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#### kaliprasad

##### Well-known member

- Mar 31, 2013

- 1,309

Prove that the last 6 digits of 7^10000 is 000001

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- #1

- Mar 31, 2013

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Prove that the last 6 digits of 7^10000 is 000001

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- Mar 5, 2012

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Prove that the last 6 digits of 7^10000 is 000001

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}

\]

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- #3

- Mar 31, 2013

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

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}

\]

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- #4

- Mar 31, 2013

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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

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:

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- #5

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.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.

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- #6

- Mar 31, 2013

- 1,309

I have done the needful.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.