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 ?

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 ?

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TopNewestYou 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. – Jimmy Kariznov · 4 years ago

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– Anup Kc · 4 years ago

which formula have u used?Log in to reply

Binomial theorem – Daniel Chiu · 4 years ago

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Thanks – Anurag Choudhary · 4 years ago

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– Daniel Chiu · 4 years ago

Well, then use the last three terms of the expansion Jimmy K. provided.Log in to reply

– Anurag Choudhary · 4 years ago

yeah its obvious :)Log in to reply