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Sub 505 Level|   Exponents|   Remainders|            
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I got D.
You can use the Mod function to solve this kind of problem.
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D it is...

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

so 7 * 1= 7
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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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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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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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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

D
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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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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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X & Y
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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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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\(3^{19}\) = \((3^2)^{9}\) * 3

=> \(\frac{(3^2) }{ 10}\) -> remainder = -1

=> \((-1)^9\) = -1

=>\( \frac{3 }{ 10}\) -> remainder = -7

- 1 * - 7 = 7

Answer D
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What is the remainder of \(3^{19}\) when divided by 10

Theory: Remainder of a number by 10 is same as the unit's digit of the number

(Watch this Video to Learn How to find Remainders of Numbers by 10)

Using Above theory Remainder of \(3^{19}\) by 10 = unit's digit of \(3^{19}\)

Now to find the unit's digit of \(3^{19}\), we need to find the pattern / cycle of unit's digit of power of 3 and then generalizing it.

Unit's digit of \(3^1\) = 3
Unit's digit of \(3^2\) = 9
Unit's digit of \(3^3\) = 7
Unit's digit of \(3^4\) = 1
Unit's digit of \(3^5\) = 3

So, unit's digit of power of 3 repeats after every \(4^{th}\) number.
=> We need to divided 19 by 4 and check what is the remainder
=> 19 divided by 4 gives 3 remainder

=> \(3^{19}\) will have the same unit's digit as \(3^3\) = 7

So, Answer will be D
Hope it helps!

MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

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I started counting from 3^0 and I got 9 as an answer, how do I know if I have to start from 3^0 or 3^1?

help
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