What is the pattern for finding the last two digits of 16 to a power?

[thereddevils]

Homework Statement

What are the last two digits of 16^198

Homework Equations

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?

 

[vela]

You missed one.

16¹⁰=1099511627776

It's 16¹² that ends in 56.

 

[thereddevils]

You missed one.

16¹⁰=1099511627776

It's 16¹² that ends in 56.

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?

 

[vela]

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

 

[thereddevils]

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

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.

 

[praharmitra]

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).

 

[praharmitra]

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.

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.

 

[thereddevils]

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.

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.

 

[vela]

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.

 

[thereddevils]

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.

yeah thanks, i figured that out.