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16 Sep 2024, 06:15
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We know to find what is the remainder when \( 3^{24} \) is divided by 5
Theory: Remainder of a number by 5 is same as the reminder 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 \(3^{24}\) 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 24 by 4 and check what is the remainder
=> 24 divided by 4 gives 0 remainder
=> 3^{24} will have the same unit's digit as \(3^{Cycle}\) = Units digit of \(3^{4}\) = 1
=> Remainder of \(3^{24}\) by 5 = Remainder of 1 by 5 = 1
So, Answer will be B
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 reminder 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 \(3^{24}\) 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 24 by 4 and check what is the remainder
=> 24 divided by 4 gives 0 remainder
=> 3^{24} will have the same unit's digit as \(3^{Cycle}\) = Units digit of \(3^{4}\) = 1
=> Remainder of \(3^{24}\) by 5 = Remainder of 1 by 5 = 1
So, Answer will be B
Hope it helps!
MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem
20 Nov 2025, 23:36
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If a & p are co-prime , then
\(\frac{a^{p-1}}{p}=1\) That is the Remainder is 1.
\(\frac{3^{24}}{5}\)=\(\frac{(3^4)^6}{5}\)=\(\frac{(1^6)}{5}\)
Therefore Remainder is 1
\(\frac{a^{p-1}}{p}=1\) That is the Remainder is 1.
\(\frac{3^{24}}{5}\)=\(\frac{(3^4)^6}{5}\)=\(\frac{(1^6)}{5}\)
Therefore Remainder is 1
