What is the tens digit of 21^3^50

Question

What is the tens digit of  21^{3^{50}} ?

A. 1

B. 2

C. 5

D. 7

E. 8

nick1816 · Answer

OUT OF SCOPE OF GMAT

21 = 21 mod 100

21^2 = 41 mod 100

21^3 = 61 mod 100

21^4 = 81 mod 100

21^5 = 01 mod 100

After this point, cycle will repeat.

So basically we have to find the remainder of 3^50 when divided by 5.

3^{50} = 9^{25} = (-1) mod 5 = 4 mod 5

3^{50} = 5k +4

21^{5k+4} = 21^{5k} * 21^4

21^{5k} * 21^4 mod 100 = 01*81 mod 100 = 81 mod 100

Last second digit of  (21^3)^{50}  is 8.

yashikaaggarwal · Answer

The tens digit of any number ending with 1 is the unit digit of the product of the unit digit of the power& tens digit of the base; with the unit digit as 1
For tens digit of number ending with 3, repeatedly square no.unit we get 1 as unit digit, here 3^4 = 81
So, (3^4)^12*3^2
=> 81^12*3^2
=> 61*9 => 49
Tens digit of 3^50 is 49
Tens digit of 21^49 = 81
Where 8 is tens Digit and 1 is units

Answer is E

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mrr0821 · Answer

nick1816 
Can you refer me a source where can I have a clear understanding about the method you employed here? I am unable to understand your solution.

yashikaaggarwal 
Why 81? Why can't we take (3*7)^3^50?

subhakimi · Answer

We need to find the last two digit of this number or the remainder when divided by 100.

N^20 leaves a remainder of 1 when it is divided by 100. Given, N is coprime to 100.

So here, we need to find out what is the remainder when 3^50 is divided by 20.
=3^50 = (3^4)^12 * 3^2 
So we can say the remainder will be 9.
Hence we can re-write it as, 21^(20k+9)
So all we need to find is the last two digit of 21^9.

(21^3)^3 = (x61)^3 = 81

So the remainder is 81 and the answer is 8.

yashikaaggarwal · Answer

Defining tens place of a number ending with 1 as unit digit no. is easier than looking after for other digits, there are some tricks to sort out such question type. I already wrote one of those in my answer. I suggest you to Go through that formulas, solving these question types will come handy to you.

Shubham1213372 · Answer

I saw one rule somewhere just trying to put in this question. Kindly correct me if you think I am wrong here:

(21)^3^50
Let's take 3^50 first
We know that 3 is the unit digit here and it's unit digit keeps on repeating after every 3^4 term i.e.
3^1 =3
3^2 =9
3^3 =7
3^4 =1
And so on

So for such cases we can divide the power by 4 and the remainder that we get will be power of value.

50/4 leaves a remainder of 2
Hence our value is now 3^2 which is equivalent to 9.

Now come to original question and let's bring 21 part here and our original question will look like this

21^9?

Here comes the interesting part to  calculate the second digit simply multiply the tens digit value with the power and the resulting unit value will be your answer  
for this case 2 is the tens digit and 9 is the power, results in 18, where 8 is the unit digit and it is our answer.

Hence E is our answer

Posted from my mobile device

nick1816 · Answer

I used modular arithmetics. 
Don't waste your time on these problems tho. You never gonna see such problem in GMAT

If you're still interested in learning modular arithmetics, Khan academy or brilliant. org are some good sources or you can PM me.

mrr0821 · Answer

Thanks, guys!

After understanding modular arithmetic, I see things crystal clear.

Thanks for introducing me to this new concept!

What is the tens digit of 21^3^50

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What is the tens digit of 21^3^50

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