8.3 Divisibility Rules

A number is divisible by 2 if it's last digit are.....

0, 2, 4, 6, or 8

does 6 go into 642?

1) 642 is even (last digit is so 642 is divisible by 2) 2) add all the digits of 642 (sum is 12), which is divisible by 3... so, 642, is divisible by 6

does 3 go into 21,496?

1) add up the digits 2) does 3 go into 22? so, 21,496 is not divisible by 3

Does 3 go into 1,620?

1) add up the digits (1 +6+2+0=9 2) does 3 go into 9? yes! so, 1620 is divisible by 3!

does 8 go into 68,011?

1) check to see if 8 goes into 192 2) if yes than the original number is divisible by 8 too...if not, then the original number is not divisible by 8

does 8 go into 53, 496, 192?

1) check to see if 8 goes into 192 2) if yes, than the original number is divisible by 8 too

does 4 go into 23,524?

1) does 4 go into 24? yes, so the original number, 23,524..

is 18 divisible by 3? (In other words does 3 go into 18? If so, how times does 2 go into 18?

3 goes into 18 six times !

Divisibility test for 5: this test states that....

A number is divisible by 5 if the last digit is 0 or 5.

more practice:

Is 689,531 divisible by 4?

Is 258,493,556 divisible by 4?

Is 73,232 divisible by 8?

does 2 go into 619? (is 619 divisible by 2?)

NO! the last digit is 9 and it's not an even number

Does 2 go into 16? (in other words is 16 divisible by 2)

Yes, since the last digit is 6. It's even

divisibility test

a method for checking or testing to see if a counting number is divisible by another counting number without actually carrying out the division

Divisibility by 10 Test: this test states that

a number is divisible by 10 if the last digit is 0 or 10

Divisibility test for 2 says that....

a number is divisible by 2 if it is even

divisibility test for 3: this test states that...

a number is divisible by 3 if the sum of its digits is divisible by 3

we can determine whether a number is divisible by 3 by

adding its digits and seeing if that number is divisible by 3

why does the divisibility test for 3 work

as with the divisibility test for 2, let's consider counting numbers as represented by base ten bundles a counting number is divisible by 3 exactly when that number of toothpicks can be divided into groups of three with none left over

Does 4 go into 572?

does 4 go into 16,421? ......NO! because 21 is not divisible by 4!

does 6 go into 629,492?

does 6 go into 462?

in general for all counting numbers....

if we combine the individual toothpicks with all the leftover toothpicks that result from dividing each bundle of 10, 100, 1000, etc into groups of 3, then the number of toothpicks we will have is given by the sum of the digits of the number. all the other toothpicks have already been put into groups of 3 with none left over, then the original number is divisible by 3. if we can't put left over toothpicks into groups of 3, with none left over, then original number is not divisible by 3. we can determine whether a number is divisible by 3 by adding its digits and seeing if that number is divisible by 3

is 972, 552 divisible by 5?

is 290 divisible by 10

is 378 divisible by 2?

is 590 divisible by 5?

is 729 divisible by 10?

is 83,497 divisible by 2?

Divisibility test for 6: two tests required. This test states that a number is divisible by 6 if ...

it is divisible by 2 and 3

express the divisibility test for 3 explanation algebraically

let's restrict our explanation to 4 digit counting numbers. such a number is of the form ABCD ABCD = A x 1000 + b x 100 + C x 10 + D = (A x 999 + b x 99 + c x 9) + ( a + b + c + d) = (a x 333 + b x 33) x 3 + (a + b + c + d)

Does 5 go into 126, 371? ( is 126,371 divisible by 5?)

no! the last digit is not a 0 or a 5

Does 6 go into 245,631?

no, it's not even so 2 doesn't go in.

ABCD = A x 1000 + b x 100 + C x 10 + D = (A x 999 + b x 99 + c x 9) + ( a + b + c + d) = (a x 333 + b x 33) x 3 + (a + b + c + d)

the first equation expresses abcd in its expanded form. after second equal sign, we have separated the leftover toothpicks that result from dividing the place value bundles into groups of 3 we have joined leftovers with individual toothpicks to make a + b + c + d toothpicks expression of third equal sign shows that the toothpicks other than these remaining a + b + c + d can be divided evenly into groups of 3, namely a x 333 + b x 33 + c x 3 groups of 3 thus original a+b+c+ d toothpicks can be divided into groups of 3 with none left over exactly the the remaining a + b + c + d toothpicks can be divided into groups of 3 with none left over

Divisibility test for 8: a number is divisible by 8 if ..

the last three digits are a number that's divisible by 8

divisibility test for 4: A number is divisible by 4 if...

the last two digits are a number that's divisible by 4

A number is divisible by 9 if....

the sum of the digits is divisible by 9

if the original number was 248

there will be 2 leftover toothpicks from the 2 bundles of 100, 4 leftover toothpicks from the 4 bundles of 10 (we could make another group of 3 , but don't to explain why we add digits of number to determine divisibility rule for 3) work with 4 digit, not 1 if we combine the leftover toothpicks with the 8 individual toothpicks, then in all, that is 2 + 4 + 8 toothpicks

consider dividing bundled toothpicks that represent a counting number into groups of 3

to divide these toothpicks into groups of 3, we could start by dividing each of the bundles of 10, 100, 1000, etc intro groups of 3. (1 left over after dividing each bundle into groups of 3) for each place from the 10s place up, each bundle of toothpicks can always be divided into groups of 3 with 1 left over collect leftover toothpicks from bundle and join them with individual toothpicks. how many toothpicks with that be?

Does 5 go into 35? (is 35 divisible by 5)

yep the last digit is 5. And we know that 35/5=7

Does 2 go into 25,014? (is 25,014 divisible by 2)

yep, since the last digit is 4 (it's even)

does 5 go into 39,460?

yes! since the last digit is a 0