# چندگانه

سرفصل: بخش ریاضی / سرفصل: خواص عدد صحیح / درس 3

## بخش ریاضی

14 سرفصل | 192 درس

### توضیح مختصر

• زمان مطالعه 0 دقیقه
• سطح خیلی سخت

### دانلود اپلیکیشن «زوم»

این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید ### فایل ویدیویی

برای دسترسی به این محتوا بایستی اپلیکیشن زبانشناس را نصب کنید.

## Multiples

Now we can talk about multiples. In a previous video in this module, we learned that the following statements are equivalent: 7 is a factor of 91, 7 is a divisor of 91, and 91 is divisible by 7. All three of these give the same mathematical information. So now we can add one more statement to this list, 91 is a multiple of seven.

So first up, I’ll just point out that these are four different ways to say exactly the same thing. And the test could expect you to hear one and know that any of the others is true. It’s very important to be comfortable changing back and forth between these statements. So throughout this video, just assume that P, Q, and r are positive integers.

If we are given that P is a multiple of r, this means I could multiply r by some positive integer, and the product would be P. In other words, r is a factor of P and a divisor of P. So multiple is kind of the inverse relationship to factor, because if r is a factor of P, then P is a multiple of r. So for example, 75 and 1250 are multiples of 5.

63 and 888 are multiples of 3. Incidentally, we can see that 888 is a multiple of 3, because if we take the sum of the digits, 8 plus 8 plus 8 is 24. Since the digits add up to something divisible by three, 888 is divisible by three, and, if 888 is divisible by 3, that means it must be a multiple of 3.

So a few ideas about multiples. Idea number one, just as 1 is a factor of every positive integer, every positive integer is a multiple of 1. It doesn’t really make sense to talk about the multiples of one. The multiples of one are all the positive integers. Just as every positive integer is a factor of itself, every positive integer is a multiple of itself.

So we could take any number, say 7. The largest factor of 7 is 7, and the smallest multiple of 7 is also 7, and that’s true of any number. Multiple idea number two. If we need, for example, the first five multiples of a number, say we needed the first five multiples of 12.

We simply multiply the original number by the numbers 1, 2, 3, 4, 5. And similarly you needed the first four or the first seven numbers, you just multiply it by the first few positive integers. So these are first five multiples of 12. Twelve times one, 12 times two, all the way up to 12 times five. Notice we can also get those by repeatedly adding 12.

Start with 12, plus 12 is 24, plus 12 is 36, plus 12 is 48, plus 12 is 60. So very important, there are different ways to think about the multiples. So this suggests that we can just add r to get more multiples. For example, if we know P is a multiple of r, then it must be true that P- r and P + r are also multiples of r.

In other words, you could add or subtract r and get more multiples of r. So for example, let’s talk about this with numbers. Suppose you are told that 2401 is a multiple of 7. Now, you wouldn’t be expected to know that, but supposed somehow the test tells you that. Okay, so we know 2107 is a multiple of 7.

Well what that means is then other multiples of 7 must include, if we add 7, and I add 7 again, and I add 7 again. Or we could go back to 2401 and subtract 7 and subtract 7 again, and subtract 7 again. Given one multiple, we can create many more by adding and subtracting. Idea number four.

If P and Q are multiples of r, then we could also subtract multiples of r and get more multiples of r. So, for example, 700 clearly is a multiple of 7. That’s obvious. 49 is a multiple of 7. That’s clear.

Therefore, other multiples of 7 must include 700 + 49. And we can add 49 again or we can subtract 700- 49 and subtract 49 again, so all of these new numbers are multiples of 7, so once again, you can add or subtract any two multiples of 7 and get another multiple of 7. Along these lines, notice the following.

Since 52 is a multiple of 13, we can add two 52’s, or three 52’s, or any number of 52’s, and it would still be a multiple of 13. Of 13. We can add 150 52’s and it would still be a multiple of 13. In other words, any multiple of 52 is a multiple of 15. We can generalize this rule.

If P is a multiple of r, then any multiple of P is also a multiple of r. That’s a big idea. And if P and Q are both multiples of r, then the product P times Q must be a multiple of r. For example, 24 and 80 are both multiples of 8. To get more multiples of 8, we could add those two multiples.

We could subtract them. We could multiply them, so 104, 56, and 1920, these are all multiples of 8. But notice we can’t divide. If we divide 80 by 24, the quotient isn’t even an integer. So we can’t do that, but we can always add, subtract, or multiply two multiples of a number to get more multiples.

Here’s a practice question. Pause the video and then we’ll talk about this. Okay, so let’s think about this. So we have a bunch of multiples of P, so P is a factor of all of these numbers. And we want to know, p could equal which of the following? Well, let’s think about this.

First of all, I’m gonna look at that last piece of information. We know that K is a multiple of P, and then they’re also telling us that 15 times K is a multiple of P. Turns out that’s redundant, because if K is a multiple of P, then 15 times K has to be a multiple of P. Any multiple of K is gonna be a multiple of P.

So that particular piece of information is completely useless and doesn’t help us at all, so we can just ignore that piece of information. But these others with the sums, these are interesting. K and K + 200 and K + 350 are all multiples of P. This means differences among these will also be multiples of P, so, for example, if I subtract (K + 200)- K, I get 200.

200 must be a multiple of P. I could also subtract (K + 350)- K, and that would give me 350. 350 must be a multiple of P. And then since those two are multiples of P, I could subtract them. 350- 200 is 150. That’s a multiple of P.

Well, if 150 is a multiple of P, and 200 is a multiple of P, I could also subtract them. So, 200 minus 150. That gives me 50. That’s a multiple of P. So, in other words, P is a factor of all four of these numbers, 50, 150, 200, and 350.

So let’s take that information and now look at our answer choices. Here are the answer choices and P has to be a factor of one of these numbers. Well, notice that it can’t be anything bigger than 50, because anything bigger than 50 can’t be a factor of 50. It cant be 20 because, well, 200 is divisible by 20, but the others aren’t divisible by 20.

But notice that all four of those number are divisible by 25. 25 is the only number on the list that divides evenly into 50, 150, 200, and 350, so it’s the only one that is clearly a factor of all four of them. Therefore, that has to be the answer. In summary, we discussed multiples and their relationship to factors and divisors.

If P is a multiple of r, we can add or subtract r any number of times and get more multiples of r. If P and Q are multiples of r, then the sum, difference, or product will be other multiples of r, and any multiple of P or Q would be yet another multiple of r.

### مشارکت کنندگان در این صفحه

تا کنون فردی در بازسازی این صفحه مشارکت نداشته است.

🖊 شما نیز می‌توانید برای مشارکت در ترجمه‌ی این صفحه یا اصلاح متن انگلیسی، به این لینک مراجعه بفرمایید.