Question bank — Division — long division, remainder, dividend - divisor - quotient vocabulary

A remainder of $0$ means the divisor divides the dividend exactly. ::: True — $0$ leftover means the dividend is a whole multiple of the divisor, i.e. $a=d\times q+0=dq$. In $17÷5=3$ rem $2$, the numbers $17$ and $5$ can be swapped without changing the meaning. ::: False — $17÷5$ shares $17$ into groups of $5$; $5÷17$ shares $5$ into groups of $17$ (giving $0$ rem $5$). Dividend $a$ and divisor $d$ play different roles. The remainder can be as large as the divisor. ::: False — it must be strictly less: $0\le r<d$. If $r=d$ you have one more whole group, which belongs in the quotient. If you divide any number by $1$, the remainder is always $0$. ::: True — every group of size $1$ fits perfectly, so $a=1\times a+0$; there is never anything left over. Dividing a smaller number by a bigger number is impossible. ::: False — it just gives quotient $0$ and the whole number as remainder: $3÷7=0$ rem $3$, since $7\times 0+3=3$. The quotient of $a÷d$ is always smaller than $a$ (for whole $a$ and divisor $d>1$). ::: True — dividing into groups larger than $1$ can only reduce the count, so $q<a$; equality would need $d=1$. Doubling both the dividend and the divisor keeps the quotient the same and doubles the remainder. ::: True — start from $a=dq+r$. Multiply the whole line by $2$: $2a=(2d)q+2r$. So dividing $2a$ by $2d$ gives the same $q$ with new remainder $2r$. It stays valid because $r<d$ means $2r<2d$, so $2r$ is still a legal remainder. Example: $17÷5=3$ rem $2$ becomes $34÷10=3$ rem $4$. A number divides itself exactly. ::: True — $a÷a=1$ rem $0$, since exactly one group of size $a$ fits into $a$.

A student writes $23÷4=4$ rem $7$. What broke the rule? ::: The remainder $7$ is larger than the divisor $4$, so another group fits. Correct it to $5$ rem $3$ (since $4\times 5+3=23$). A student computes $6048÷6$ and writes quotient $18$. What step was skipped, and where does the $0$ go? ::: They forgot a placeholder $0$. Column by column: $6÷6=1$ (write $1$, rem $0$); bring down $0\to 0$, and $0÷6=0$, so write $0$ in the quotient (this is the skipped digit); bring down $4\to 4$, $4÷6=0$, write another $0$; bring down $8\to 48$, $48÷6=8$. Quotient $=1008=1008$, not $18$. The two middle $0$s hold the hundreds and tens places. A student checks $728÷6=121$ rem $2$ by computing $121÷6$ and comparing. Is that a valid check? ::: No — the check is $\text{divisor}\times \text{quotient}+\text{remainder}=6\times 121+2=728$, matching the dividend. Re-dividing the quotient tests nothing. A student says "$0÷5=5$." Why is that nonsense? ::: Zero groups of anything is zero: $0÷5=0$ rem $0$, since $5\times 0+0=0$. Sharing no sweets gives each kid nothing. A student says "$5÷0=0$." What's the real problem? ::: Division by $0$ is undefined — no quotient $q$ makes $0\times q=5$, because $0\times q$ is always $0$. The equation has no solution, which is why the domain forbids $d=0$. A student concludes $945÷7$ has a remainder because $945$ is odd. Is that reasoning sound? ::: No — oddness only tells you about dividing by $2$. In fact $945=7\times 135$ exactly, remainder $0$. Parity says nothing about divisibility by $7$.

Why must the remainder be strictly less than the divisor? ::: If it equalled or exceeded the divisor, another full group could be removed, so the quotient wasn't the true count of whole groups. See Modular Arithmetic & Remainders. Why does long division start from the biggest place value, not the smallest? ::: A leftover in a big place (say a leftover hundred) is worth ten of the next place down, so it must be carried downward. Starting big lets each leftover flow into the next column cleanly. Why does "bring down the next digit" actually work? ::: A leftover in one column is really that many tens of the next column; bringing the next digit down glues the leftover-tens to the new units, forming the exact number to divide next. It's pure place value. Why is division called the inverse of multiplication? ::: Because $a=d\times q+r$ rebuilds the dividend by multiplying the divisor and quotient back up. With remainder $0$, division perfectly undoes multiplication — see Multiplication — repeated addition. Why can the digit-sum test check divisibility by $3$ but not by $7$? ::: Because $10\equiv 1(\mathrm{mod}3)$ (i.e. $10$ leaves remainder $1$ mod $3$), each place value contributes just its digit's remainder; for $7$, powers of $10$ leave a messy repeating pattern of remainders, so no clean digit-sum shortcut exists. Why does a remainder of $0$ let us write the division as a whole-number fraction? ::: Because $a=dq$ means $a/d=q$ exactly, a whole number — see Fractions as Division. A non-zero remainder leaves a genuine fractional part.

What is the quotient and remainder of $0÷8$? ::: Quotient $0$, remainder $0$: no whole to share means no groups and nothing left, and $8\times 0+0=0$. What happens to the remainder when the dividend is a multiple of the divisor? ::: The remainder is exactly $0$ — the whole amount splits into complete groups with nothing left, the definition of "divides exactly." Divide $7$ by $7$: what are quotient and remainder, and why is this the boundary? ::: Quotient $1$, remainder $0$. It's the smallest dividend for which divisor $7$ makes even one whole group. Divide $6$ by $7$: how does the algorithm handle a divisor larger than the dividend? ::: Quotient $0$, remainder $6$ — zero full groups fit, so the entire dividend is left over, and $7\times 0+6=6$. What is the largest possible remainder when dividing by $9$? ::: $8$ — the remainder can run from $0$ up to $\text{divisor}-1=8$, but never reach $9$ itself. If a number leaves remainder $0$ when divided by both $4$ and $6$, must it leave remainder $0$ when divided by $24$? ::: Not necessarily — it need only be a multiple of the LCM, which is $12$ (e.g. $12$ works but isn't a multiple of $24$). See HCF and LCM. Can the quotient ever be larger than the dividend (for whole $a$, divisor $d\ge 1$)? ::: No — rearrange the identity to $q=\frac{a-r}{d}$. Since $r\ge 0$ we have $a-r\le a$, and since $d\ge 1$ dividing by $d$ can only shrink or hold: $q=\frac{a-r}{d}\le \frac{a}{1}=a$. So $q\le a$, with equality only when $r=0$ and $d=1$. (A divisor below $1$ — a fraction — is what would let $q$ exceed $a$, but that leaves the whole-number domain.) How is $-17÷5$ handled, and why isn't the remainder $-2$? ::: The identity keeps $0\le r<d$, so we step past $-17$ to the next lower multiple of $5$, which is $-20$: $-17=5\times (-4)+3$, giving quotient $-4$, remainder $3$. A "$-2$" would break the $0\le r<d$ rule. When the divisor is negative, e.g. $17÷(-5)$, what keeps the remainder valid? ::: The remainder must satisfy $0\le r<|d|$, using the size of the divisor. Here $17=(-5)\times (-3)+2$: quotient $-3$, remainder $2$ — still in $0\le r<5$. And the fourth sign case: $-17÷(-5)$ — both negative — what are quotient and remainder? ::: Keep $0\le r<|d|=5$. Step down to the multiple of $-5$ at or below $-17$, namely $-20=(-5)\times 4$: $-17=(-5)\times 4+3$, so quotient $4$ (positive, since two negatives multiply up) and remainder $3$. This completes all four sign quadrants: (+,+), (−,+), (+,−), (−,−).