful.

11^1 = 11
11^2 = 121
11^3 = 1331
11^4 = 14641

As one can recognize then, the tens digit is equal to the exponents digit. 11^11 is equal to 11^1 tens digit wise, therefore 11^13 has the same tens digit as 11^3, which would be 3.

Answer C

scorpionkapoor77 · Answer

11^2  has 2 in Tens digit
 11^3  has 3 in Tens digit
 11^9  has 9 in Tens digit
11^11 has 1 in Tens digit
Since the Tens digit of the number is same as the units digit of the power of 11, Therefore, 11^13 must have 3 in Tens digit.

Answer is C 3. IMO

kunal555 · Answer

11^1  = 11
 11^2  = 121
 11^3  = 1331

When the power of 11 is increased by 1, the 10's digit will increase by 1.
This will continue till  11^9 .
For  11^{10} , we will get 0 at ten's place.
Same cycle will again continue for next powers of 11.
So,
10's digit of  11^{13}  will be same as 10's digit of  11^3  i.e 3

Answer:- C

anudeep133 · Answer

Solution:

11^n will have units digit of n as its tens digit. 
Ex. 11^ 2  = 1 2 1 , 11^ 3  = 13 3 1... so on.
11^1 3  will have 3 as tens digit.

Option C

harishbiyani8888 · Answer

11 ^13 can be written as 11^8 x 11^4 x 11^1

Now, last two digits of 11 raise to power are

11 = 11
11^2 = 21
11 ^4 = 21 * 21 = 41
11 ^ 8 = 41 * 41 = 81

SO 11 ^13 = 81 * 41 * 11 = last two digits = 31

So the tens digit is 3.

Correct answer is C.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

This question may well test the upper level of GMAT difficulty, but can be solved using the large-number-with-exponents strategy of finding patterns. Look at the progression of the easier-to-calculate powers of 11:

11^{1}= 11

11^{2}= 121

11^{3} = 1331

11^{4}= 14641

As you’re looking at digits places, you should first note that the units digit of each is 1, which will hold true anytime you’re multiplying a units digit of 1 by a units digit of 1. Now look at the tens places. They maintain the value of the exponent and progress from 1 to 2 to 3 to 4. They simply increase by one each time. Now, you may well just assume that the pattern will continue to to hold, and after this many in a row with your knowledge that exponents are very pattern-driven, you’re likely to be correct. But for the sake of thoroughness, let’s look at why that rule holds. Each time you multiply a number by 11, you’re multiplying it by 10+1. Therefore, you can multiply it by 1, multiply it by 10, and add those products together. Multiplying by 1 is easy (it just stays the same) and multiplying by 10 is the same as multiplying by 1 and just adding a 0 at the end. Therefore,  11^{3} , or  11^{2}*11 , is:

121*1 = 121

121*10 = 1210

121 + 1210 = 1331

Watching the tens place, you should notice that the *10 portion just takes the previous units digit (which is always 1) and shifts it to the tens place, and that the *1 portion keeps the previous tens place. Essentially, we use the *10 portion to add 1 to the tens digit each time, ensuring that the pattern will hold, and the  tens digit of any power of 11 will simply be the units digit of the exponent .

Therefore, the answer is C.

Omkar1527 · Answer

Hi Bunuel,

Thank you for the in detail explanation.

About this - "tens digit of any power of 11 will simply be the units digit of the exponent"

do we have any other numbers with such interesting facts ?

Thanks,

walker · Answer

11^13 = (10 + 1)^13 = 10^13+...+1 3 0+1

It's C (3)

Posted from my mobile device

MathRevolution · Answer

11^1 = 11

11^2 = 121 :  Increase the power of 11 by 1 and thus increase Ten's digit of the previous result by 1.

Continue till  11^9 = xxxxxxxxxx91

For  11^10  = Add 1 to Ten's digit (9) and thus it becomes 0.

=>  11^{11} = 0 + 1  =  1

=>  11^{12}  =  1  + 1 =  2

=>  11^{13} =  2  + 1 =   3

Answer C

BrushMyQuant · Answer

We need to find the tens digit of 11^13

Last Two digits of 11 follow following pattern

Last Two digits of 11^1 = 11 
 Last Two digits of 11^2 = 21 
 Last Two digits of 11^3 = 21*11 = 31 
 Last Two digits of 11^4 = 31*11 = 41 
 Last Two digits of 11^5 = 41*11 = 51 
 . 
 . 
 . 
 This will go on till 11*10 where the pattern will be set to 01 
 And for 11^11 it will again become 11

=> We have a cycle of 10 (11, 21, 31,....,01) 
 => 13 = 10 + 3
  => 11^13 will have the same last two digits as 11^3 = 31 
 => Tens' digit = 3

So, Answer will be C 
Hope it helps!

Link to Theory for Last Two digits of exponents here    .

Link to Theory for Units' digit of exponents here    .

What is the tens digit of 11^{13}?

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What is the tens digit of 11^{13}?

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