How can I find the last two numbers of 2^2005-2^1999 ??
I tried everything I can but I can't think of how to approach this problem
Any help is appreciated.,,,,,
How can I find the last two numbers of 2^2005-2^1999 ??
I tried everything I can but I can't think of how to approach this problem
Any help is appreciated.,,,,,
Now we need to find the last two digits in , and the problem is solved.
The given number is divisible by 4, hence its last two digits are multiples of 4.
So the digits are either 04,08,12,16,......96.
We can see that the last digit in clearly is 8.
Now since the above number is divisble by four, the, possible values for the last two digits are 08,28,48,68,88.
Now, consider the powers of 2 which have the last digit 8
8
128
2048
32768
524288
8388608
We can see, that the digits 08,28,48,68,88 repeat at regular intervals
Hence, we can obtain the last two digits in the number
which are 88
So the last two digits in the number are the last two digits in the number 63X88,
i.e. 44
Indeed: Namely for any n we write for the set residu-classes modulo n.This is a more formal method I guess...
A class consists of all elements in that have the same residu by dividing through n. One may write the class
Then for numbers we write mod n when they appear in the same class, that is: have the same residu modulo n.
One can prove that you can sum and multiply residuclasses. Thus . And
What this means is: given numbers . Say that divided by n have residu respectively, then has residu mod n when divided by n. Likewise: has residu mod n when divided by n.
So, one can do many handy calculations by looking at residu-classes.