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What is the remainder of 3^19 when divided by 10?

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

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New post 27 May 2006, 22:53
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What is the remainder of 3^19 when divided by 10?

A. 0
B. 1
C. 5
D. 7
E. 9
[Reveal] Spoiler: OA

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New post 27 May 2006, 23:17
I got D.
You can use the Mod function to solve this kind of problem.

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New post 27 May 2006, 23:17
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3^19=3^4*3^4*3^4*3^4*3^3
3^4=81/10 has rmainder 1
3^3=27/10 has remainder 7
1*1*1*1*7=7

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New post 27 May 2006, 23:21
D it is...

3 ^ 16 will give remainder 1
3 ^ 3 will give remainder 7

so 7 * 1= 7

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New post 31 May 2006, 18:12
I am getting 7 - D

brute force to find the pattern in this case 39713971 and then just count to the 19th power we get 7

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How do you solve? [#permalink]

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New post 07 Feb 2014, 17:06
what is the remainder when 3^19 is divided by 10

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

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Re: How do you solve? [#permalink]

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New post 07 Feb 2014, 17:15
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Luning1 wrote:
what is the remainder when 3^19 is divided by 10

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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New post 07 Feb 2014, 19:29
what is the remainder when 3^19 is divided by 10?

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

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

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New post 10 Sep 2014, 21:52
we are interested in last digit of 3^19, so we should find circulation

3*1=3
3*2=9
3*3=7
3*4=1
3*5=3
circulation is 4, that means 19/4=4 with 3 remainder, so 7 is last digit. Dividing 10 gives us 7/10, so 7 is remainder

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

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New post 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

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

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New post 30 Sep 2014, 23:23
For \(3^1\), remainder = 3

\(3^2\), remainder = 9

\(3^3\), remainder = 7

\(3^4\), remainder = 1

& so on........

For \(3^{19}\), remainder = 7

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

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

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New post 14 Nov 2015, 12:21
X & Y wrote:
What is the remainder of 3^19 when divided by 10?

A. 0
B. 1
C. 5
D. 7
E. 9


We know \(\frac{3^4}{10}\) = Reminder 1

3^19 = \(\frac{(3^4)^4 *(3^3)}{10}\)

\(\frac{(3^4)^4}{10}\) will have remainder 1

We need to find the reminder of \(\frac{(3^3)}{10}\)

\(\frac{(3^3)}{10} = [m]\frac{(20 + 7)}{10}\) => Reminder is 7
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What is the remainder of 3^19 when divided by 10? [#permalink]

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New post 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

Answer is D
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What is the remainder of 3^19 when divided by 10?   [#permalink] 30 Jun 2017, 20:49
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