Divisibility Rules

Mike McGarry
Lesson by Mike McGarry
Magoosh Expert
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Divisibility rules. First of all let's start with a couple of easy questions. Is 56 divisible by 7? Is 50 divisible by 13? Theoretically those are questions you should be able to answer with little problem.

The answer to the first of course is yes. 56 equals 7 times 8. The answer to the second of course is no. 13 does go into 52, but it does not go into 50. So first of all, I'll just mention if these are not questions that you can answer just by looking at them, probably you need to practice your times table a little bit more.

You really need to have your basic times tables down, so that questions like this are in fact very easy. Now a slightly harder question, is that long number divisible by three? Well no one expects you to do this division in your head. We can't do the full division and find the quotient, but we can use a divisibility rule to answer the question.

And the test loves these divisibility rules. So first of all, divisibility by two. Of course, all even numbers are divisible by two. To tell whether a large number is even, all we have to do is look at the last digit. If the ones digit is even, then the number is even.

So we have these large numbers. We can ignore the rest of the number and just look at the ones place. The ones place we see 5, 1, 7, 3. Those are all odd, but the 6 is even. That means that the middle one, with the one digit as six, that's the only even number.

The only number divisible by two. So that's divisibility by two. Divisibility by five. This is another divisibility rule that involves looking at the last digit. If the last digit is a five or a zero, then the number is divisible by five, otherwise it isn't.

So, again we have numbers like this. We can ignore the rest of the number, and just look at that last digit. Indeed, we have a last digit of five, so that number is divisible by five. But the others do not end in five or zero. So they are not divisible by 5. So both of those rules, divisibility by 5.

And divisibility by two, those just involve looking at the 1's place and nothing else. Another rule for 4, this rule is similar, here we look at the last two digits, the ten's place and the one's place. We have to look at two digits. If the last two digits form a two-digit number divisible by four, then the entire number is divisible by four.

So again our same list of long numbers, look at those last two digits, and think of them as two-digit numbers. 55, 41, 96, 37, 33. The only one among those that is divisible by 4 is 96. 96 is a number divisible by 4, so that means that that whole middle number is divisible by 4.

Now divisibility for 3, the test loves this rule. This rule is a little different. Here we add up all the digits of the number, if the sum of the digits is divisible by 3, then the number is divisible by 3. And if the sum of the digits is not divisible by 3, the number is not divisible by three.

So, for example, with 135, we add 1 + 3 + 5, that's 9. Since 9 is divisible by 3, 135 must be divisible by 3. With 734, We add 7 + 3 + 4. That equals 14. Since 14 is not divisible by three, we know that 734 is not divisible by three.

1296 we add1+2+9+6=18, and since 18 is divisible by 3, then 1296 must be divisible by 3. So it works the same way for large numbers. Here's the question we had the very beginning of the module. So is that large number divisible by 3? Well here's what I notice, I notice that that middle 3+3+4=10, I can take the 5+5=10, and that only leaves a one, a zero, a two, and another one and those add up to four.

So that means that everything adds up to 24. The sum of the digits is 24, which is divisible by three, so the original number must be divisible by three. So all we have to do is add up the digits, and then see if that's divisible by three, and then that tells us whether the number overall is Is divisible by 3. The divisibility rule for 9.

This is exactly like the rule for 3. Add all the digits. If the sum of the digits is divisible by 9, then the number is divisible by 9. If the sum of the digits is not divisible by 9, then the number is not divisible by 9. So for example, 1296, we found in the last few slides that the sum of this was 18, the sum is divisible by 9.

So 1296 must be divisible as well. 1372, we add these digits up we get a sum of 12, so the sum of the digits is divisible by three, but not nine. So the number is divisible by three, but not nine. Notice incidentally 1372, the last two-digit number 72 is divisible by four. So that's a number that would be divisible by three and by four.

Which would mean it's divisible by 12. We can also use the divisibility rule for 9, for larger numbers. So is this large number divisible by 12? But we already found out that the sum of the digits is 24. So 24 is divisible by 3, but not by 9. So the original number, that 9 digit number, is divisible by 3, but not by 9.

The divisibility rule for 6. Here we're gonna have a combination. In order to be divisible by 6, a number must be, a, divisible by 2, and b, divisible by 3. We check the divisibility by 2, by looking at the last digit, making sure that it's even, and we check divisibility by 3, by finding the sum of the digits.

So any even number divisible by 3, has to be divisible by 6. Is 1296 divisible by 6? Well, first of all we know it's even, the sum of the digits we found already is 18, which is divisible by 3. Therefore 1296 is divisible by 6. Is this long number divisible by 6?

Well, clearly it's even, that's easy to determine. We add the digits, add the first three they add up to 15. The second three digits add up to 14. The last three digits add up to 17. 15 Plus 17 plus 14 is 46. The sum of the digits is not divisible by 3.

So the original number is not divisible by 3 or by 6. In this video we discussed the most common divisibility rules, the rules for 2, 5, 4, 3, 9 and 6.

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