ultimate congruency

7⁴≡401mod1000

7⁸≡801mod1000

7¹⁶≡601mod1000

7³²â‰¡201mod1000

7⁶⁴≡401mod1000

and the pattern continues

so we can say 7⁹⁹⁹⁶≡601mod1000

7³â‰¡343mod1000

so 7⁹⁹⁹⁹≡943mod1000

so last 3 digits are 943

ans. is 143 , and you can check it in mathematica .

sorry to say , but congruency will make things more difficult along with the calculations too , but

nonetheless a solution I never thought of .

My solution , first by congruency -

suppose 7⁹⁹⁹⁹ = x (mod 1000)

Multiply each side by 7

so 7¹⁰⁰⁰⁰ = 7x ( mod 1000 ) -------------- 1

but φ(p) for 1000 = 400

so 7⁴⁰⁰ = 1 ( mod 1000 )

again 7¹⁰⁰⁰⁰ = 1 ( mod 1000 ) --------------------- 2

so from 1 and 2 , either 7x = 1 which is absurd ,

or 7x = 1 ( mod 1000 )

from here , x = 143 ( as 7 x 143 = 1001 )

I bet you liked this :P

Another method -------------

As 7¹⁰⁰⁰⁰ = 1 ( mod 1000 )

So 7¹⁰⁰⁰⁰ = 001 (mod 1000 )

So last 3 digits of 7¹⁰⁰⁰⁰ are 001.

So last 3 digits of 7⁹⁹⁹⁹ should be such that multiplying them by 7 , we should a get a

number ending with 001 .

Clearly last 3 digits are 143 only .