AFCAT Numerical Ability · Number System

Remainders and Quotients: Division, Remainder Arithmetic and Powers

Every division is dividend = divisor × quotient + remainder; remainders add and multiply like the numbers themselves, and the remainders of powers repeat in a cycle.

All Number System notes

Why this matters

Half the page is plain division: a quotient, a remainder or the number itself. The HARD questions ask for the remainder of a power far too large to work out, and the way in is to reduce the base first and then find where the remainders start to repeat.

Concept 1 of 3: Dividend = divisor × quotient + remainder

Dividing

N

d

takes away as many whole

d

s as fit, the quotient, and leaves a remainder smaller than

d

. The same equation runs backwards: from the divisor, quotient and remainder you rebuild the number.

Definition

$N = dq + r$ with $0 \le r < d$ .
To find $q$ , estimate with a round multiple and adjust: $27 \times 40 = 1080$ , $27 \times 45 = 1215$ .
The remainder is always smaller than the divisor, so an option with $r \ge d$ is wrong whatever its quotient.
A remainder of 0 means $d$ divides $N$ exactly.

Division algorithm

N = d \times q + r,\qquad 0 \le r < d

Worked example

Find the quotient and remainder when 1,234 is divided by 27.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Number System · Remainders and Quotients

When

749

is divided by

25

, what are the quotient and remainder?

The remainder is smaller than the divisor

Dividing 500 by 16, the answer 30 remainder 20 is wrong because 20 is more than 16: one more 16 fits, giving 31 remainder 4.

Multiply back to check

d \times q + r

must give the dividend exactly.

16 \times 31 + 4 = 500

confirms the division above.

Drill 13 more on dividend = divisor × quotient + remainder

Concept 2 of 3: Remainders of sums, multiples and products

Only the remainder matters. If

N

leaves 6 on division by 11, then

N

is a multiple of 11 plus 6, and the multiple of 11 drops out of anything you add or multiply. So work with the remainders and reduce at the end.

Definition

Notation: $N \equiv r \pmod d$ is short for ' $N$ leaves remainder $r$ on division by $d$ '.
Sums: the remainder of $a + b$ is the remainder of $r_a + r_b$ . Products: the remainder of $a \times b$ is the remainder of $r_a \times r_b$ .
So $kN$ leaves the remainder of $k \times r$ , and $N + c$ the remainder of $r + c$ .
A product with a multiple of $d$ in it leaves 0; adding 1 to it leaves 1.
On division by 9 (or 3), a number leaves the same remainder as its digit sum, the fact behind the rule on Divisibility Rules.

Working with remainders

a \equiv r_a,\ b \equiv r_b \pmod d \;\Rightarrow\; a + b \equiv r_a + r_b,\quad ab \equiv r_a r_b \pmod d

Worked example

N

leaves remainder 6 on division by 11. Find the remainder when

4N + 7

is divided by 11.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Number System · Remainders and Quotients

When

N

is divided by

6

, remainder is

5

. What is the remainder when

3N

is divided by

6

Reduce the answer

N \equiv 7 \pmod 9

, then

2N

gives

14

, which is not a remainder on 9. Reduce it:

14 - 9 = 5

The digit-sum shortcut is for 3 and 9 only

123 has digit sum 6, and it does leave 6 on division by 9. On division by 7 it leaves 4, not 6.

Drill 5 more on remainders of sums, multiples and products

Concept 3 of 3: Remainders of large powers

Reduce the base first, then list its powers' remainders until one of them is 1. From there the list repeats, so a huge exponent only matters through its remainder on the length of the cycle. A base that leaves 1 or one less than the divisor needs no cycle at all.

Definition

Reduce the base: $38^{n}$ on 37 is $1^{n} = 1$ , because $38 \equiv 1 \pmod{37}$ .
A base one less than the divisor is $-1$ : its even powers leave 1 and its odd powers leave $d - 1$ .
Otherwise list $a, a^2, a^3, \ldots$ reduced until you reach 1. If $a^k \equiv 1$ , split the exponent as $kq + s$ ; the answer is the remainder of $a^s$ .
Fermat: if $p$ is prime and does not divide $a$ , then $a^{p - 1} \equiv 1 \pmod p$ . So the cycle length divides $p - 1$ .

Splitting the exponent

a^k \equiv 1 \pmod d \;\Rightarrow\; a^{kq + s} \equiv a^s \pmod d

Worked example

Find the remainder when

3^{41}

is divided by 7.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Number System · Remainders and Quotients

What is the remainder when

2^{50}

is divided by

7

Minus 1 means one less than the divisor

12^{25} \equiv -1 \pmod{13}

, and the remainder to choose is 12. A negative number is never the answer.

Use the full cycle length

The powers of 3 on 7 repeat every 6, not every 3. Splitting 41 by 3 gives

3^2

and the wrong remainder 2; by 6 it gives

3^5

and 5.

Drill 6 more on remainders of large powers

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Dividend = divisor × quotient + remainder
Division algorithm
$N = d \times q + r,\qquad 0 \le r < d$
Remainders of sums, multiples and products
Working with remainders
$a \equiv r_a,\ b \equiv r_b \pmod d \;\Rightarrow\; a + b \equiv r_a + r_b,\quad ab \equiv r_a r_b \pmod d$
Remainders of large powers
Splitting the exponent
$a^k \equiv 1 \pmod d \;\Rightarrow\; a^{kq + s} \equiv a^s \pmod d$

Watch out for (6)

The remainder is smaller than the divisor
→ Dividend = divisor × quotient + remainder
Multiply back to check
→ Dividend = divisor × quotient + remainder
Reduce the answer
→ Remainders of sums, multiples and products
The digit-sum shortcut is for 3 and 9 only
→ Remainders of sums, multiples and products
Minus 1 means one less than the divisor
→ Remainders of large powers
Use the full cycle length
→ Remainders of large powers

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