MHB Last 6 Digits of 7^10000: 000001

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The discussion centers on proving that the last six digits of 7^10000 are 000001. Participants are encouraged to share their solutions directly in the thread rather than linking to external sites, promoting ease of access for all members. One user plans to post their solution a week later, inviting others to contribute their methods first. The conversation emphasizes clarity and accessibility in mathematical problem-solving. Overall, the focus remains on collaborative proof and sharing of solutions within the forum.
kaliprasad
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Prove that the last 6 digits of 7^10000 is 000001
 
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kaliprasad said:
Prove that the last 6 digits of 7^10000 is 000001

\[\begin{array}{}
7^4 &=& 2401 &\equiv& 1 \pmod{400} \\
7^{100} &=& (7^4)^{25} &=& (400k+1)^{25} &=& ...\ +\ 25 \cdot 400k + 1 &\equiv& 1 \pmod{10000} \\
7^{10000} &=& (7^{100})^{100} &=& (10000m + 1)^{100} &=& ...\ +\ 100\cdot 10000m + 1 &\equiv& 1 \pmod{1000000} \\
\blacksquare
\end{array}
\]
 
I like Serena said:
\[\begin{array}{}
7^4 &=& 2401 &\equiv& 1 \pmod{400} \\
7^{100} &=& (7^4)^{25} &=& (400k+1)^{25} &=& ...\ +\ 25 \cdot 400k + 1 &\equiv& 1 \pmod{10000} \\
7^{10000} &=& (7^{100})^{100} &=& (10000m + 1)^{100} &=& ...\ +\ 100\cdot 10000m + 1 &\equiv& 1 \pmod{1000000} \\
\blacksquare
\end{array}
\]

neater than my solution. I shall post mine one week later so that others can post
 
here is my solution

we know 7^4 = 2401

so 7^10000 = (2401)^2500

now 2401^2500 = (2400+1)^2500

if we collect nth term it it (2500 C n) (2400)^n

for n > 2 2400^n is divisible by 10^6

so we need to look for n = 2 and n =1

n=0 gives 1 and 1-1 = 0

n = 2 => (2500C2)(2400)^2 = 2500*1200*2400 so 10^6 is a factor
n =1 =>(2500)(2400) = 6000000 so 10^6 is a factor

so all the elements except last that is 1 is divisible by 10^6 and last element is 1

so last 6 digits are 000001 or 7^10000 mod 10^6 = 1
 
Last edited:
Hello Kali,

In our guidelines, we ask:

Please do not give a link to another site as a means of providing a solution, either by the author of the topic posted here, or by someone responding with a solution.

For the convenience of our members, we prefer that they not have to follow links, but that the solution(s) be posted here. This makes it easier on the majority. :D
 
MarkFL said:
Hello Kali,

In our guidelines, we ask:
For the convenience of our members, we prefer that they not have to follow links, but that the solution(s) be posted here. This makes it easier on the majority. :D

I have done the needful.
 
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