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.

_________________

Best,

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