look at numbers ending in 7, such as
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649
7^7 = 823543
notice that the unit digits forms a cyclic pattern:
7, 9, 3, 1, 7, 9 3, 1, ....
the same would be true for something like 37^n
try it.
Also:
2017^1 = 2017 --- 4 digits
2017^2 = 4068289 --- 7 digits
2017^3 = 8205738913 --- 10 digits
2017^4 = 1.655097539.. x 10^13
= 1655097539...3 --- 14 digits
2017^5 = 3.338...1 x 10^16 ---- 17 digits
2017^6 = 6.7334...7 x 10^19 --- 20 digits
...
2017^15 = 3.7203... x 10^49 --- 50 digits
..
2017^30 = 1.38... x 10^99 ---- 100 digits
You could use the first pattern to establish what the last digit of 2017^(2017^2017) is.
As to "solving" 2017^2017 without a calculator, or even with a calculator would be silly.
This would fall into the category of "true" pure math.
Please help,
1) How do you solve 2017^2017 without a calculator?
2) What is the last digit of the solution to this problem: 2017^(2017^2017)?
Thank you!
1 answer
1 answer