Q7. Show that 1¹⁹⁹⁷ + 2¹⁹⁹⁷ + ... + 1996¹⁹⁹⁷ is divisible by 1997.
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Short memo:-
The second solution is deleted as there is a mistake.
Solution A:-
Step 1. Polynomial identity
Step 2. Grouping the terms
Hence, the given number is divisible by 1997.
More information:-
Why does ?
Let's prove this right now. This is an identity for an odd exponent . Let's try synthetic division. (Refer to the attachment.)
Answer:
According to fermat's Little Theorem, Since 1997 is a prime number,
∴ 1¹⁹⁹⁷+ ,2¹⁹⁹⁷... ,+ 1996 ¹⁹⁹⁷
are divided by1997 then,remainder will be respectively 1,2,..,1996
1+2+..+1996 =
= 998 × 1997
Which is divisible by 1997.
Hence remainder =0