Remainder

Is it C?

the additon of units digit of given numbers revels 47.

either way can someone tell a quicker way..

cheers.

1^1 + 2^2 + 3^3 + 2^8 + 5^5 + 6^6 + 7^7 + 2^24 + 3^18 + 10^10

1^1 + 2^2 = 5 remainder 0
5^5 and 10^10 will also have remainder 0

so we are left with,

3^3 + 2^8 + 6^6 + 7^7 + 2^24 + 3^18

the units digit for the above expression,

7 + 6 + 6 + 3 + 6 + 9

37, which when divided by 5 leaves remainder 2

Choose C

sacx wrote:1^1 + 2^2 + 3^3 + 2^8 + 5^5 + 6^6 + 7^7 + 2^24 + 3^27 + 10^10

1^1 + 2^2 = 5 remainder 0
5^5 and 10^10 will also have remainder 0

so we are left with,

3^3 + 2^8 + 6^6 + 7^7 + 2^24 + 3^27

the units digit for the above expression,

7 + 6 + 6 + 3 + 6 + 7

35, which is evenly divisible by 5

Choose A

Nicely done sacx.. only one mistake though.. I have pointed that out in bold too

9^9 = 3^18.. so this gives the last digit as 9...

You have taken 9^9 = 3^27. This is incorrect.

So units digit for the expression is 7 + 6 + 6 + 3 + 6 + 9 = 37

and so remainder = 2

sacx wrote:1^1 + 2^2 + 3^3 + 2^8 + 5^5 + 6^6 + 7^7 + 2^24 + 3^27 + 10^10

1^1 + 2^2 = 5 remainder 0
5^5 and 10^10 will also have remainder 0

so we are left with,

3^3 + 2^8 + 6^6 + 7^7 + 2^24 + 3^27

the units digit for the above expression,

7 + 6 + 6 + 3 + 6 + 7

35, which is evenly divisible by 5

Choose A

3^27=9^9
why go thru the headache of using powers of 3 which is more error prone as shown by the error you committed. All powers of 9 alternate between 1 and 9
9^0=1
9^1=9
9^2=81
9^3=729
.....
If odd it is a 9. How did you get a 7?

Powers of 3 repeat 1 3 9 7 every four terms

The series begin with 1, not a 3, b/c 3^0=1

27/4=6 remainder 3. So next 3 is 1, 3, and 9.

3^27 is more difficult than 9^9 since the latter has a period of 2.

dumb.doofus wrote:

sacx wrote:1^1 + 2^2 + 3^3 + 2^8 + 5^5 + 6^6 + 7^7 + 2^24 + 3^27 + 10^10

1^1 + 2^2 = 5 remainder 0
5^5 and 10^10 will also have remainder 0

so we are left with,

3^3 + 2^8 + 6^6 + 7^7 + 2^24 + 3^27

the units digit for the above expression,

7 + 6 + 6 + 3 + 6 + 7

35, which is evenly divisible by 5

Choose A

Nicely done sacx.. only one mistake though.. I have pointed that out in bold too

9^9 = 3^18.. so this gives the last digit as 9...

You have taken 9^9 = 3^27. This is incorrect.

So units digit for the expression is 7 + 6 + 6 + 3 + 6 + 9 = 37

and so remainder = 2

thanks dumb.doofus for pointing it out. I have edited my original post