Fundamentals - The last 3 digits of 7[p]9999[/p] are

62

(10-3)⁹⁹⁹⁹

Now try binomial theorem and use the first three terms.. other terms will not be involved in the first 3 digits.. why?

3

the answer is cuming 143...it it correct??

1

Let's TAKE , 7 ⁹⁹⁹⁹ = x ( mod 1000 )

so , 7 ¹⁰⁰⁰⁰ = 7 x ( mod 1000 )

But see , according to Euler's totient function theorem , 7 ^{f ( p )} = 1 ( mod 1000 ) ,

Here , f ( p ) = 1000 ( 1 - 1 / 2 ) ( 1 - 1 / 5 ) = 400

So , 7 ¹⁰⁰⁰⁰ = 1 ( mod 1000 )

Hence , 7 x = 1 ( mod 1000 ) ,

Clearly , x = 143 gives ,

7 x 143 = 1001 = 1 ( mod 1000 )

The last 3 digits of 7[p]9999[/p] are