# Alternative solution to problem :Last 3 digit of 3^4798

I want to find Last 3 digits of 3^4798 I found one solution that can be explained by Euler's phi function.

Somebody suggest any alternative solution ?

Note by Anurag Choudhary
4 years, 11 months ago

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You could write $$3^{4798} = 9^{2399} = (10-1)^{2399} = \displaystyle\sum_{k = 0}^{2399} \dbinom{2399}{k}\cdot 10^k \cdot (-1)^{2399-k}$$, and then use the fact that only term that is not a multiple of $$10$$ is the $$k = 0$$ term which is $$-1$$. So, the last digit is $$9$$.

EDIT: I misread the question as the last digit instead of the last 3 digits. As Daniel C. pointed out below, this can be fixed by using the $$k = 0,1,2$$ terms.

- 4 years, 11 months ago

which formula have u used?

- 4 years, 11 months ago

- 4 years, 11 months ago

last digit is easy one and may alternative solutions are there for that We need last 3 digits

Thanks

- 4 years, 11 months ago

Well, then use the last three terms of the expansion Jimmy K. provided.

- 4 years, 11 months ago

yeah its obvious :)

- 4 years, 11 months ago