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1^1+2^2+3^3+...+10^10 is divided by 2. Find the remainder.

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1^1+2^2+3^3+...+10^10 is divided by 2. Find the remainder.  [#permalink] New post 08 Oct 2003, 23:37
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1^1+2^2+3^3+...+10^10 is divided by 2. Find the remainder.
The same is divided by 3. Find the remainder.
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 [#permalink] New post 09 Oct 2003, 04:37
The sum of first 10 squares comes to 385 with the formula n*(n+1)*(2n+1)/6
therfore the remainder is 1 with both 2 and 3

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 [#permalink] New post 09 Oct 2003, 04:50
am doomed in that case.. hahah.. just kidding .. For 2 I would say the series goes like odd, even , odd even and so on. The sum of and odd and an even is always odd SO with 2 I can say the answer is always 1. With 3 I am not too sure. would need to think a bit mate.
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 [#permalink] New post 09 Oct 2003, 05:02
actually 3 in not so difficult either
just consider remainders of each component, and the common remainder
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 [#permalink] New post 09 Oct 2003, 06:01
I say remainders are 1 and 2 for 2 and 3, respectively.

My approach:

For 2, there are five even numbers and five odd numbers, so when you sum them to gether you end up with even + 1 (which is odd). And that 1 is the remainder.

For 3, all multiples of 3 are out. All rest of the numbers leave 1 as remainder when raised to even power and, when raised to odd power, they leave the same remainder as they would in their simple form. i.e 5 mod 3 = 2, (5^3) mod 3 = 2. So in this case just use that method to figure our remainders, add them and devide by 3 to find remainder.
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 [#permalink] New post 09 Oct 2003, 06:12
but dontt u think 385/3 leaves 1 as remainder? some flaw in ur technique
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 [#permalink] New post 09 Oct 2003, 06:21
ngulati wrote:
The sum of first 10 squares comes to 385 with the formula n*(n+1)*(2n+1)/6
therfore the remainder is 1 with both 2 and 3

cheers
neeraj


Neeraj,

Your formula is right but the question does not ask for the sum of squares. The question infact asks for the sum of the first 10 numbers raised to their own powers.

Amar.
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 [#permalink] New post 09 Oct 2003, 06:23
oh yeah I overlookedthat... thanks mate...
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 [#permalink] New post 09 Oct 2003, 06:46
ngulati wrote:
but dontt u think 385/3 leaves 1 as remainder? some flaw in ur technique


My formula works even when you're only squar'ing the numbers. :shock: 8-)
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 [#permalink] New post 08 Jan 2004, 06:42
to find the remainder of 1^1+2^2....+10^10 divided by two I followed this

if you add two odd numbers then you get an even number and it is divisible by 2. In the above equation all even numbers and derived even number are divisible by 2
The add numbers are
1^1+3^3+5^5+7^7+9^9
These five numbers are with two pairs of odd numbers and one remaining odd number.
Take any odd number as residue and when you divide it you will get 1 as the remainder.
So 1 is the remainder.
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Re: PS: TRIBUTE TO REMAINDERS [#permalink] New post 14 Jan 2004, 06:32
stolyar wrote:
1^1+2^2+3^3+...+10^10 is divided by 2. Find the remainder.
The same is divided by 3. Find the remainder.


This can be solved rather quickly using logic and basic number theory...

1) There are 5 odd and 5 even numbers in the sum. Hence, the sum is ODD and the remainder is 1 when divided by 2.

2) Add up the remainders of individual terms. Ignore 3, 6, and 9 since remainders will be zero.

1^1 = 1
2^1 = 4 = remainder of 1
4^4 = (3+1)^4 which means every term in the expansion is divisible by 3 except the 1*1 term hence remainder 1.
5^5 = 25^2 * (4+1) = (24+1)^2 *5 = (24^2 + 48 + 1)*5 = all terms divisible by 3 expect 5*1 so remainder is 2.
7^7 = (6+1)^7 = all terms in expansion divisible by 3 except 1*1 so remainder is 1.
8^8 = 64^4 = (63+1)^4 = all terms in expansion divisible by 3 except 1*1 so remainder is 1.
10^10 = 999999999 + 1 so remainder is 1.

sum of all remainders is 8. 8 mod 3 = 2 hence remainder is 2 when the entire expression is divided by 3.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Re: PS: TRIBUTE TO REMAINDERS   [#permalink] 14 Jan 2004, 06:32
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