Very tricky and interesting modular question

**elemental** · July 9th 2010, 12:29 PM

This is a very tricky question:
Find the last three digits of
2008^2007^2006^2005^...^1

If you're having trouble thinking about how the exponents are arranged, just imagine every consecutive number a power of the previous.

Good luck, I know the answer, and will tell you if you get it!

**Also sprach Zarathustra** · July 9th 2010, 12:57 PM

Originally Posted by elemental

This is a very tricky question:
Find the last three digits of
2008^2007^2006^2005^...^1

If you're having trouble thinking about how the exponents are arranged, just imagine every consecutive number a power of the previous.

Good luck, I know the answer, and will tell you if you get it!

James Bond +1 ?

**undefined** · July 9th 2010, 02:07 PM

Originally Posted by elemental

This is a very tricky question:
Find the last three digits of
2008^2007^2006^2005^...^1

If you're having trouble thinking about how the exponents are arranged, just imagine every consecutive number a power of the previous.

Good luck, I know the answer, and will tell you if you get it!

Spoiler:

**melese** · July 9th 2010, 02:13 PM

Originally Posted by elemental

This is a very tricky question:
Find the last three digits of
2008^2007^2006^2005^...^1

If you're having trouble thinking about how the exponents are arranged, just imagine every consecutive number a power of the previous.

Good luck, I know the answer, and will tell you if you get it!

Assuming you meant $2008^{2007^{2006^{2005^{2004...^{1}}}}}$ , call it $P$ and let $a=2007^{2006^{2005^{2004...^{1}}}}$ and $b=2006^{2005^{2004...^{1}}}$ .

I start with the fact that $2008\equiv 3\equiv -2(mod\ 5)$ .

$b=2k$ for some integer $k$ , so $a=2007^b=2007^{2k}\equiv 3^{2k}\equiv 9^k\equiv 1(mod\ 4)$ .
$a=4t+1$ for some integer $t$ , so $(-2)^a=(-2)^{4t+1}=16^t(-2)\equiv-2(mod\ 5)$

Now, $P=2008^a\equiv (-2)^a\equiv(-2)(mod\ 5)$ . Also, obviously $P$ is even, hence $P\equiv -2(mod\ 2)$ . These two congruences give $P\equiv -2\equiv 8(mod\ 10)$ .

The last digit of $2008^{2007^{2006^{2005^{2004...^{1}}}}}$ is $8$ .

**elemental** · July 9th 2010, 05:06 PM

Well done guys, 008 is indeed correct.

**mr fantastic** · July 9th 2010, 07:03 PM

Challenge questions belong in the Challenege Questions subforum (read the rules at that subforum before posting in it).

**simplependulum** · July 9th 2010, 07:41 PM

We wish to find out the remainder when $2006^{2005^{...}}$ is divided by $40$

Let $x$ be $2006^{2005^{...}}$

then

$x \equiv 1 \bmod{5}$ and

$x \equiv 0 \bmod{8}$ so

$\boxed{ x \equiv 16 \bmod{40} }$

Then consider $7^x$ in modulo $100$

it is $7^{40k + 16} \equiv 7^{16 }\bmod{100}$

Let $y = 7^{16 }$ we have

$y \equiv (-1)^{16} \equiv 1 \bmod{4}$ and

$y = 7^{16} = (49)^8 \equiv (-1)^8 \equiv 1 \bmod{25}$ so

$y \equiv 1 \bmod{100}$

Then $8^y$ in modulo $125$

Since $(8,125) = 1$ we have $8^{100} \equiv 1$

Therefore , $8^y = 8^{100k + 1 } \equiv 8 \bmod{125}$

Together with $8^y \equiv 0 \bmod{8}$ we obtain $8^y \equiv 8 \bmod{1000}$

Actually , the answer is exactly $8 \equiv 8^y \equiv (2008)^y \bmod{1000}$

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