Find the remainder when the difference between and is

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Find the remainder when the difference between and is divided by . Note that and . So, Thus, so 1 is the remainder when the difference is divided by . (Perform the subtraction yourself, divide by , and see!) When , and are integers and is a positive integer such that the following is always true: . Modular arithmetic provides an even larger advantage when multiplying than when adding or subtracting. Let's take a look at a problem that demonstrates the point. Jerry has 44 boxes of soda in his truck. The cans of soda in each box are packed oddly so that there are 113 cans of soda in each box. Jerry plans to pack the sodas into cases of 12 cans to sell. After making as many complete cases as possible, how many sodas will Jerry have leftover? Proof of the addition rule Subtraction Problem Solution Subtraction rule Multiplication Problem Solution
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2/17/2018 Art of Problem Solving 6/8 First, we note that this word problem is asking us to find the remainder when the product is divided by . Now, we can write each and in terms of multiples of and remainders: This gives us a nice way to view their product: Using FOIL, we get that this equals We can already see that each part of the product is a multiple of , except the product of the remainders when each and are divided by 12. That part of the product is , which leaves a remainder of when divided by . So, Jerry has sodas leftover after making as many cases of as possible. First, we note that Thus, meaning there are sodas leftover. Yeah, that was much easier. When , and are integers and is a positive integer such that The following is always true: . Since exponentiation is just repeated multiplication, it makes sense that modular arithmetic would make many problems involving exponents easier. In fact, the advantage in computation is even larger and we explore it a great deal more in the intermediate modular arithmetic article. Note to everybody: Exponentiation is very useful as in the following problem: What is the last digit of if there are 1000 7s as exponents and only one 7 in the middle? We could solve this problem using mods. This can also be stated as . After that, we see that 7 is congruent to -1 in mod 4, so we can use this fact to replace the 7s with -1s, because 7 has a pattern of repetitive period 4 for the units digit. is simply 1, so therefore , which really is the last digit. Solution using modular arithmetic Multiplication rule Exponentiation Problem #1 Problem #2
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2/17/2018 Art of Problem Solving 7/8 What are the tens and units digits of ?
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