modular arithmetic

How would one use modular arithmetic to find the units digit of 7^7^7?

tens digit of 2^65 ?

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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

FOR TENS DIGIT OF 2^65, FIND ITS MOD 100.

NOTE THAT 2^10=1024 = 24 (MOD 100) sO 2^20 = 576(MOD 100) = -24(MOD100) 2^40= (-24)^2 (MOD 100) = -24 (MOD100) AGAIN! 2^60 = 2^20 *2^40 (MOD 100) = 576(MOD100) = -24(MOD100)

ALSO, 2^5 = 32(MOD100) SO 2^65=2^60 * 2^5(MOD 100) = (-24)(32)(MOD100) = 32(MOD100)

SO THE TENS DIGIT IS 3 AND THE UNITS DIGIT IS 2.

Shourya Pandey - 8 years, 3 months ago

The first problem :

$7^{7^{7}} \equiv 7^{3} \equiv 3 \pmod {10}$ and hence the unit digit is $3$ . (The powers of $7$ form a cycle mod $10$ . )

Zi Song Yeoh - 8 years, 3 months ago

how can you equate power 7 as power 3. I guess this is wrong. Please clarify.

Namra Aziz - 8 years, 3 months ago

no..its perfectly alright..

Nishanth Hegde - 8 years, 3 months ago

$7^{7} \equiv 7^{3}\cdot7^{4} \equiv 7^{3} \equiv 3 \pmod{10}$ . Since $7^2 \equiv -1 \pmod {10} \Rightarrow 7^{4} \equiv 7^{2}\cdot7^{2} \equiv 1 \pmod{10}$

Zi Song Yeoh - 8 years, 3 months ago

@Zi Song Yeoh – Just use euler's theorum for $7^4$ , since phi(10) = 4

Harshit Kapur - 8 years, 3 months ago

@Harshit Kapur – Yes.

Zi Song Yeoh - 8 years, 3 months ago

But unit digit would be 7.

Ram Prakash Patel - 8 years, 3 months ago

Zi Song is right. The answer is 3.

Rohan Rao - 8 years, 3 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}`	$\boxed{123}$