# Find last two digit of a number

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• Feb 15th 2010, 02:31 PM
firebio
Find last two digit of a number
Find the last two digits of $\displaystyle 3^{83} + 21^{401}$

so:

$\displaystyle 3^{83} \equiv M (mod 100)$
$\displaystyle 21^{401} \equiv N (mod 100)$

So last two digits is M+ N

I dont really know how to continue...

Any help would be appreciated
Thanks in advance
• Feb 15th 2010, 02:48 PM
icemanfan
By calculating $\displaystyle 3^{20}$ and $\displaystyle 21^5$ we can see that

$\displaystyle 3^{20} \equiv 1 \mod 100$ and

$\displaystyle 21^5 \equiv 1 \mod 100$.

Therefore,

$\displaystyle 3^{83} = 3^{80} \cdot 3^3 \equiv 3^3 \mod 100$

$\displaystyle 21^{401} = 21^{400} \cdot 21 \equiv 21 \mod 100$

$\displaystyle 27 + 21 = 48 \equiv 48 \mod 100$

Answer is 48.
• Feb 16th 2010, 12:26 PM
firebio
Quote:

Originally Posted by icemanfan
$\displaystyle 27 \cdot 21 = 567 \equiv 67 \mod 100$

Answer is 67.

thanks

But answer should be $\displaystyle 27 + 21 = 48$?

Is there any simpler way to compute this without calculating 3^20?
• Feb 16th 2010, 12:30 PM
icemanfan
Quote:

Originally Posted by firebio
thanks

But answer should be $\displaystyle 27 + 21 = 48$?

Is there any simpler way to compute this without calculating 3^20?

You are correct, I made a silly mistake at the end. If you can determine that $\displaystyle 3^{20} \equiv 1 \mod 100$ without actually calculating $\displaystyle 3^{20}$, then yes. Personally, I don't know how you would do it. In most of these types of problems, the cycle is much shorter than 20.
• Feb 16th 2010, 12:34 PM
firebio
Quote:

Originally Posted by icemanfan
You are correct, I made a silly mistake at the end. If you can determine that $\displaystyle 3^{20} \equiv 1 \mod 100$ without actually calculating $\displaystyle 3^{20}$, then yes. Personally, I don't know how you would do it. In most of these types of problems, the cycle is much shorter than 20.

i see.. i asked because this was a test question and calculating 3^20 seems a bit harsh.

Thanks so much.