Problem for the next perfect sqare year!
3 ( 1 2 + 4 2 + 9 2 + 1 6 2 + … + 2 0 2 5 2 ) 2 + 2 ( 1 2 + 4 2 + 9 2 + 1 6 2 + … + 2 0 2 5 2 ) + 1 Find the remainder when the number above is divided by ( 4 2 + 9 2 + 1 6 2 + … + 2 0 2 5 2 ) .
The answer is 6.
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2 solutions
Moderator note:
Yes. Because 4 2 + 9 2 + 1 6 2 + . . . + 2 0 2 5 2 pops up everywhere. It might be useful to consider a symbol to represent it. Thus making it so much simpler.
I did not see 1 at the 'end':)
Let 4 2 + 9 2 + 1 6 2 + . . . 2 0 2 5 2 = n then 3 ( 1 2 + 4 2 + 9 2 + 1 6 2 + . . . 2 0 2 5 2 ) 2 + 2 ( 1 2 + 4 2 + 9 2 + 1 6 2 + . . . 2 0 2 5 2 ) + 1 = 3 ( 1 + n ) 2 + 2 ( 1 + n ) + 1 = 3 n 2 + 8 n + 6 = ( 3 n + 8 ) n + 6 , so 3n+8 is the quotient and 6 is the remainder.
Let ( 1 2 + 4 2 + 9 2 + 1 6 2 + . . . + 2 0 2 5 2 ) = x . Therefore we want to know the remainder of x − 1 3 x 2 + 2 x + 1 . By remainder theorem what we want to know is 3 ( 1 ) 2 + 2 ( 1 ) + 1 = 6