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## Remainders

Remainders. This is an important topic. So far in discussing division, we have talked mostly about multiples and their factors. When we divide any integer by one of its divisors, the quotient is always an integer and in fact, that integer is always another one of the number’s factors.

So for example, 21 divided by 3 is 7. Well 7 of course is a factor of 21. In fact, it is related the fact that 21 divided by seven equals 7, both of them are connected to the fact that the product of three and seven is 21, those are both factors of 21.

What happens when we divide one number by another number that is not one of the the factors, one of the divisors of the first number? For example, what happens if we divide 17 by 5? Now we talked about this a little bit in a previous video, but we’re gonna review here. One approach would be to change the fractions.

So a perfectly valid answer to that question, what is 17 divided by 5? Appropriately, valid answer is the improper fraction 17 over 5. That number is 17 divided by 5. Sometimes we need to change that to a mixed numeral. Here we can write this as a mixed-numeral 3 in 2 5ths and of course, what that mixed-numeral means, is that there’s implicit addition.

What that really means, is that it’s 3 plus 2 5ths. This is a perfectly mathematically correct answer and in some contexts, this is exactly what’s required. Sometimes the test will ask you for some kind of division and the numbers will be listed as mixed-numerals or a decimals and so, this is the way you have to go.

Sometimes though, we want to keep everything in terms of integers. For examples, suppose we have some, we have seventeen identical items. And we can form some kind of important set from fi, groups of five of these item, so we’re grouping identical items. In that case, we’d want to know how many full sets can we create and we’d want to know, how many extra will be left over after we create those full sets.

Here clearly, we can form three full sets of five. And then, we would have two extra items left over. In other words, the quotient would be 3 and the remainder would be 2. In this video, I will use the term mixed-numeral quotient and integer quotient.

To distinguish between these two types of division. So, we’re talking about 17 divided by 5. The mixed-numeral quotient is 3 and 2 5ths. So I could also write that mixed-numeral quotient as a decimal, if I wanted. The integer quotient, I’m gonna state it this way, verbally. 5 goes into 17 three times with a remainder of 2, that three is the integer quotient.

So notice, that I’m not writing this as an equation. I’m writing that in verbal form. Some of you might have seen in school something along the lines of this. 17 divided by 5 equals 3 remainder 2. That’s a slick little notation. I’m not gonna use that only because, first of all, some people find that confusing.

And second of all, the test never uses that. That you will not see on the test. The test always states this stuff in verbal form, so I’ll be stating it in verbal form here. In these problems, find the integer quotient and the remainders. Just try and do this without a calculator.

Pause the video, and then I’ll talk about these. 20 divided by 6. Well obviously 6 goes into 18 three times. And we have two leftover. 95 divided by 7.

This is a little bit trickier, you have to know that 91 equals 7 times 13. And so it goes in 13 times with 4 left over. 56 divided by 8, of course, 8 goes into 56 an even number of time, it goes in 7 times and we have no remainder. So for that, we can actually just write the equal sign, and note that we have no remainder at all.

Let’s talk about the terminology here. 20 divided by 6 yields 3 with a remainder of 2. What are the terms for all these roles? The 20, the number divided is called the dividend. The 6, the number by which we’re dividing is called the divisor. Now, be very careful here.

We’re using divisor in a slightly different sense than we’ve used in previous videos. In previous videos, we were talking about divisor as a synonym for factor. The factors of a number. The divisors of a number. So we’re not using divisor in it’s factor sense here, because of course 6 is not factor of 20.

We’re just using divisor in the sense of the number by, which we’re dividing. So, this is divisor in the division sense. A slightly different use of the word. Of course, 3 is the integer quotient and 2 is the remainder. Well, a few things to notice. First of all, notice that the remainder is always less than the divisor.

The remainder has to be a positive number, it could be zero, it could be positive, but it has to be less then the divisor. If it were greater than the divisor then, the divisor could go into it one more time. So that’s, why it’s always less then the divisor. If D is the is the dividend S is the divisor Q is the quotient and R is the remainder, then we can write things in this form D divided by S dividend divided by divisor equals the quotient the integer quotient plus the fraction remainder over divisor.

So this is an important formula, because it allows us to make a connection between the integer quotient and the mixed-numeral quotient. Q is the integer quotient, but if we then, hm, add that fraction, remainder over divisor, that gives us the mixed-numeral quotient. Now it’s very important to realize also Q is an integer, so any non-integer part of the mixed-numeral quotient is gonna be that fraction.

This means, that the quotient is written in mixed-numeral or decimal form, the non-integer part of the quotient is remainder over divisor. That’s a really subtle and important point. Supposed we’re told, we divide two numbers and we get that horrible long number with the decimal park. We know that the remainder divided by the divisor will equal that non-integer part, that decimal 0.375 which happens to equal the fraction 3 8ths.

We don’t necessarily know that r equals 3 and s equals 8. It could be that have a remainder of 3 and the divisor of 8. That could be true, but r over S could also equal any fraction equivalent to 3 over 8. So it could be 6 over 16, 9 over 24, 12 over 32 et cetera. So, all we have here is ratio information.

We know that the ratio of r over s equals 3 over 8, and if we knew either the remainder or the divisor, we could find the other. One important skill, is generating examples of possible dividends that, when divided by a certain divisor, yield a specific remainder. For example, what numbers when divided by 12, have a remainder of 5? Well clearly, the simplest one, would just be 5 plus 12, which is 17.

When we divide 17 by 12, it goes in once, with a remainder of 5. Similarly, we could take any multiple of 12, and simply add 5. So all of those are numbers that when we divide by 12, it will go in a certain number of times and then it will have a remainder of 5. And notice, we can get from one to the next simply by adding or subtracting 12. So 17 plus 12 is 29 plus 12 is 41 plus 12 is 53.

So much in the same way as we can get from one multiple to the next, by adding the number, we can also get from one of these numbers to the next by adding the numbers. If I tell you that 1997 is a number that when you divide by 12 has a remainder of 5, then you could find other numbers of this set by adding or subtracting 12.

So, we can add 12, and that would be, that would be 2009. We could subtract 12 that would be 1985. Incidentally, what we’re doing here, we would be finding all the years that, in the Chinese calendar, were the year of the Ox. So, I’m mentioning this only cuz it’s a really unexpected application of remainders and dividers.

There’s all kinds of subtle ways, that remainder and divisor questions can show up on the test. Another tricky question is, what is the smallest positive integer that when divided by 12 is a remainder of 5. For you may think that it’s 17, but it’s actually 5. Wait a second, what’s going on here?

12 is bigger than 5, so if we divide 5 by 12, it goes into it zero times, an integer quotient of zero, and the remainder is 5. In general, if the divisor is larger than the dividend. Then the integer quotient is zero and the remainder equals the dividend. The test absolutely loves to test questions about this. Finally, a very important topic is rebuilding the dividend.

What do I mean by this? Above we have that fraction equation. If we multiply both sides of the equation by S, the divisor, we clear all the fractions, and we get the equation, dividend equals divisor times quotient plus remainder. That formula is immensely important.

That is called the rebuilding the dividend equation, and that is an immensely important in formula for problem solving. Let me give an example. Here’s a question. Pause the video, and then I’ll talk about a solution to this. When positive integer N is divided by positive integer P, the quotient is 18, with a remainder of 7.

N when N is divided by P plus 2 the quotient is 15 and the remainder is one so we’re being asked to find the dividend. So using the rebuilding the dividend formula for the first sentence. We get N equals 18 times P plus 7 using the formula for the second sentence we get. N equals 15 times P plus 2 plus 1 the remainder.

And of course I can simplify this, to 15 times P plus 31. Those are two expressions for N, so I can set them equal to each other. Subtract 15 times P, subtract 7. We get 3 times P equals 24, P equals 8. And then we can plug in, and we can solve for the dividend which is 151. So using that formula is a very powerful problem solving tool.

In summary, dividing by a number that is not a factor of the dividend, we could get either a mixed-numeral quotient or an integer-quotient and a remainder. The remainder is always greater than or equal to zero and less than the divisor. R over S is the non-integer part of a mixed-numeral or decimal quotient. That’s the remainder divided by the divisor, which we can always set it equal to that non-integer part.

One important skill is generating examples of possible dividends for a given divisor and remainder. And finally, we talked about the very important rebuilding the dividend formula.

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