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Updated on: 29 Mar 2023, 12:44
Originally posted by CrackverbalGMAT on 26 May 2021, 04:34.
Last edited by CrackverbalGMAT on 29 Mar 2023, 12:44, edited 2 times in total.
Last edited by CrackverbalGMAT on 29 Mar 2023, 12:44, edited 2 times in total.
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Solution:
Every 3*3 = 9 on dividing with 5 gives -1 as remainder
We have 243 3s =>121 3s paired and one additional unpaired 3
=> 3^243 / 5 ----> 9^121 * 3 /5 ----->(-1)^121 * 3/5-----> -1*3/5 -------> -3/5
= -3 as negative remainder
= +2 as a positive remainder
(option c)
Devmitra
Every 3*3 = 9 on dividing with 5 gives -1 as remainder
We have 243 3s =>121 3s paired and one additional unpaired 3
=> 3^243 / 5 ----> 9^121 * 3 /5 ----->(-1)^121 * 3/5-----> -1*3/5 -------> -3/5
= -3 as negative remainder
= +2 as a positive remainder
(option c)
Devmitra
13 Aug 2021, 05:58
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If 3^243 + 2^243 is divided by 5, then the remainder is (I WANT SOLUTION TO THIS)
Updated on: 16 Sep 2024, 06:16
Originally posted by BrushMyQuant on 15 Sep 2022, 09:51.
Last edited by BrushMyQuant on 16 Sep 2024, 06:16, edited 2 times in total.
Last edited by BrushMyQuant on 16 Sep 2024, 06:16, edited 2 times in total.
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We know to find what is the remainder when \(3^{243}\) is divided by 5
Theory: Remainder of a number by 5 is same as the Remainder of the unit's digit of the number by 5
(Watch this Video to Learn How to find Remainders of Numbers by 5)
Using Above theory , Let's find the unit's digit of \(3^{243}\) first.
We can do this by finding 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 243 by 4 and check what is the remainder
=> 243 divided by 4 gives 3 remainder
=> \(3^{243}\) will have the same unit's digit as \(3^3\) = 7
=> Unit's digits of \(3^{243}\) = 7
But remainder of \(3^{243}\) by 5 cannot be more than 5
=> Remainder = Remainder of 7 by 5 = 2
So, Answer will be C
Hope it helps!
MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem
Theory: Remainder of a number by 5 is same as the Remainder of the unit's digit of the number by 5
(Watch this Video to Learn How to find Remainders of Numbers by 5)
Using Above theory , Let's find the unit's digit of \(3^{243}\) first.
We can do this by finding 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 243 by 4 and check what is the remainder
=> 243 divided by 4 gives 3 remainder
=> \(3^{243}\) will have the same unit's digit as \(3^3\) = 7
=> Unit's digits of \(3^{243}\) = 7
But remainder of \(3^{243}\) by 5 cannot be more than 5
=> Remainder = Remainder of 7 by 5 = 2
So, Answer will be C
Hope it helps!
MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem
