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In the question here

Find the last two digits of the given number

The method for finding the last two digits of a single integer is fairly clear. However, when confronted with an expression like the one in this question, I am unsure of how to proceed. Ideally, I would like to avoid using any techniques asides from congruence.

SolidSnackDrive
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    Hint: $3^2\times11=99\equiv-1\pmod{100}$. – YiFan Tey Jul 18 '18 at 22:19
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    "The method for finding the last two digits of a single integer is fairly clear." Good. Find the last two digits of the three integers $808^2, 3^{305}$ and $11^{151}$ separately. What do you get? What do you think you should do afterwards? There are shortcuts like the one hinted at above, and while it makes the calculations easier, it's not really necessary. At least until you feel you have the basic approach nailed down. – Arthur Jul 18 '18 at 22:19
  • Maybe I don't understand after all. There doesn't appear to be a pattern in the last two digits for $3^n$ like there is for $7^n$. Another method shown involves $01^n$ which is very simple since it's just 1. I don't understand how doing something like $50-1$ like in the linked question, or $(10-1)$ like some of the answers below is helpful. – SolidSnackDrive Jul 18 '18 at 23:31

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Use binomial theorem along with $3^2 = 10 -1$ and $11 = 10 + 1$.

So $808 = (800 + 8) = 800^2 + 2*8*800 + 8^2\equiv 8^2=64 \mod 100$

And $3^{305} = 3*(3^2)^{152} = 3*(10 - 1)^{152} = 3(...... + 152*10*(-1)^{151} + (-1)^{152})\equiv 3(-20 + 1) =-57\equiv 43\mod 100$.

And $11^{151} = (10 + 1)^{151} = ...... + {151\choose 2}10^2 + 151*10 + 1\equiv 11\mod 100$.

So $808^2 + 3^{305} \cdot 11^{151} \equiv 64 + 43 \cdot 11 = 64 + 473\equiv 37 \mod 100$.

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Oh, for goodness sake!

$3^2*11 = (10-1)(10+1) =100-1\equiv -1$

So $3^{2k}11^k =(99)^k \equiv (-1)^k\mod 100$

so

$3^{305}\cdot 11^{151}= 3^3*3^{302}\cdot 11^{151}=$

$27*(-1)^{151} \equiv -27\mod 100$.

So $808^2 + 3^{305}11^{151}\equiv 64-27= 37\mod 100$.

fleablood
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