find the last two digits of $2^{250}$.

Question

Suppose we want the last two digits of $3^{250}$, one can use the theorem $a^{\phi(n)}\cong 1(\mod n)$ whenever $(3,n)=1$. But instead, if i have $2^{250}$, how do i solve this problem, because here $(2,100)\neq 1$.

notation $(a,b)=$the gcd of $a$ and $b$

Look at the last two digits of $2^1, 2^2, 2^3,\ldots$ and see if you can find a pattern. — vadim123, Nov 22 '14 at 00:21
you are right... this is ok. but if you are having a number like $22$ or some other number which is not relatively prime to 100 what can we do? that is my aim of this question. — David, Nov 22 '14 at 01:14
You can always determine the last two digits of a number by calculating it modulo $4$ and modulo $25$. (This is an example of the Chinese remainder theorem.) — Greg Martin, Nov 22 '14 at 01:19
http://en.wikipedia.org/wiki/Modular_exponentiation and everything else is a contrived gimmick. — DanielV, Jan 12 '15 at 23:25
Writing down the last two digits of a^0, a^1, a^2, ... always works. It has to repeat within 100 steps. sometimes earlier. Of course it can take as much as 100 steps, so you use a spreadsheet. — gnasher729, Nov 25 '15 at 00:31

score 3 · Accepted Answer · answered Nov 22 '14 at 03:50

As $(2^{250},100)=2^2,\left(2^{250-2},\dfrac{100}{2^2}\right)=1$

Let us find $2^{248}\pmod{25}$

As $(2,25)=1$ and $\phi(25)=20,2^{20}\equiv1\pmod{25}$ and $248\equiv8\pmod{20}$

$\implies2^{248}\equiv2^8\pmod{25}\equiv6$ as $2^8=256$

$\implies2^{248}\cdot2^2\equiv6\cdot2^2\pmod{25\cdot2^2}$

Done!

score 2 · Answer 2 · edited Nov 22 '14 at 03:50

2

Well, the last two digits are divisible for $4$, so

$$2^{250}\equiv 4a \pmod{100}\iff 2^{248}\equiv a \pmod{25}$$

and $(2,25)=1$

edited Nov 22 '14 at 03:50

Aditya Hase

8,851

answered Nov 22 '14 at 00:31

Exodd

10,844

1

Is this true for every number other than 2?@Exodd – David Nov 22 '14 at 00:49
@David there's a more general rule: if you're trying to evaluate $a$ modulus $n$, and $(a,n)=b$, then the result will be a multiple of $b$, and you can divide $$a\equiv c ;(mod; n) \iff a/b\equiv c/b ;(mod; n/b)$$ – Exodd Nov 22 '14 at 18:25

score 2 · Answer 3 · answered Nov 22 '14 at 02:44

Nowhere near as neat as Exodd's solution but I thought I'd post this because I think it gives a handy insight into the behaviour of some powers of $2$ modulo $100$.

I exploited the fact that $2^{14} = 16384 \equiv -16 = -2^4 \pmod{100}$ and $2^{16} = 65536 \equiv -64 = -2^6 \pmod{100}$, giving $2^{30} = 2^{14} \cdot 2^{16} \equiv (-2^4)\cdot (-2^6) = 2^{10} \pmod{100}$.

So $2^{250} = 2^{30\cdot 8 + 10} \equiv (2^{10})^8\cdot 2^{10} = 2^{90} = (2^{30})^3 \equiv (2^{10})^3 = 2^{30} \equiv 2^{10} = 1024 \equiv 24 \pmod{100}$.

As I said, not as elegant, and perhaps not very general, but I put it down here in case it gives the original asker or others a little more insight.

There is a general observation that since there are only finitely many numbers modulo $n,$ if we keep multiplying by a fixed constant $k$ then eventually we will start repeating. — David K, Nov 22 '14 at 03:36

find the last two digits of $2^{250}$.

3 Answers3