Last digits of form the sequence
n 1 2 3 4 5 . . .
last digits of 7 9 3 1 7 . . .
This sequence is periodic with the period length of 4.
How many periods of the length 4 are in the sequence 1, 2, 3, 4, . . . , 72 ?
The number of periods is exactly = 18, so the last digit of the number is 1.
ANSWER. has the last digit 1.
Look at the pattern in the final digits of 7^n for increasing powers n. Since we are only interested in the final digit, we only need to keep the last digit after each multiplication.
7^1 final digit 7
7^2 final digit 9 (7*7 = 49)
7^3 final digit 3 (9*7 = 63)
7^4 final digit 1 (3*7 = 21)
7^5 final digit 7 (1*7 = 7)
...
It should be clear that the sequence of final digits repeats ever 4 powers.
Since the power 72 is a multiple of 4, the final digit of 7^72 is the same as the final digit of 7^4, which is 1.
