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Lesson · Grade 6 · Mathematics

Divisibility Rules for 2, 3, 4, 5, 6, 9 and 10

  • Grade 6
  • Mathematics
  • 1 min 37 s
  • British Columbia
  • BC6-MA-08

One number, 4170, is tested against all seven divisibility rules. The last digit settles 2, 5 and 10, the last two digits (70 ÷ 4 = 17 remainder 2) settle 4, and the digit sum (4 + 1 + 7 + 0 = 12) settles 3 and 9. Passing both 2 and 3 settles 6, which 4170 ÷ 6 = 695 confirms, and the video ends on the answer (divisible by 2, 3, 5, 6 and 10, but not 4 or 9) and a one-line recap of which digits each rule checks.

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In one sentence

The question this lesson answers

Is 4170 divisible by 2, 3, 4, 5, 6, 9 or 10?

Divisible means it divides with no remainder.

Transcript

What the lesson says

Is 4170 divisible by 2, 3, 4, 5, 6, 9 or 10?

Divisible means it divides with no remainder.

For 2, the last digit must be even, and 0 is even, so yes.

For 5, the last digit must be 0 or 5, and it is 0, so yes.

For 10, the last digit must be 0, so yes again.

For 4, look at the last two digits, which make 70.

70 divided by 4 is 17, remainder 2, so no.

For 3 and 9, add the digits: 4 plus 1 plus 7 plus 0 is 12.

12 divided by 3 is 4, with no remainder, so yes.

12 divided by 9 is 1, remainder 3, so no.

For 6, it must be divisible by both 2 and 3, and it is.

Check: 4170 divided by 6 is 695, with no remainder.

So 4170 is divisible by 2, 3, 5, 6 and 10, but not by 4 or 9.

Use the last digits for 2, 4, 5 and 10, the digit sum for 3 and 9, and both 2 and 3 for 6.

Curriculum

Where this fits

British Columbia, Grade 6

Mathematics, Number: factors and multiples: greatest common factor and least common multipleBC6-MA-08

Previous: Prime and Composite Numbers: Count the FactorsNext: Prime factorization with a factor tree

The outcomes and strands follow the British Columbia curriculum.

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