Question

I WILL GIVE MEDAL
What is the last digit of

Answer 1

\[7^{217}\]

Answer 2

im getting

Answer 3

you should look for a pattern (of the last digit)
for example for \[7^0 7^1 7^2 7^3 7^4 \]
you get the numbers 1 7 9 3 1
so the last digit repeats after 4 steps

Answer 4

in other words, 7^4 ends in a 1
and if you check, you will see 7^8 also ends in a 1

Answer 5

amazing

Answer 6

so you should figure out what is the biggest number <= 217 that is divisible by 4
and you know that will end in 1

Answer 7

7^40 is also ending in 1

Answer 8

find the remainder when you divide 217 by 4

Answer 9

if you know some "divisibility rules" , you know a number is divisible by 4 if its last two digits are. in other words 216 is divisible by 4
so you can write the problem as
7^216 * 7^1

Answer 10

and you know 7^216 ends in 1
and 7^1 ends in 7
so the last digit will be 1*7

Answer 11

as welsh mentioned, the "fast way" is to divide 4 into 417, and get the remainder
and it turns out the last digit will be the same as the last digit of 7^remander
i.e. 7^1 in this case.

Answer 12

but when im doing its not ending at 7

Answer 13

7^217 is a huge number and calculators usually don't use enough digits
if you use wolfram
http://www.wolframalpha.com/input/?i=7%5E217
it shows the entire number

Answer 14

oh thankyou so much
i was very worried about this