Question

Q7. Show that 1¹⁹⁹⁷ + 2¹⁹⁹⁷ + ... + 1996¹⁹⁹⁷ is divisible by 1997.​

Answer 1

Short memo:-

The second solution is deleted as there is a mistake.

Solution A:-

Step 1. Polynomial identity

$x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+...-xy^{n-2}+y^{n-1})$

Step 2. Grouping the terms

$(1^{1997}+1996^{1997})+(2^{1997}+1995^{1997})+...+(998^{1997}+999^{1997})$

$=(1+1996)(...)+(2+1995)(...)+...+(998+999)(...)$

$=1997(...)$

Hence, the given number is divisible by 1997.

More information:-

Why does $x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+...-xy^{n-2}+y^{n-1})$?

Let's prove this right now. This is an identity for an odd exponent $n$. Let's try synthetic division. (Refer to the attachment.)

Answer 2

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 =

$\frac{1}{2 } \times 1996 \times 1997$

= 998 × 1997

Which is divisible by 1997.

Hence remainder =0