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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
1
Bookmarks
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?
[#permalink]
20 Jun 2024, 21:29
1
3 follows the pattern as 3 , 9, 27, 81 , 243 so after multiplication 4 times the pattern of units place digit repeats so 283 divided 4 will have remainder 3 means the number will have units place digit as 7 because as we follow the pattern of 3, 9, 27, 81 ... the 3rd place number's units digit is 7. Now dividing this number by five will have remainder 2.
gmatclubot
Re: What is the remainder when 3^283 is divided by 5? [#permalink]