CDS Mathematics · Number System

Division, Parity & Consecutive Integers

The four primitives the whole chapter rests on: the division algorithm, odd-even bookkeeping, the fact that an odd square always leaves remainder 1 on division by 8, and the divisibility that a run of consecutive integers hands you for free.

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Why this matters

Seventeen PYQs, four of them HARD — but the real weight of this page is indirect. The odd-square fact turns up in five questions spread across the chapter, the consecutive-integer rule silently solves a dozen more, and a quarter of every CDS Maths paper is written in one of three statement formats that you can learn to attack mechanically. Read this page as the toolkit you carry into the other eleven.

Concept 1 of 7

The division algorithm

Intuition

Dividing one whole number by another always produces exactly two things: a quotient and a remainder, and the remainder is always strictly smaller than the divisor. That last clause is the one that earns marks — a remainder on division by 8 can be 0 through 7 and nothing else, so an option offering 9 is dead on sight.

Definition

For integers $N$ and $d>0$ there are unique integers $q$ and $r$ with

$N = dq + r$ , and
$0 \le r < d$ .

Here $d$ is the divisor, $q$ the quotient and $r$ the remainder. Uniqueness matters: there is only one such pair, so any consistent story about a division pins the number down completely.

Division algorithm

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

Nthe number being divided (the dividend)
dthe divisor
qthe quotient
rthe remainder, strictly less than d

Worked example

A number leaves quotient 23 and remainder 5 when divided by 12. What is the number, and what remainder does it leave on division by 5?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Number SystemEASY

When

N

is divided by 17, the quotient is equal to 182. The difference between the quotient and the remainder is 175. What is the value of

N

[Q24 · CDS (I) 2019 — Elementary Mathematics · 2019]

A remainder can never equal or exceed the divisor

Every year an option list on a remainder question includes a value at least as big as the divisor — 9 as a remainder on division by 9, say. It is free elimination. Cross those options out before you compute anything, because

0 \le r < d

is part of the definition, not a rule of thumb.

Concept 2 of 7

Parity bookkeeping for sums and products

Intuition

Odd and even behave like a two-element arithmetic of their own, and it is almost always faster to track parity than to track the numbers. A sum is odd only when an odd number of its terms are odd; a product is even the moment one single factor is even.

Definition

Write every integer as even ( $2k$ ) or odd ( $2k+1$ ). Then parity alone decides the parity of any sum, difference or product, whatever the actual values.

The parity of a sum depends on how many odd terms there are, not on their size.
A product is odd only when every factor is odd.
Subtraction behaves exactly like addition for parity, since $-1$ and $+1$ are both odd.

Expression	Result	Why
odd + odd	even	$(2a+1)+(2b+1)=2(a+b+1)$
odd + even	odd	one unpaired unit is left over
even + even	even	both are multiples of 2
odd − odd	even	same as odd + odd for parity
odd × odd	odd	no factor of 2 anywhere
odd × even	even	one factor of 2 is enough
even × even	even	at least two factors of 2
n(n+1)	always even	consecutive integers, so one of them is even This row does the most work in the chapter. Any expression of the form $q^2+q$ is even without exception, which is what collapses several CDS parity questions to a single line.
2k ± any even	even	evens are closed under addition

Track parity, not values. Nine rows that settle most of what CDS asks about odd and even.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Number SystemMODERATE

for the items that follow : Each item contains a Question followed by two Statements. Answer each item using the following instructions :

Question : Is

p^2 + q^2 + q

odd, where p, q are positive integers ? Statement I :

2p + q

is odd. Statement II :

q - 2p

is odd.

[Q76 · CDS (I) 2023 — Elementary Mathematics · 2023]

An even product does not mean both factors are even

2 \times 3 = 6

is even although 3 is odd. One even factor is enough. The correct statement is the contrapositive: if a product is odd then every factor is odd — and that direction is the one CDS actually uses to force a prime to be 2.

A statement about parity often says nothing about the variable you want

In the data-sufficiency format, "

2p+q

is odd" tells you only that

q

is odd, because

2p

is even whatever

p

is. If the question turns on

p

, that statement is useless — and so is the one that looks different but carries the same single fact. Two statements saying the same thing are jointly insufficient too.

Drill 2 more on parity bookkeeping for sums and products

Concept 3 of 7

The square of an odd number leaves remainder 1 on division by 8

Intuition

Square any odd number — 9, 25, 49, 81, 121 — and subtract 1: you get 8, 24, 48, 80, 120, every one a multiple of 8. This is not a coincidence and it is worth memorising, because it is the single most reused fact in this chapter.

Definition

For every odd integer $m$ , $m^2 \equiv 1 \pmod 8$ ; equivalently $m^2$ has the form $8n+1$ . The one-line proof is the consecutive-integer rule in disguise: writing $m=2k+1$ ,

m^2 = 4k^2+4k+1 = 4k(k+1)+1,

and

k(k+1)

is a product of consecutive integers, hence even. So

4k(k+1)

is a multiple of 8.

Every even power of an odd number also leaves remainder 1, since $m^{2t}=(m^2)^t \equiv 1^t = 1$ .
The difference of the squares of two odd numbers is therefore a multiple of 8.

Odd square modulo 8

m \text{ odd} \;\Longrightarrow\; m^2 = 8n+1 \quad\text{i.e.}\quad m^2 \equiv 1 \pmod 8

Worked example

Show that the difference of the squares of any two odd integers is divisible by 8, and find the remainder when

37^2 - 15^2

is divided by 8.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Number SystemMODERATE

Let

P = N^2

where N is an odd integer. What is the remainder when P is divided by 8 ?

[Q31 · CDS (I) 2026 — Elementary Mathematics · 2026]

Remainder 1 modulo 8 is a stronger claim than remainder 1 modulo 4

An odd square is

\equiv 1

modulo 8, which of course also gives

\equiv 1

modulo 4 — but not the reverse. If a question offers both 4 and 8, the 8 statement is the sharp one and the one that decides borderline options, as in the

\frac{m^4+4m^2+11}{16}

question where you need the mod-8 fact twice over.

The rule is for ODD bases only

4^2=16 \equiv 0

and

6^2=36 \equiv 4 \pmod 8

, so nothing survives if the base is even. Before using the fact, confirm the base is odd — in a statement question that condition is usually stated once, at the top, and easy to skim past.

Drill 3 more on the square of an odd number leaves remainder 1 on division by 8

Concept 4 of 7

A product of consecutive integers is divisible by the factorial of how many there are

Intuition

Among any two consecutive integers one is even. Among any three, one is a multiple of 3 and at least one is even. The pattern keeps going, and it means an expression that factorises into a run of consecutive integers carries guaranteed divisors before you substitute anything.

Definition

The product of $k$ consecutive integers is divisible by $k!$ .

$k=2$ : $n(n+1)$ is divisible by $2$ .
$k=3$ : $n(n+1)(n+2)$ is divisible by $3! = 6$ .
$k=5$ : a run of five is divisible by $5! = 120$ .

The examination skill is spotting the run, because it is usually disguised as a polynomial: $n^3-n=(n-1)n(n+1)$ and $n^5-5n^3+4n=(n-2)(n-1)n(n+1)(n+2)$ .

Consecutive-run divisibility

k! \;\big|\; n(n+1)(n+2)\cdots(n+k-1)

Worked example

Show that

n^3 - n

is divisible by 6 for every integer

n

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Number SystemMODERATE

Consider the following statements : 1.

n^3 - n

is divisible by 6. 2.

n^5 - n

is divisible by 5. 3.

n^5 - 5n^3 + 4n

is divisible by 120. Which of the statements given above are correct ?

[Q37 · CDS (I) 2023 — Elementary Mathematics · 2023]

The rule gives a guarantee, not the largest divisor

A run of three is divisible by 6 — but a particular run may be divisible by much more, and a question asking for the largest number that always divides an expression needs the smallest case tested.

n=1

n=2

usually settles it in one line, which is exactly how the 48-versus-24 question above is decided.

Drill 2 more on a product of consecutive integers is divisible by the factorial of how many there are

Concept 5 of 7

Centring a run of consecutive integers on its middle term

Intuition

If three consecutive numbers are called

n-1, n, n+1

instead of

n, n+1, n+2

, the algebra gets dramatically shorter: the odd powers cancel in pairs and the sum of squares collapses to

3n^2+2

. Always put the middle term at the centre of your notation.

Definition

For three consecutive integers written symmetrically as $n-1,\, n,\, n+1$ :

their sum is $3n$ , so it is always a multiple of 3;
their sum of squares is $3n^2+2$ ;
their sum of cubes is $3n^3+6n = 3n(n^2+2)$ , always a multiple of 9 for integer $n$ .

For an even-length run the midpoint is a half-integer, so use $n$ and $n+1$ around the centre instead. The principle is the same: symmetry kills the cross terms.

Sum of squares of three consecutive integers

(n-1)^2 + n^2 + (n+1)^2 = 3n^2 + 2

Worked example

The sum of the squares of three consecutive natural numbers is 77. What is their sum?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Number SystemMODERATE

If the sum of the squares of three consecutive natural numbers is 110, then the sum of their cubes is

[Q12 · CDS (II) 2018 — Elementary Mathematics · 2018]

What is true for three consecutive integers is not true for four

Three consecutive integers always sum to a multiple of 3, and it is tempting to generalise. But four consecutive integers sum to

4n+6

, which is never a multiple of 4. The rule is that a run of odd length has a genuine middle term and so a clean multiple; an even-length run does not.

Drill 2 more on centring a run of consecutive integers on its middle term

Concept 6 of 7

Using the parity of a total to count the odd terms

Intuition

If several whole numbers add to an even total, the count of odd ones among them must itself be even. That single observation converts a question about four unknown integers into a question about three possible cases — and

(-1)^a

is just a way of reading off parity as a number.

Definition

Suppose $a_1+\cdots+a_k = S$ . Then the number of odd $a_i$ is even when $S$ is even, and odd when $S$ is odd.

$(-1)^a = +1$ when $a$ is even and $-1$ when $a$ is odd, so a sum of such terms counts parities directly.
With $k$ terms, if $j$ of them are odd then $\sum (-1)^{a_i} = (k-j) - j = k-2j$ .
Having found the possible $j$ , check each one is actually attainable under the question's constraints before counting it.

Sum of parity signs

\sum_{i=1}^{k} (-1)^{a_i} = k - 2j \quad (j = \text{how many } a_i \text{ are odd})

Worked example

Three positive integers

a, b, c

satisfy

a+b+c = 50

. How many values can

T = (-1)^a + (-1)^b + (-1)^c

take?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6Number SystemHARD

Let a, b, c and d be four positive integers such that

a + b + c + d = 200

. If

S = (-1)^a + (-1)^b + (-1)^c + (-1)^d

, then what is the number of possible values of S?

[Q31 · CDS (I) 2023 — Elementary Mathematics · 2023]

Count the values that are ATTAINABLE, not the cases that are arithmetically allowed

Parity narrows

j

to a short list, but each surviving case still has to be realisable under the question's own constraints — usually "positive integers". Write one explicit example per case, as above. A case you cannot exhibit does not count, and a case you forgot to exhibit is the commonest way this question is marked wrong.

Drill 1 more on using the parity of a total to count the odd terms

Concept 7 of 7

The three statement formats CDS uses, and how to attack each

Intuition

Fifty-five of this chapter's 218 questions — a full quarter — are not plain "compute this" questions. They are one of three fixed formats, and each has a mechanical attack. Recognising the format before reading the mathematics is worth real time in a 1.2-minute-per-question paper.

Definition

The formats and their attacks:

Consider the following statements (numbered 1, 2, 3): judge each statement independently, then match your verdict pattern to the option codes. One counterexample kills a statement outright.
Data sufficiency (Statement-I, Statement-II): do not solve the problem. Ask only whether each statement pins the answer down uniquely, first alone and then together.
Which one of the following is correct / is not correct: this is elimination, not derivation. Look for the option that overreaches with words like always, every, or must.

Format	What it really asks	The attack
Consider the following statements 1, 2, 3	Is each statement true, separately?	Test each on its own; hunt one counterexample per statement
Statement-I / Statement-II (data sufficiency)	Is the answer UNIQUE, not what the answer is	Check I alone, then II alone, then both; stop at uniqueness The commonest error is solving the problem instead of testing sufficiency. If a statement leaves two possible values, it is insufficient even when both are easy to find.
Which one is correct	Three options are false	Eliminate by counterexample rather than proving the survivor
Which one is NOT correct	Three options are true	Read the word NOT twice; the wrong answer is usually the true statement you liked CDS sets both polarities and prints them in the same typeface. Underline the word NOT before you start.
Two statements that say the same thing	Whether either adds anything	If both carry one fact, together they are still insufficient
Option offering none of the above	Whether your value is really absent	Recompute once; this option is occasionally the intended answer

A quarter of this chapter arrives in one of these shapes. Name the format first, then do the mathematics.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 7Number SystemMODERATE

Consider the following for the items that follow : Each item contains a Question followed by two Statements. Answer each item using the following instructions :

Let a, b, c and d be positive integers. Question : Which one of a, b, c, d is closest to the product abcd ? Statement-I : a > b > c Statement-II : c is not the smallest.

[Q61 · CDS (I) 2024 — Elementary Mathematics · 2024]

In data sufficiency, insufficient plus insufficient is not always sufficient

The option codes tempt you into assuming that if neither statement works alone, both together must. They need not: the self-check above leaves two candidates even with both statements in hand, and CDS prints exactly that case. Always run the combined test explicitly instead of inferring it.

A statement question is not an all-or-nothing question

When three statements are offered, the option list usually includes several partial combinations such as "1 and 3 only". Students who decide the question feels true and pick "all of them" lose marks to a single planted counterexample — most often a statement that holds for every case except

n=0

, or except the prime 2.

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 (5)

The division algorithm
Division algorithm
$N = d\,q + r, \qquad 0 \le r < d$
The square of an odd number leaves remainder 1 on division by 8
Odd square modulo 8
$m \text{ odd} \;\Longrightarrow\; m^2 = 8n+1 \quad\text{i.e.}\quad m^2 \equiv 1 \pmod 8$
A product of consecutive integers is divisible by the factorial of how many there are
Consecutive-run divisibility
$k! \;\big|\; n(n+1)(n+2)\cdots(n+k-1)$
Centring a run of consecutive integers on its middle term
Sum of squares of three consecutive integers
$(n-1)^2 + n^2 + (n+1)^2 = 3n^2 + 2$
Using the parity of a total to count the odd terms
Sum of parity signs
$\sum_{i=1}^{k} (-1)^{a_i} = k - 2j \quad (j = \text{how many } a_i \text{ are odd})$

Reference tables (2)

Parity bookkeeping for sums and products9 rows

Expression	Result	Why
odd + odd	even	$(2a+1)+(2b+1)=2(a+b+1)$
odd + even	odd	one unpaired unit is left over
even + even	even	both are multiples of 2
odd − odd	even	same as odd + odd for parity
odd × odd	odd	no factor of 2 anywhere
odd × even	even	one factor of 2 is enough
even × even	even	at least two factors of 2
n(n+1)	always even	consecutive integers, so one of them is even This row does the most work in the chapter. Any expression of the form $q^2+q$ is even without exception, which is what collapses several CDS parity questions to a single line.
2k ± any even	even	evens are closed under addition

Track parity, not values. Nine rows that settle most of what CDS asks about odd and even.

The three statement formats CDS uses, and how to attack each6 rows

Format	What it really asks	The attack
Consider the following statements 1, 2, 3	Is each statement true, separately?	Test each on its own; hunt one counterexample per statement
Statement-I / Statement-II (data sufficiency)	Is the answer UNIQUE, not what the answer is	Check I alone, then II alone, then both; stop at uniqueness The commonest error is solving the problem instead of testing sufficiency. If a statement leaves two possible values, it is insufficient even when both are easy to find.
Which one is correct	Three options are false	Eliminate by counterexample rather than proving the survivor
Which one is NOT correct	Three options are true	Read the word NOT twice; the wrong answer is usually the true statement you liked CDS sets both polarities and prints them in the same typeface. Underline the word NOT before you start.
Two statements that say the same thing	Whether either adds anything	If both carry one fact, together they are still insufficient
Option offering none of the above	Whether your value is really absent	Recompute once; this option is occasionally the intended answer

A quarter of this chapter arrives in one of these shapes. Name the format first, then do the mathematics.

Watch out for (10)

A remainder can never equal or exceed the divisor
→ The division algorithm
An even product does not mean both factors are even
→ Parity bookkeeping for sums and products
A statement about parity often says nothing about the variable you want
→ Parity bookkeeping for sums and products
Remainder 1 modulo 8 is a stronger claim than remainder 1 modulo 4
→ The square of an odd number leaves remainder 1 on division by 8
The rule is for ODD bases only
→ The square of an odd number leaves remainder 1 on division by 8
The rule gives a guarantee, not the largest divisor
→ A product of consecutive integers is divisible by the factorial of how many there are
What is true for three consecutive integers is not true for four
→ Centring a run of consecutive integers on its middle term
Count the values that are ATTAINABLE, not the cases that are arithmetically allowed
→ Using the parity of a total to count the odd terms
In data sufficiency, insufficient plus insufficient is not always sufficient
→ The three statement formats CDS uses, and how to attack each
A statement question is not an all-or-nothing question
→ The three statement formats CDS uses, and how to attack each

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