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 ?

#NumberTheory #MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Jimmy Kariznov - 7 years, 10 months ago

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

Thanks

Anurag Choudhary - 7 years, 10 months ago

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

Daniel Chiu - 7 years, 10 months ago

@Daniel Chiu – yeah its obvious :)

Anurag Choudhary - 7 years, 10 months ago

which formula have u used?

anup kc - 7 years, 10 months ago

Binomial theorem

Daniel Chiu - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$