# Last two digits of 16^198

1. Jul 24, 2010

### thereddevils

1. The problem statement, all variables and given/known data

What are the last two digits of 16^198

2. Relevant equations

3. The attempt at a solution

I begin by looking for a pattern.

16^2 = 256
16^4 = ...36
16^6 = ...16
16^8 = ...96
16^10 = ...56

so i see that the pattern repeats itself every 4 times. Then, i grouped the power as follows,

(2,4,6,8) , (10,12,14,16) , (18,20,22,24) , ... , (194,196,198,200)

The last number of each group has formula of 8n and that's how i got the last group.

198 is third in the group which corresponds to the number having 16 as its last two digits.
But i think i went wrong somewhere because the answer given is 96. Am i totally off track?

2. Jul 25, 2010

### vela

Staff Emeritus
Re: indices

You missed one.

1610=1099511627776

It's 1612 that ends in 56.

3. Jul 25, 2010

### thereddevils

Re: indices

Thanks Vela, i am using a 570 calculator and it gives 1.099511628 x 10^12 for 16^10 so how do i know its last two digits?

4. Jul 25, 2010

### vela

Staff Emeritus
Re: indices

Just keep the last two digits in your calculations. The higher order digits don't matter.

5. Jul 25, 2010

### thereddevils

Re: indices

yes, i know. The problem is how do i know the last two digits from 1.099511628 x 10^12. It looks to be 00. That's the maximum degree of accuracy my calculator can measure to.

6. Jul 25, 2010

### praharmitra

Re: indices

Think about this... u'd get the last two digits of 16^198 as a remainder if u divide it by 100. Correct?

So, u have to find 16^198 (mod 100). To do this you can use, general theorem

a*b (mod c) = a(mod c) * b(mod c).

7. Jul 25, 2010

### praharmitra

Re: indices

if you want to use ur calculator for solving this, but can't because of limitations in accuracy, here is what u can do.

16^10 = 1.099511628... X 10^12 is clearly an approximation. To find the last two digits, subtract from this approximate number the exact number 1.099511628 X 10^24. This will give you the remaining digits of the answer. If that is too long again, repeat. Finally, you should get the last two digits.

However, I suggest the method given above by me. It is easier.

8. Jul 25, 2010

### thereddevils

Re: indices

thanks i figured out another way of getting the last two digits.

Btw, i am not familiar with modular arithmetic but will pick up very soon.

9. Jul 25, 2010

### vela

Staff Emeritus
Re: indices

You can toss any higher digits even on intermediate calculations.

162 = 256 ⇒ 56
164 ⇒ 56*56 = 3136 ⇒ 36
166 ⇒ 56*36 = 2016 ⇒ 16
168 ⇒ 56*16 = 896 ⇒ 96
1610 ⇒ 56*96 = 5376 ⇒ 76
1612 ⇒ 56*76 = 4256 ⇒ 56

Even if you don't know modular arithmetic, this should make sense to you. Any digit in the 100s position or higher can only contribute some multiple of 100 to the product, so it won't affect the two least significant digits. For example, consider 256*256. We can write

2562=(200+56)(200+56) = 2002 + 2(200)(56) + 562

The first two terms will be some multiple of 200, so their bottom two digits have to be zero. Only the last term, 56*56, contributes to the two lowest-order digits.

10. Jul 25, 2010

### thereddevils

Re: indices

yeah thanks, i figured that out.