What is the digit in the ten's place of 2^1001?

Question

What is the digit in the ten's place of  2^{1001} ?

A) 2
B) 4
C) 5
D) 6
E) 8

pkleo · Answer

Last digit would be the same as in 2^1, i.e. 2

Answer A

rsrighosh · Answer

I am not sure if my approach is correct, but i noticed that tens digits were always odd except 2^1. the only odd option was 5. So chose C.

2^1 = 2
2^5 = 32
2^9 = 512
2^13 = 8192
.
.
.

versewriter · Answer

This is how we can solve this:

We have to simplify the given equation
2^1001
(2^10)^100 * 2
(1024)^100 * 2

Now we know that cyclicity 24 is two and the last two digit repeats after every two step (we can also calculate this manually if we don't know that) ie. the last digit is either 76 or 24.

So in our equation 76 is the last two digit i.e. (...76) * 2. So the last digit will be 15.

Option C

Vishwas1998 · Answer

=((2^10)^100) * 2
= (1024)^100 *2
-->(24)^100 *2
-->24 raised to any even power gives 76 as last two digits
=76*2
=152

So, the tens digit is 5

Hence, C

Abhishek009 · Answer

2^{1001}= 2^{10*10}*2

2^10  will end in  24  &  24  raised to an even power always ends with  76

ie,  24*24*2 = 1152 , Answer must be (C) 5.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82758 ___ Harsha Passionate about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com GMAT-Club-Forum-liie8cbj.png

BrushMyQuant · Answer

We need to find the tens' digit of  2^{1001}

2^{1001}  =  2^{1000 + 1}  =  2^{10*100} * 2^1  =  (2^{10})^{100} * 2 
=  1024^{100} * 2 
=  1024^{Even} * 2

Now, we know that 
 
  Odd power of 1024 has last two digits as 24  
  Odd power of 1024 has last two digits as 76

=> Last two digits of  1024^{Even} * 2  = 76 * 2 = 52
= >Tens' digit = 5

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 digit in the ten's place of 2^1001?

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What is the digit in the ten's place of 2^1001?

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