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What is the remainder when 3^48 is divisible by 19?

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What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Feb 2020, 18:40
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Question Stats:

38% (02:26) correct 62% (02:32) wrong based on 42 sessions

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GMATBusters’ Quant Quiz Question -7


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What is the remainder when 3^48 is divisible by 19?
A. 8
B. 9
C. 10
D. 11
E. 12

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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Feb 2020, 18:42
Remainder questions are based on a pattern. Let us check the Pattern
When 3 is divided by 19, remainder = 3
When 3^2 is divided by 19, remainder = 9
When 3^4 is divided by 19, remainder = 5
When 3^8 is divided by 19, remainder = 5*5= 25 = remainder of 6
When 3^16 is divided by 19, remainder = 6*6= 36 = remainder of 17 or -2
When 3^48 is divided by 19, remainder = -2*-2*-2 = -8 , converting to Positive remainder= -8+19 = 11

The remainder = 11

Answer D

Concept of Negative Remainder:
By definition, the remainder cannot be negative. But in certain cases, you can assume that for your convenience/ease the calculation.
But a negative remainder in real sense means that you need to add the divisor in the negative remainder to find the real remainder. Always convert the final remainder in POSITIVE.

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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Feb 2020, 19:24
What is the remainder when 3^48 is divisible by 19?

Cyclicity of 3 = 3, 9, 7 and 1

\(3^{48}\) can be written as \(3^4+3^5+3^5+3^5
\)

\(3^4\) = 81 , when 81/19 , we get remainder as 5
\(3^5\) = 243 , when 243/19 , we get remainder as 15

\(3^4+3^5+3^5+3^5\) = 5+15+15+15 = 50
When we divide 50/19 , we get the remainder 12

When \(3^{48}\) / 19 we get the remainder as 12

Ans E
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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Feb 2020, 20:07
Qs; What is the remainder when 3^48 is divisible by 19?

Sol: in such qs we need to find the remainder with increasing power of numerator.
when 3/19 remainder is 3
when 3^2/19 remainder is 9
when 3^4 /19 then remainder = 9*9=81 but we need to adjust the excess remainder as remainder is greater than divisor, so Remainder (81/19)= 5 [from previous division we know remainder is 9 when 3^2/19]

when 3^8 then remainder = 5*5= 25 adjusting, remainder (25/19)= 6 [from previous case we know that 3^4 divide by 19 will result in remainder of 5].

when 3^16 then remainder =6*6=36 adjusting the remainder (36/19)= 17 or -2 (extra 2 is required to divide the 36 completely by 19). this is for easy calculation.

we can simplify the given equation 3^48 = 3^(16*3)
so when 3^48 is divided by 19 then remainder will be = (-2)^3= -8 adjusting for positive values -8+19= 11.
so the remainder would be 11.

Answer is D.
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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Feb 2020, 20:26
the solution is based in a series of small divisions:
19*=multiple of 19
3^48=(3^4)^12=81^12
the remainder of dividing 81 by 19 is 5. Then,
81^12=(19*+5)^12=19*+5^12=19*=125^4
But 125=19*-8 Then,
125^4=(19*-8)^4=19*+8^4=19*+(64)(64)=19*+(19*+7)^2=19*+49=19*+11
Answer D
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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Feb 2020, 22:57
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Solution:
3^1 = 3,
3^2 = 9,
3^3 = 27,
3^4= 81,

so since units digit of 3^4 is 1, so following the similar pattern, units digit of 3^48 = (3^4)^12 = 1, so the the remainder will be the digit the ending with 1 = 11.


Answer D.
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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Apr 2020, 05:54
GMATBusters wrote:

GMATBusters’ Quant Quiz Question -7


For questions from previous quizzes click here

What is the remainder when 3^48 is divisible by 19?
A. 8
B. 9
C. 10
D. 11
E. 12


VeritasKarishma hi, would you mind solving this through binomial expansion? please.
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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Apr 2020, 09:05
bm2201 wrote:
Solution:
3^1 = 3,
3^2 = 9,
3^3 = 27,
3^4= 81,

so since units digit of 3^4 is 1, so following the similar pattern, units digit of 3^48 = (3^4)^12 = 1, so the the remainder will be the digit the ending with 1 = 11.


Answer D.



I don't find any logic in your approach. Its the quotient's units digit. How did u relate it to remainder's digit ?
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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Apr 2020, 09:14
Binomial approach is not recommended here.

exc4libur wrote:
GMATBusters wrote:

GMATBusters’ Quant Quiz Question -7


For questions from previous quizzes click here

What is the remainder when 3^48 is divisible by 19?
A. 8
B. 9
C. 10
D. 11
E. 12


VeritasKarishma hi, would you mind solving this through binomial expansion? please.

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Re: What is the remainder when 3^48 is divisible by 19?  [#permalink]

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New post 01 Apr 2020, 11:36
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Euler no of 19 is 18 ( Euler no of a prime no p is p-1 )
so , 3^48 / 19 will be reduced to 3^12 /19 ( 12= 48-36 = 48- 18*2 )

now, 3^12 /19 = (3^4)^3 /19 = 81^3/19
now, rem is 5 when 81/19
therefore, actual remainder is 5^3 = 125 = 125- 6*19 = 11 ( remainder must be <19)

correct answer is D
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Re: What is the remainder when 3^48 is divisible by 19?   [#permalink] 01 Apr 2020, 11:36

What is the remainder when 3^48 is divisible by 19?

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