What will be the last 2 digits of 16 to the power 198? How to calculate it n other such cases quick n easy?

Question

How can you calculate the answer of any number(16 in this case) raised to any power(198 here) quickly and easily without cumbersome failure-prone multiplication? Particularly in this case, at least how can we calculate the last two digits? Does anyone know a straight formula?

Answer 1

Last 2 digits of 16^198.

As others have pointed out, we want to get the remainder when
we divide 16^198 by 100.

We can quickly find out the answer by first finding the smallest exponent p such that 16^p has remainder 16 when divided by 100.

16 ^2 = 256 has remainder 56 when divided by 100.
16^3 has same remainder as 16 * 56 = 896 has remainder 96.
16^4 has same remainder as 16 * 96 = 1536 has remainder 36.
16^5 has same remainder as 16 * 36 = 576 has remainder 76.
16^6 has same remainder as 16 * 76 = 1216 has remainder 16.

6 - 1 = 5
So we see that the patteren repeats every 5 steps.
198 has remainder 3 when divided by 5.
so 16^198 has same last two digits as 16^3 which is
96.

Answer 2

save in options, there's a trend with the superb digit and the flexibility of any quantity that kinds a cycle. thus that is: 2^a million=2 2^2=4 2^3=8 2^4=16 2^5=32 2^6=sixty 4 2^7=128 2^8=256 ............. Now, it is going with the help of the cycle 10 circumstances at the same time as it receives to 2^40. Then, it starts with the first step of the cycle. The very last digit is two.

Answer 3

I don't know a formula, but here's a method that could be programmed

16^198 reduced mod 100 will give you the last two digits by using iterative squaring

16 mod 100 = 16
16^2 mod 100 = 16^2 mod 100= 56
16^4 mod 100 = 56^2 mod 100 = 36
16^8 mod 100 = 36^2 mod 100 = 96
16^16 mod 100 = 16
16^32 mod 100 = 56
16^64 mod 100 = 36
16^128 mod 100 = 96
(The pattern keeps repeating)

198 = 2^7+2^6+2^2+2^1 (you get this by writing 198 in binary form)

So 16^(198) mod 100
=16^( 2^7+2^6+2^2+2) mod 100
=16^(2^7) * 16^(2^6) * 16^(2^2) * 16^2
=96*36*36*56 mod 100 = 96

Answer 4

you simply watch for patterns, in fact, immediately i know that the last digit will always be a 6, watch this. 16 x 16 = 256, the last digit is 6. if you are multiplying two digits where the last digit of both is 6, the last number in the answer will be 6. (e.g 6x6=36x6=216 etc) so we know what the last digit will be. as for the second to last digit, i multiplied 16 x 16 , then multiplied that by 16 etc, to look for patterns in the second digit, and i found one. the pattern is 1, 5 , 9, 3, 7, 1, 5 , 9 ,3 , 7. that is the pattern in the second to last digit forever. so watch this, i know that 16^1 = 16. that lines up with what ive said so far, the last digit is always 6 and the second to last will be 1 first. 16^2 is 256 again it lines up. it lines up forever. now we just do like this. you wanted to know the last two digits of 16^198. we divide 198 by 5 and take the remainder (3) and just count out which number it will be, the digits go 1 , 5 , 9. so there you have it, the last two digits are 9 and 6.

Answer 5

What you want to solve for is

16^(198) mod 100

I'm going to use the "==" sign to mean "is congruent to"

This is all modulo 10.

16^(198) == (256)^(99) == 56^(99) == 56 * 56^(98)
== 56 * 3136^(49) == 56 * 36^49 == 56 * 36 * 36^48
== 2016 * 36^48 == 16 * 36^48 = 16 * 1296^24 == 16 * 96^24
== 16 * (9216)^12 == 16 * 16^12 == 16 * (256)^6 == 16 * (56)^6
== 16 * 3136^3 == 16 * 36^3 == 2094336 == 36

Therefore, the answer is 36.

What I did above is known as modulo arithmetic.

Answer 6

Hint:

Without using number theory, you can try binomial expansion.

16^198 = (10+6)^198

Answer 7

Hello :)

You can use logarithm for that but log tables are required.

c = a^b can be computed as
c = antilog(b log a)

There are other methods (Ex: Addition-chain exponentiation, Exponentiation by squaring) for reducing the number of steps in computation but they are more mechanical and is better left to computers which are good at them.

Hope it helps :)