December 31st, 2017, 07:40 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27  Last digit of y
Find last digit of $\displaystyle \; y=8^{888}$

December 31st, 2017, 07:57 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,506 Thanks: 1741 
$8^{888} = 4096^{222}$ and every positive integer power of 4096 has 6 as its last digit.

December 31st, 2017, 08:09 AM  #3 
Senior Member Joined: Oct 2009 Posts: 439 Thanks: 147 
A very general method for solving this: Note first that you need to find $8^{888}$ in modulo 10. A very easy and quick way has been invented to compute very high powers modulo a number. https://en.wikipedia.org/wiki/Modular_exponentiation 
December 31st, 2017, 09:38 AM  #4 
Member Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 
Skipjack has given you the answer. Here's a little more detail: You want $n=8^{888}\pmod{10}$ Clearly n is 0 mod 2. Now $n\equiv 3^{888}\pmod{5}$ and since $3^4\equiv1\pmod{5}$, $n\equiv1\pmod{5}$. So by the Chinese remainder theorem or by inspection, $n\equiv6\pmod{10}$ 
December 31st, 2017, 12:08 PM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,119 Thanks: 464 
Here is a brute force way. Simple arithmetic experimentation will show that the final digits of positive powers of 8 seem to cycle 8, 4, 2, 6, 8. Moreover 888 is a multiple of 4. Can we prove: $n \in \mathbb N^+ \implies \exists \ m_n \in \mathbb Z \text { such that } 10m_n + 6 = 8^{(4n)}.$ $n = 1 \implies m_1 = 409 \in \mathbb Z \text { and } 10(409) + 6 = 4096 = 8^4 = 8^{(4n)}.$ $\therefore \exists \ k \in \mathbb N^+ \text { such that } \exists \ m_k \in \mathbb Z \text { such that } 8^{(4k)} = 10m_k + 6.$ $8^{\{4(k+1)\}} = 8^{(4k+4)} = 8^{(4k)} * 8^4 = (4090 + 6)(10m_k + 6) =$ $40900m_k + 4090 * 6 + 60m_k + 36 = 40960m_k + 24570 + 6 = 10(4096m_k + 2457) + 6.$ $\text {Let } m_{k+1} = 4096m_k + 2457 \implies$ $8^{\{4(k+1)\}} = 10m_{k+1} + 6 \text { and } m_{k+1} \in \mathbb Z \ \because m_k \in \mathbb Z.$ Thus the final digit of $8^{888}$ is 6. 
January 1st, 2018, 09:09 AM  #6 
Member Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 
Here's a short addendum to my previous post number 4. For $n=8^{888}$, the last two decimal digits of n are 16; i.e. $n\equiv 16\pmod{100}$. Clearly 4 divides n. Since $\phi(25)=20$, for any $x$ prime to 25, $x^{20}\equiv1\pmod{25}$. Hence $n=8^{20\cdot 44+8}\equiv 8^8\pmod{25}=(2^3)^8=2^{24}\equiv 2^4\pmod{25}=16$. The Chinese remainder theorem thus implies $n\equiv16\pmod{100}$. 

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