Divisible by 12: Use 3 and 4

Is $7128$ divisible by $12$?

Split 12 into coprime factors: $12 = 3 \times 4$, and 3 and 4 have no common factor except 1 = Test 3 and 4

Test 3 with the digit sum: $7 + 1 + 2 + 8 = 18$, and $18$ is divisible by 3 = Divisible by 3

Test 4 with the last two digits: The last two digits form $28$, and $28 \div 4 = 7$ = Divisible by 4

Combine and check: Both tests pass, so 12 divides it: $7128 \div 12 = 594$ = Divisible by 12

Answer: Yes, $7128$ is divisible by $12$ ($7128 = 12 \times 594$).

The 2 and 6 Trap

$18$ is divisible by $2$ and by $6$. Is it divisible by $2 \times 6 = 12$?

Check 2 and 6: $18 \div 2 = 9$ and $18 \div 6 = 3$ = Both divide 18

Check 12 directly: $18 = 12 \times 1 + 6$ = Remainder 6, not divisible

Find the reason: $2 = 2$ and $6 = 2 \times 3$ share the factor 2, so the 2 is counted twice. Divisible by 2 and 6 only means divisible by $\text{lcm}(2, 6) = 6$ = 2 and 6 are not coprime

Answer: No. $18$ passes the tests for 2 and 6 but is not divisible by 12. Use the coprime pair 3 and 4 instead.

Divisible by 11: Alternating Sum

Is $82{,}918$ divisible by $11$?

Write the digits from the ones digit to the left: $8, 1, 9, 2, 8$ = Start with the ones digit

Add and subtract in turn: $8 - 1 + 9 - 2 + 8 = 22$ = Alternating sum: 22

Test the result: $22 = 2 \times 11$ = Divisible by 11

Check by dividing: $82{,}918 \div 11 = 7538$ = No remainder

Answer: Yes, $82{,}918$ is divisible by $11$ because its alternating sum $22$ is a multiple of 11.

Quick Factor Search

Which of the numbers $2$ to $12$ divide $1980$?

Last-digit tests: Ends in 0: divisible by 2, 5 and 10 = 2, 5, 10

Digit sum tests: $1 + 9 + 8 + 0 = 18$: divisible by 3 and by 9 = 3, 9

Last two and three digits: $80 \div 4 = 20$: divisible by 4. $980 \div 8 = 122.5$: not divisible by 8 = 4 yes, 8 no

Alternating sum for 11: $0 - 8 + 9 - 1 = 0$ = 11

Combine the coprime pairs: $6 = 2 \times 3$ and $12 = 3 \times 4$ both pass. For 7 there is no simple rule: $1980 = 7 \times 282 + 6$ = 6, 12; not 7

Answer: $1980$ is divisible by $2, 3, 4, 5, 6, 9, 10, 11$ and $12$, but not by $7$ or $8$.

Missing Digit Under Two Rules

Which digits $d$ make $38d4$ divisible by $12$?

Rule for 4: the last two digits $d4$: $04, 24, 44, 64, 84$ are multiples of 4 = $d$ = 0, 2, 4, 6 or 8

Rule for 3: the digit sum: $3 + 8 + d + 4 = 15 + d$ must be a multiple of 3 = $d$ = 0, 3, 6 or 9

Keep the digits that pass both: Common to both lists: 0 and 6 = $d = 0$ or $d = 6$

Check: $3804 \div 12 = 317$ and $3864 \div 12 = 322$ = Both work

Answer: $d = 0$ or $d = 6$: the numbers $3804$ and $3864$ are both divisible by $12$.

Mistake: Testing 12 with the rules for 2 and 6

Why: 2 and 6 share the factor 2, so passing both only proves divisibility by 6. Counterexample: $18$ (and also $30$) is divisible by 2 and 6 but not by 12.

Correct: Use coprime factors: $12 = 3 \times 4$. Check the digit sum for 3 and the last two digits for 4.

Mistake: Testing 36 with 3 and 12

Why: 3 and 12 share the factor 3. $24$ is divisible by 3 and by 12, but $24 \div 36$ is not a whole number.

Correct: Use $36 = 4 \times 9$: last two digits divisible by 4 and digit sum divisible by 9.

Mistake: Thinking one passed test is enough

Why: A combined rule needs BOTH tests. $25$ ends in 5 but its digit sum is 7, so it is not divisible by 15.

Correct: Check every factor of the pair. If one test fails, the number is not divisible by the product.

Mistake: Reading an alternating sum of $0$ or $-11$ as not divisible by 11

Why: $0$, $-11$ and $-22$ are multiples of 11 too. For $9174$: $4 - 7 + 1 - 9 = -11$, and indeed $9174 = 11 \times 834$.

Correct: The number is divisible by 11 whenever the alternating sum is $0$ or any multiple of 11, positive or negative.

Sharing Equally

Teachers, coaches and organizers often need to know whether items can be shared out with nothing left over.

A school has 1350 stickers for 15 classes. $15 = 3 \times 5$: the last digit is 0 (divisible by 5) and $1 + 3 + 5 + 0 = 9$ (divisible by 3). So the stickers can be shared equally.

Timetables

Transport planners want departures that repeat in the same pattern every day. A day has $24 \times 60 = 1440$ minutes.

A bus every 12 minutes fits a day exactly: $1 + 4 + 4 + 0 = 9$ (divisible by 3) and $40 \div 4 = 10$, so 1440 is divisible by 12, giving 120 departures.

The 12-Digit Check

Many calculators show only 8 to 10 digits, so they cannot divide a 12-digit number exactly. Divisibility rules work on numbers of any length.

For $123{,}456{,}789{,}012$ the digit sum is $48$ (divisible by 3) and the last two digits form $12$ (divisible by 4), so the number is divisible by 12.

Q: Why must the two factors be coprime?

A: A number divisible by $a$ and by $b$ is a common multiple of $a$ and $b$, so it is divisible by $\text{lcm}(a, b)$. When $a$ and $b$ are coprime, $\text{lcm}(a, b) = a \times b$. When they share a factor, the lcm is smaller: $\text{lcm}(2, 6) = 6$, not 12.

Q: Why does the alternating sum work for 11?

A: $10 = 11 - 1$, so 10 leaves the remainder $-1$ when divided by 11, $100 = 99 + 1$ leaves $+1$, $1000 = 1001 - 1$ leaves $-1$, and so on. So each digit counts with the sign $+, -, +, -, \dots$ from the ones digit, and the number has the same remainder as its alternating sum.

Q: Does it matter whether I start the alternating sum from the left or the right?

A: For the yes or no answer it does not: the two results are equal or opposite numbers, and $-22$ is a multiple of 11 exactly when $22$ is. Starting at the ones digit with $+$ also keeps the remainder of the number (add or subtract 11 until you land between 0 and 10), so it is the safer habit.

Q: Can I test for 24 with 4 and 6?

A: No, 4 and 6 share the factor 2. $12$ is divisible by 4 and 6 but not by 24. Use the coprime pair $24 = 3 \times 8$: digit sum divisible by 3 and last three digits divisible by 8.