Is it C?
the additon of units digit of given numbers revels 47.
either way can someone tell a quicker way..
cheers.
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Remainder
Source: Beat The GMAT — Problem Solving |
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
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
Last edited by sacx on Thu May 28, 2009 2:25 am, edited 1 time in total.
SACX
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dumb.doofus
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Nicely done sacx.. only one mistake though.. I have pointed that out in bold toosacx 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
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
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3^27=9^9sacx 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
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.
thanks dumb.doofus for pointing it out. I have edited my original postdumb.doofus wrote:Nicely done sacx.. only one mistake though.. I have pointed that out in bold toosacx 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
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
