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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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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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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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GMATBusters

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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bm2201
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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Binomial approach is not recommended here.

exc4libur
GMATBusters

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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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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