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last N digits

kaliprasad

Well-known member
Mar 31, 2013
1,309
Prove that the last 6 digits of 7^10000 is 000001
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
Prove that the last 6 digits of 7^10000 is 000001
\[\begin{array}{}
7^4 &=& 2401 &\equiv& 1 \pmod{400} \\
7^{100} &=& (7^4)^{25} &=& (400k+1)^{25} &=& ...\ +\ 25 \cdot 400k + 1 &\equiv& 1 \pmod{10000} \\
7^{10000} &=& (7^{100})^{100} &=& (10000m + 1)^{100} &=& ...\ +\ 100\cdot 10000m + 1 &\equiv& 1 \pmod{1000000} \\
\blacksquare
\end{array}
\]
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
\[\begin{array}{}
7^4 &=& 2401 &\equiv& 1 \pmod{400} \\
7^{100} &=& (7^4)^{25} &=& (400k+1)^{25} &=& ...\ +\ 25 \cdot 400k + 1 &\equiv& 1 \pmod{10000} \\
7^{10000} &=& (7^{100})^{100} &=& (10000m + 1)^{100} &=& ...\ +\ 100\cdot 10000m + 1 &\equiv& 1 \pmod{1000000} \\
\blacksquare
\end{array}
\]
neater than my solution. I shall post mine one week later so that others can post
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
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:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
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.