Divisibility by 4 with the last two digits

Is $7316$ divisible by $4$?

Split off the last two digits: $7316 = 73 \times 100 + 16$ = Last two digits: $16$

Remember why the hundreds do not matter: $100 = 4 \times 25$, so $7300$ is a multiple of $4$ = Only $16$ decides

Test the last two digits: $16 = 4 \times 4$ = $16$ is divisible by $4$

Conclude and check: $7316 \div 4 = 1829$ = Divisible, remainder $0$

Answer: Yes. The last two digits form 16, which is divisible by 4, so 7316 is divisible by 4 (7316 ÷ 4 = 1829).

The tens digit shortcut

Which of $3572$, $6134$ and $9048$ are divisible by $4$?

Check $3572$: tens digit $7$ is odd: Odd tens digit: the last digit must be $2$ or $6$. It is $2$ = $72 = 4 \times 18$, divisible

Check $6134$: tens digit $3$ is odd: Odd tens digit: the last digit must be $2$ or $6$. It is $4$ = $34 = 4 \times 8 + 2$, not divisible

Check $9048$: tens digit $4$ is even: Even tens digit: the last digit must be $0$, $4$ or $8$. It is $8$ = $48 = 4 \times 12$, divisible

Answer: 3572 and 9048 are divisible by 4. 6134 is not: it leaves remainder 2.

Divisibility by 8 with the last three digits

Is $45{,}136$ divisible by $8$?

Split off the last three digits: $45{,}136 = 45 \times 1000 + 136$, and $1000 = 8 \times 125$ = Only $136$ decides

Divide $136$ by $8$ by halving three times: $136 \to 68 \to 34 \to 17$ = Three whole halvings: $136 = 8 \times 17$

Conclude and check: $45{,}136 \div 8 = 5642$ = Divisible, remainder $0$

Answer: Yes. The last three digits form 136 = 8 × 17, so 45,136 is divisible by 8 (45,136 ÷ 8 = 5642).

Divisibility by 6: two tests, both must pass

Which of $5742$ and $3754$ is divisible by $6$?

Test $5742$ for $2$: Last digit $2$ is even = Divisible by $2$

Test $5742$ for $3$: $5 + 7 + 4 + 2 = 18$, and $18 = 3 \times 6$ = Divisible by $3$, so by $6$: $5742 \div 6 = 957$

Test $3754$ for $2$: Last digit $4$ is even = Divisible by $2$

Test $3754$ for $3$: $3 + 7 + 5 + 4 = 19$, and $19 = 3 \times 6 + 1$ = Not divisible by $3$, so not by $6$

Answer: 5742 is divisible by 6. 3754 is even but fails the test for 3, so it is not divisible by 6.

Leap years

A year is a leap year if it is divisible by $4$, except century years (ending in $00$), which must be divisible by $400$. Which of $2024$, $1900$, $2000$ and $2100$ are leap years?

Test $2024$: Not a century year. Last two digits $24 = 4 \times 6$ = Leap year

Test $1900$: Century year: $1900 \div 400 = 4.75$ = Not a leap year

Test $2000$: Century year: $2000 \div 400 = 5$ = Leap year

Test $2100$: Century year: $2100 \div 400 = 5.25$ = Not a leap year

Answer: 2024 and 2000 are leap years. 1900 and 2100 are divisible by 4 but are century years not divisible by 400, so they are not leap years.

Mistake: Checking only the last digit for $4$: "$2134$ ends in $4$, so it is divisible by $4$"

Why: The rule for $4$ reads the last **two** digits, because $10$ is not a multiple of $4$. Here $34 = 4 \times 8 + 2$.

Correct: Test the last two digits: $34$ leaves remainder $2$, so $2134$ is not divisible by $4$.

Mistake: Checking only the last two digits for $8$: "$3124$ ends in $24 = 8 \times 3$, so it is divisible by $8$"

Why: $100$ is not a multiple of $8$ ($100 = 8 \times 12 + 4$), so the hundreds digit still matters. The rule for $8$ needs the last **three** digits.

Correct: Test $124$: $124 = 8 \times 15 + 4$. So $3124$ is not divisible by $8$ (remainder $4$).

Mistake: Thinking that divisible by $2$ and by $4$ means divisible by $8$

Why: $12$ is divisible by $2$ and by $4$, but $12 \div 8 = 1.5$. Combining rules only works when the two numbers are coprime, like $2$ and $3$ for $6$. $2$ and $4$ share the factor $2$.

Correct: For $8$ use its own rule (last three digits). For $6$ you may combine $2$ and $3$.

Mistake: Testing only one part of the rule for $6$: "$27$ is divisible by $3$, so it is divisible by $6$"

Why: $6$ needs **both** tests. $27$ is odd, so it fails the test for $2$. Likewise $26$ is even but its digit sum $8$ fails the test for $3$.

Correct: A number is divisible by $6$ only if it is even **and** its digit sum is divisible by $3$, for example $5742$.

Leap years in a century

The calendar adds a day (29 February) in every year divisible by 4, but skips century years that are not divisible by 400.

$2024$ is a leap year ($24 = 4 \times 6$), $2000$ was one ($2000 = 400 \times 5$), but $1900$ was not ($1900 \div 400 = 4.75$).

Packing in sixes

Bakeries, egg farms and drink makers pack in 6s. The rules for 2 and 3 tell them before counting whether nothing will be left over.

$4320$ eggs fill boxes of $6$ exactly: $4320$ is even and $4 + 3 + 2 + 0 = 9$ is divisible by $3$, so $4320 \div 6 = 720$ boxes.

Crates of eight

Bottles and cans often travel in crates of 8. Only the last three digits of a large order decide whether the crates come out even.

$1560$ pencils fit in boxes of $8$: the last three digits $560 = 8 \times 70$, so $1560 \div 8 = 195$ boxes.

Q: Why do the last two digits decide divisibility by 4?

A: Any whole number is a number of hundreds plus its last two digits: $7316 = 73 \times 100 + 16$. Since $100 = 4 \times 25$, every whole number of hundreds is divisible by $4$. The remainder of the whole number divided by $4$ is therefore the same as the remainder of the last two digits. The same idea with $1000 = 8 \times 125$ gives the rule for $8$.

Q: If a number is divisible by 4, is it also divisible by 8?

A: Not always. $12$, $20$ and $28$ are divisible by $4$ but not by $8$. The chain only goes downward: every multiple of $8$ is a multiple of $4$ (because $8 = 4 \times 2$), and every multiple of $4$ is even.

Q: Can I build other rules the way 6 = 2 and 3 works?

A: Yes, but only with coprime numbers (no common factor other than $1$). $6 = 2 \times 3$ works because $2$ and $3$ are coprime. $8 = 2 \times 4$ does not work, because $2$ and $4$ share the factor $2$: $12$ passes both tests and is still not divisible by $8$. The lesson Combined Divisibility Rules shows this for $12$, $15$, $18$ and $36$.

Q: Why must century years be divisible by 400 to be leap years?

A: A year lasts about $365.2422$ days. Adding a leap day every $4$ years gives an average of $365.25$ days, a bit too long. Skipping $3$ leap days every $400$ years ($1700$, $1800$, $1900$, but not $2000$) leaves $97$ leap years in $400$ years: $365 + \frac{97}{400} = 365.2425$ days, very close to the real value.