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

\(11^1\) = 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

Solution:

11^n will have units digit of n as its tens digit.
Ex. 11^2 = 121 , 11^3 = 1331... so on.
11^13 will have 3 as tens digit.

Option C

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.

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.

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,

11^13 = (10 + 1)^13 = 10^13+...+130+1

It's C (3)

Posted from my mobile device

\(11^1 = 11\)

\(11^2 = 121\): 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[Ten's digit]

=> \(11^{12}\) = 1 + 1 = 2[Ten's digit]

=> \(11^{13}\)= 2 + 1 = 3[Ten's digit]

Answer C

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