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Re: What is the remainder when 3^283 is divided by 5?
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18 Dec 2017, 02:11
1
For this question, we must know the pattern of 3. 3^1 = 3 3^2 = 9 3^3 = 7 3^4 = 1 3^5 = .... 3 So we have a pattern of 4. When we divide 383 by 4 we got 70,75. So the 280th ending must be a one. Then 283th must be a 7. (Something ending with 7 / 5 = Quotient wth a remainder of 2. So C.
Re: What is the remainder when 3^283 is divided by 5?
[#permalink]
20 Oct 2022, 09:42
We need to find What is the remainder when \(3^{283}\) is divided by 5
Theory: Remainder of a number by 5 is same as the unit's digit of the number
(Watch this Video to Learn How to find Remainders of Numbers by 5)
Using Above theory , Let's find the unit's digit of \(3^{283}\) 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 283 by 4 and check what is the remainder => 283 divided by 4 gives 3 remainder
=> \(3^{283}\) will have the same unit's digit as \(3^3\) = 7 => Unit's digits of \(3^{283}\) = 7
But remainder of \(3^{283}\) by 5 cannot be more than 5 => Remainder = Remainder of 7 by 5 = 2
So, Answer will be C Hope it helps!
Watch the following video to learn the Basics of Remainders
Re: What is the remainder when 3^283 is divided by 5?
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29 Apr 2024, 08:14
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