the last four digits?

Question

how to find the last four digits of 7^(7^7)?

Answer 1

First, the notation in the problem is correct. 7^(7^7) = 7^823543 which is a number with almost 700,000 digits so no simple calculator can give the answer to the precision required to determine the last 4 digits.

There is no need to deal with such long numbers, only the last four digits of any intermediate multiplication is needed since they are the only ones that determine the last four digits in successive multiplications. Begin by calculating some values of 7 to low powers (or at least the last four digits):

7^10 = 282475249 ends in 5249
7^20 = (7^10)^2 ends the same as 5249^2 which is 2001
7^40 = (7^20)^2 ends the same as 2001^2 which is 4001
7^80 = (7^40)^2 ends the same as 4001^2 which is 8001
7^100 = 7^80 * 7^20 ends the same as 8001 * 2001 which is 0001

This shows that 7^100 ends in 0001. It follows immediately that 7^(n*100) ends in 0001 for alll positive integers n.

The original problem of 7^(7^7) is 7^823543 = 7^823500 * 7^43. The factor 7^823500 must end in 0001 since the exponent is a multiple of 100. Therefore 7^(7^7) ends the same as 7^43. From above, 7^40 ends in 4001 so 7^43 ends the same as 4001 * 7^3 which ends in 2343. So 7^(7^7) ends in 2343.

Answer 2

There are 42 digits in the result. If we divide that by 10^38, the last four digits would be a remainder. The function that does that is called the modulo function and the answer is

7^7^7modulo(10^38)

My mathcad program won't handle such large numbers.

EDIT: For four digits in the remainder, divide by 10^4, so the answer should be (7^7^7)modulo(10^4)

Answer 3

7^7 is 823543

If you work through 7^1, 7^2 all the way to 7^100, you find that the last 4 digits have a pattern. It goes something like:

7,49,343,2401,6807,7649,3543,4801,3607,5249,
6743,7201,0407,2849,9943,9601,7207,0449,3143,2001(20th term),.............. 4001(40th term),....... 6001(60th term),....8001(80th term),.....7143(99th term), 0001(100th term) and then it repeats itself.

So the cycle repeats every 99 terms.
Taking 7^7 and dividing it by 99, you get a remainder of 61. Hence you need the last 4 digits of the 62nd term, which is 4049

i worked it out using excel. When the number gets too large, i carried on calculating the series using only the last 4 digits and multiplying it by 7, since our concern is only the last 4 digits. Rinse and repeated it till 100 to see the pattern. Hope this helped.

Edit: i just read Pretzels answer after submitting mine. Trust me, the cycle repeats every 99 terms. Even pretzels says that 7^100 gets you to 0001 again, meaning that 7^1 till 7^99 is the cycle, and 7^100 is the new cycle. Hence the cycle is 99 terms, and not 100 terms.

Answer 4

The last four digits repeat themselves every time 7 is multiplied by itself 100 times.
7^(7^7) = 7^823543.
The last four digits of 7^823543 is the same as the last four digits of 7^43 which are 2343.

Answer 5

this is 7^823543 no store bought calc will tell you the last 4 digits most will just say x10^41 there must be a simple trick goodluck

Answer 6

343

Answer 7

solve this by :
let x =7^7^7
taking log both sides
logx=7^7log7
again taking log both sides
log(logx)=log(7^7log7)
log(ab)=loga+logb
log(logx)=7log7+log(log7)
now find the log values with help of log table
u will get log(logx)
then remove log by taking mantissa n all
thus u will find the value of x which is reqired.


hope u got ur answer:-))

Answer 8

well 7 to the seventh power is ?

then take that answer and make it the exponet to the first 7...
. Viola! there"s your answer..

Answer 9

7??

Answer 10

0000, the full answer is 2.5692357752105887808867741122424 *10^41