# Math Help - An interesting Korean math olympiad problem about exponents

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.,,,,,

2. $2^{2005}-2^{1999}$

$=2^{1999}(2^{6}-1)$
$=2^{1999}(63)$

Now we need to find the last two digits in $2^{1999}$, 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 $2^{1999}$ 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 $2^{1999}$

which are 88

So the last two digits in the number $2^{2005}-2^{1999}$ are the last two digits in the number 63X88,

i.e. 44

3. We can calculate it by looking at residu-classes modulo 100
Observe that
$2^{10}\equiv 24$mod 100
$2^{20}\equiv 24^2 \equiv 76$mod 100
$2^{40}\equiv 76^2 \equiv 76$ mod 100

Thus we observe that $2^{20k + a} \equiv 2^{a}(76)$mod 100

Thus $2^{2005}-2^{1999} \equiv (2^5-2^{19})76 = 2^5(1-2^{14})76 \equiv 32(1-84)76\equiv -56\cdot 76 \equiv 44$mod 100

Thus 44 are the last 2 digits.

4. Originally Posted by Dinkydoe
We can calculate it by looking at residu-classes modulo 100
Observe that
$2^{10}\equiv 24$mod 100
$2^{20}\equiv 24^2 \equiv 76$mod 100
$2^{40}\equiv 76^2 \equiv 76$ mod 100

Thus we observe that $2^{20k + a} \equiv 2^{a}(76)$mod 100

Thus $2^{2005}-2^{1999} \equiv (2^5-2^{19})76 = 2^5(1-2^{14})76 \equiv 32(1-84)76\equiv -56\cdot 76 \equiv 44$mod 100

Thus 44 are the last 2 digits.
This is a more formal method I guess...

5. This is a more formal method I guess...
Indeed: Namely for any n we write $(\mathbb{Z}/n\mathbb{Z})=\left\{\overline{0},\overline{1},\cdo ts \overline{n-1}\right\}$ for the set residu-classes modulo n.
A class $\overline{a}$ consists of all elements in $\mathbb{Z}$ that have the same residu by dividing through n. One may write the class $\overline{a} = a+ n\mathbb{Z}= \left\{a+ nb| b\in \mathbb{Z}\right\}$

Then for numbers $k_1,k_2$ we write $k_1\equiv k_2$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 $\overline{a}+\overline{b} = \overline{a+b}$. And $\overline{a}\cdot\overline{b} = \overline{ab}$

What this means is: given numbers $d_1, d_2$. Say that $d_1,d_2$ divided by n have residu $r_1,r_2$respectively, then $d_1+d_2$ has residu $r_1+r_2$mod n when divided by n. Likewise: $d_1d_2$ has residu $r_1r_2$mod n when divided by n.

So, one can do many handy calculations by looking at residu-classes.