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What is the remainder when 3^48 is divisible by 19?
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01 Feb 2020, 18:40
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GMATBusters’ Quant Quiz Question 7 For questions from previous quizzes click hereWhat 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?
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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?
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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?
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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?
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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?
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01 Feb 2020, 22:57
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?
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01 Apr 2020, 05:54
GMATBusters wrote: GMATBusters’ Quant Quiz Question 7 For questions from previous quizzes click hereWhat 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?
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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?
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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 hereWhat 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?
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01 Apr 2020, 11:36
Euler no of 19 is 18 ( Euler no of a prime no p is p1 ) so , 3^48 / 19 will be reduced to 3^12 /19 ( 12= 4836 = 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




Re: What is the remainder when 3^48 is divisible by 19?
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01 Apr 2020, 11:36




