what is the last digit of 7^10000 (thats 7 to the 10000th power)?

Answer 1

You'll find a general pattern that repeats itself.
7^(4n+1) ends in 7
7^(4n+2) ends in 9
7^(4n+3) ends in 3
7^(4n) ends in 1

It follows from the multiplication of the last digit. 7x7 = 49.
9x7 = 63. 3x7 = 21. 1x7 = 7. Then you start over.

7^10000 should end in a "1"

Answer 2

When multiplying two numbers, only the last digits of the multipliers affect the last digit of the product. By continuously multiplying by 7, you will get a repeating sequence 7, 9, 3, 1 for this last digit. Each multiply by 7 takes one step through this loop. After 10000 times, you will go through this loop 2500 times, eventually ending on the 1 at the end. That is the last digit of 7^10000.

Answer 3

Last digit of 7^1=7
7^2->9
7^3->3
7^4->1
7^5->7
the cycle repeats every 4 times, so the last digit of 7^10000 is the same as the last digit of 7^ remainder(10000/4), so it is 1.

Answer 4

The last digit of 7^n cycles through the values 7, 9, 3, and 1.
The last digit of 7^10000 is 1.

Answer 5

considering a million = a^(2^5), mod (2^6) ==> a million = a^ten thousand, mod 2^6 for a?0 mod 2 considering a million = a^(4*5^5), mod(5^6) ==> a million = a^ten thousand, mod 5^6 for a?0 mod 5 thus, a million=a^ten thousand mod 10^6 for atypical a no longer divisible with the aid of 5. So 7^ten thousand mod 10^6 = a million ---------- --------- --------- ---------- clarification: Euler's Theorem (a generalization of Fermat's Little Theorem) says that: a million = a^(?(n)) mod n the place n is particularly top to a and the place Euler's totient function, ?(n), counts the style of valuable integers decrease than n that are particularly top to n. If n=p^ok for p top, this turns into: a million = a^((p-a million)p^(ok-a million)) mod p^ok for a?0 mod p ***** ***** ***** stable capture Zanti! Dang 0 - I completely misinterpret the subject. i'm able to patch up my evidence, despite the fact that that's not lovable anymore, and that i will in all probability eliminate it in a on an identical time as. considering a million = a^(2^5), mod (2^6), for a?0 mod 2 ==> a^ten thousand = (a^32)^312 * (a^sixteen) = a^sixteen mod (2^6) yet 7^2 = 40 9 and (a million+40 8)^4 is obviously a million extra effective than a diverse of sixty 4 so 7^8 = a million mod (2^6) so as that 7^sixteen = a million mod (2^6) thus, 7^ten thousand = a million mod (2^6) considering a million = a^(4*5^5), mod(5^6), for a?0 mod 5 and because 1047^5 = 7 mod (5^6) ==> 7^ten thousand mod (5^6) = (1047^5)^ten thousand mod (5^6) = (1047^(5*(5^4)*4))^8 mod (5^6) = (a million)^8 mod (5^6) = a million mod (5^6) So 7^ten thousand mod 10^6 = a million

Answer 6

the answer is 0

morningfoxnorth is wrong...

Use your calculator on your computer, and set it to "scientific". Hit the "7" button, then the button that looks like this = "x^y", and then type in 10000, and then hit the "=". The last digit is 0.

Answer 7

The last digit is 0

Answer 8

The last 6 digits are 000001