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Square of 102-- Square of 101+ Square of 100- Square of 99+ Square of 98- Square of 97.. Square of +2- Square of 1 is the remainder divided by 103 This is a factorization problem
A square-b square=(a+b)(a-b)
102+101+100+99+2+1=(102+101)(102-101)+(100+99)(99-98)...+(2+1)(2-1)
=102+101+100+.+1
Sum by isodyne Sn=n (a1+an)/2
Sn=100(1+102)/2=100*103/2
So it can be divided by 103, the rest is 0
Given S=12-22+32-42+...+992-1002+1012, the remainder of S divided by 103 is ______.
Original formula=12+(3+2)(3-2)+(5+4)(5-4)+...+(101+100)(101-100)
=1+5+9+…+201
=51(1+201)
2
=5151.
Because 5151
103=50…1,
So the remainder of S divided by 103 is 1.
Therefore, the answer is:1.
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