Cyclicity in Remainders

Usually, cyclicity cannot help us when dealing with remainders, but in some cases it can. Let’s look at the cases in which it can, and we will see why it helps us in these cases. First let’s look at a pattern:   20/10 gives us a remainder of 0 (as 20 is exactly divisible by 10) 21/10 gives a remainder of 1 22/10 gives a remainder of 2 23/10 gives a remainder of 3 24/10 gives a remainder of 4 25/10 gives a remainder of 5 and so on… In the case of this pattern, 20 is the closest multiple of 10 that goes completely into all these numbers and you are left with the units digit as the remainder. Whenever you divide a number by 10, the units digit will be the remainder. Of course, if the units digit of a number is 0, the remainder will be 0 and that number will be divisible by 10 — but we already know that. So remainder when 467,639 is divided by 10 is 9. The remainder when 100,238 is divided by 10 is 8 and so on… Along the same lines, we also know that every number that ends in 0 or 5 is a multiple of 5 and every multiple of 5 must end in either 0 or 5. So if the units digit of a number is 1, it gives a remainder of 1 when divided by 5. If the units digit of a number is 2, it gives a remainder of 2 when divided by 5. If the units digit of a number is 6, it gives a remainder of 1 when divided by 5 (as it is 1 more than the previous multiple of 5). With this in mind: 20/5 gives a remainder of 0 (as 20 is exactly divisible by 5) 21/5 gives a remainder of 1 22/5 gives a remainder of 2 23/5 gives a remainder of 3 24/5 gives a remainder of 4 25/5 gives a remainder of 0 (as 25 is exactly divisible by 5) 26/5 gives a remainder of 1 27/5 gives a remainder of 2 28/5 gives a remainder of 3 29/5 gives a remainder of 4 30/5 gives a remainder of 0 (as 30 is exactly divisible by 5) and so on… So the units digit is all that matters when trying to get the remainder of a division by 5 or by 10. Let’s take a few questions now: What is the remainder when 86^(183) is divided by 10? Here, we need to find the last digit of 86^(183) to get the remainder. Whenever the units digit is 6, it remains 6 no matter what the positive integer exponent is. So the units digit of 86^(183) will be 6. So when we divide this by 10, the remainder will also be 6. Next question: What is the remainder when 487^(191) is divided by 5? Again, when considering division by 5, the units digit can help us. The units digit of 487 is 7. 7 has a cyclicity of 7, 9, 3, 1. Divide 191 by 4 to get a quotient of 47 and a remainder of 3. This means that we will have 47 full cycles of “7, 9, 3, 1” and then a new cycle will start and continue until the third term. 7, 9, 3, 1 7, 9, 3, 1 7, 9, 3, 1 7, 9, 3, 1 … 7, 9, 3 So the units digit of 487^(191) is 3, and the number would look something like ……………..3 As discussed, the number ……………..0 would be divisible by 5 and ……………..3 would be 3 more, so it will also give a remainder of 3 when divided by 5. Therefore, the remainder of 487^(191) divided by 5 is 3. Last question: If x is a positive integer, what is the remainder when 488^(6x) is divided by 2? Take a minute to review the question first. If you start by analyzing the expression 488^(6x), you will waste a lot of time. This is a trick question! The divisor is 2, and we know that every even number is divisible by 2, and every odd number gives a remainder 1 when divided by 2. Therefore, we just need to determine whether 488^(6x) is odd or even. 488^(6x) will be even no matter what x is (as long as it is a positive integer), because 488 is even and we know even*even*even……(any number of terms) = even. So 488^(6x) is even and will give remainder 0 when it is divided by 2.