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The answer is 7 but I don't understand why. Is it as simple as 10-3?

Just look at how power of 3 works..

3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 3^5 =243 .....

You'd notice that unit digit repeats itself after every 4 numbers.. In other words, 3^19 will have same unit number as 3^15, or 3^11 or 3^7 or 3^3....Which tells us, 3^19 has unit digit as 7..

Now, we know that any number with unit digit as 7 when divided by 10 will give u 7....

PS: Wrong section... Moving it to PS..
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Units digit of 3^n (n = 1,2,3....) follows the following pattern: 3,9,7,1,3,9,7,1... When divided by 10, the remainders would be 3,9,7,1,3,9,7,1... 3^19/10 -> remainder = 7

What is the remainder of 3^19 when divided by 10? [#permalink]

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12 Sep 2014, 11:48

I got D, and this is how I solved: I looked for patterns: ^2 - units digit 9 ^3 - units digit 7 ^4 - units digit 1 ^5 - units digit 3

hence, we can see that when raised to a power which is multiple of 4, the units digit is 1, and when to an even power not multiple of 4, the units digit is 9 and we can then see: ^16 - units digit 1, or ^18 - units digit 9 and ^19 - units digit 7

therefore, when divided by 10, the remainder must be 7

Re: What is the remainder of 3^19 when divided by 10? [#permalink]

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What is the remainder of 3^19 when divided by 10? [#permalink]

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30 Jun 2017, 20:49

The question is essentially asking - What is the unit digit of \(3^{19}\)

Quick check on cycilicity of unit digit says

\(3^1 = 3\)

\(3^2 = 9\)

\(3^3 = 7\)

\(3^4 = 1\)

Therefore, the unit digit repeats after every 4th power of 3. For unit digit of \(3^{19}\) First find \(Remainder \ of (\frac{19}{4})= 3\). The unit digit is 3rd term in the above sequence = 7