L2 · Divisibility by 3 and 9

Learning goal

“I can test divisibility by 3 and by 9 using the sum of the digits - and I can explain, with tens and hundreds, why the digit sum works.”

Digit sum — the sum of all the digits of a number. The digit sum of 4,215 is 4 + 2 + 1 + 5 = 12.

Multiple — a number in a counting-by sequence: 12 is a multiple of 3 because 3 x 4 = 12.

Remainder — what is left over after dividing as evenly as possible.

Divisible — divides exactly, remainder 0 (from L1).

Worked example

Is 4,215 divisible by 3? By 9? Step 1: Add the digits: 4 + 2 + 1 + 5 = 12. Step 2: Is 12 a multiple of 3? Yes (3 x 4 = 12). So 4,215 is divisible by 3. Step 3: Is 12 a multiple of 9? No. So 4,215 is NOT divisible by 9. Step 4 (sanity check): 4,215 / 3 = 1,405 exactly; 4,215 / 9 = 468.33... - matches. Say it plainly: for 3 and 9 you do not look at the last digit - you add ALL the digits and test the sum. If the sum is still big, add again: the digit sum of 999,999 is 54, and 5 + 4 = 9, a multiple of both 3 and 9.

Check yourself — explain it

To test 7,632 for divisibility by 3, the first move is...

Paired problem — example still visible

Is 5,106 divisible by 3? By 9? -> Digit sum 5 + 1 + 0 + 6 = 12 -> divisible by 3, not by 9.

Faded completion — last step is yours

"8,214: digit sum = 8 + 2 + 1 + 4 = ___, so it is divisible by ___" (one of 3/9) -> 15; 3 only (15 is a multiple of 3 but not of 9)

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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