Number System — CDS Mathematics Formulas and Traps

Division, Parity & Consecutive Integers

The division algorithm

Division algorithm

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

$N$ the number being divided (the dividend)
$d$ the divisor
$q$ the quotient
$r$ the remainder, strictly less than 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})

Parity bookkeeping for sums and products

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 each

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.

Common traps

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.

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.

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.

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

or

n=2

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

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.

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.

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.

Place Value & Digit Problems

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Writing a number in expanded algebraic form

Expanded form

\overline{ab} = 10a+b, \qquad \overline{abc} = 100a+10b+c

$a$ leading digit, never 0
$b, c$ following digits, 0 to 9

Reversing a two-digit number: the 11 and 9 identities

Reversal sum and difference

N+N' = 11(a+b), \qquad N-N' = 9(a-b)

Reversing a three-digit number and swapping just two digits

Three-digit reversal difference

\overline{XYZ} - \overline{ZYX} = 99\,(X-Z)

The cyclic sum of a three-digit number is 111 times its digit sum

Cyclic sum identity

\overline{XYZ} + \overline{YZX} + \overline{ZXY} = 111\,(X+Y+Z)

Numbers built by repeating a block of digits

Repeated-block constants

\overline{abcabc} = \overline{abc}\times 1001, \qquad \overline{XYXYXY} = \overline{XY}\times 10101

Strings of repeated ones and nines

Repunit closed form

R_n = \underbrace{11\cdots1}_{n\text{ ones}} = \frac{10^{n}-1}{9}

Only the last few digits decide the last few digits

Last k digits

\text{last } k \text{ digits of } AB \;=\; (A \bmod 10^{k})(B \bmod 10^{k}) \bmod 10^{k}

Solving equations whose unknowns are single digits

Complement trick for near-round multipliers

999 \times n = 1000n - n, \qquad 99 \times n = 100n - n

Common traps

The digit constraints are part of the problem, not an afterthought

An equation like

b=2a

has infinitely many integer solutions and exactly four digit solutions. Most wrong answers on this concept come from solving the algebra correctly and then forgetting that

a \ne 0

and

b \le 9

. Impose both before you count.

Decide which way the difference runs before using it

9(a-b)

is positive when the leading digit is larger. If the question says the number increases on reversal, the quantity you know is

9(b-a)

, and getting the sign backwards produces the reversal of the intended answer — which is usually also in the option list.

A difference that is not a multiple of 9 means no such number exists

Since

N-N'=9(a-b)

always, a question offering a difference of 20 or 15 has no solution. Occasionally CDS uses this as the point of the question, so treat it as information rather than a misprint.

99, 90 and 9 are three different swaps

Only the full reversal gives 99. Swapping the first two digits gives 90 and swapping the last two gives 9. Students who memorise "the answer is a multiple of 99" get the 90-case wrong every time, and CDS sets both.

Divisible by 3 does not upgrade to divisible by 9

111

carries exactly one 3. The statement "

S

is always divisible by 9" is the standard planted falsehood in this question, and

100

(giving

S=111

) is the one-line counterexample. Keep 37 in mind too — it is the divisor students never think to check, and it is always there.

1001 and 10101 factorise differently

A repeated three-digit block gives

1001 = 7\times11\times13

— which contains 11. A repeated two-digit block over six digits gives

10101 = 3\times7\times13\times37

— which does not contain 11 but does contain 3 and 37. Reaching for the wrong constant is the whole failure mode here; count the block length first.

A repunit is not a power of ten

R_n = \frac{10^n-1}{9}

, not

10^n

and not

10^{n-1}

. The number of ones is

n

, and the number of digits of

10^n-1

is also

n

— but

10^n

itself has

n+1

digits. Off-by-one here silently changes the answer on every question of this type.

Keep as many digits as the question asks for, and no fewer

For the last three digits you must keep three digits of each factor. Keeping two and multiplying gives the last two digits correctly but the hundreds digit wrongly, because a carry from the dropped place can reach it. Match

k

to the question exactly.

Hundreds place is not the hundredth digit

CDS has printed a question asking for "the digit at the 100th place" of a number with only 95 digits. Read whether the paper means a place value (hundreds) or a position counted from one end; if the position does not exist, the intended reading is the place value.

Maximising one digit means minimising the others, within bounds

Once you reach something like

P+R+Q = 9

, the largest

Q

needs the other letters as small as their own constraints permit — which is 0 for an interior digit but 1 for a leading digit. Assuming 0 everywhere is the standard slip and inflates the answer by one.

Divisibility Rules & Missing Digits

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Recovering a hidden digit from a divisibility condition

Hidden-digit condition

P \equiv -\!\!\sum(\text{known digits}) \pmod{9}, \qquad 0 \le P \le 9

Divisors for which only the tail of the number matters

Tail rule

N \equiv \left(N \bmod 10^{k}\right) \pmod{2^{k}} \quad\text{and}\quad \pmod{5^{k}}

What to do when the divisor has no usable test

The 1001 grouping fact

1001 = 7 \times 11 \times 13

The divisibility test table

Divisor	Test	Reason
2	last digit is even	$10 \equiv 0 \pmod 2$
3	digit sum divisible by 3	$10 \equiv 1 \pmod 3$
4	last two digits divisible by 4	$100 \equiv 0 \pmod 4$
5	last digit is 0 or 5	$10 \equiv 0 \pmod 5$
6	passes both the 2 and 3 tests	$6 = 2\times 3$ , coprime parts
8	last three digits divisible by 8	$1000 \equiv 0 \pmod 8$
9	digit sum divisible by 9	$10 \equiv 1 \pmod 9$
10	last digit is 0	$10 \equiv 0 \pmod{10}$
11	alternating digit sum divisible by 11	$10 \equiv -1 \pmod{11}$ Alternate the signs from the units digit leftwards. A result of $0$ counts as divisible.
16	last four digits divisible by 16	$10^4 \equiv 0 \pmod{16}$
25	last two digits are 00, 25, 50 or 75	$100 \equiv 0 \pmod{25}$
12	passes the 4 and 3 tests	$12 = 4\times 3$ , not $2\times 6$ Testing 2 and 6 is wrong: 18 passes both and is not a multiple of 12.
7 and 13	no short test worth learning	$10$ has order 6 modulo both

Thirteen rows. The reason column is not decoration — it is what tells you how many trailing digits a power-of-2 test needs.

Common traps

The smallest odd composite number is 9, not 1, 3 or 15

CDS hides the divisor behind a description.

1

is neither prime nor composite, and

3, 5, 7

are all prime — so the smallest odd composite is

9

. Get that wrong and you run a perfectly correct digit-sum test against the wrong divisor.

For a composite divisor, split into COPRIME parts

To test 12, use 4 and 3 — not 2 and 6. The parts must be coprime and must multiply to the divisor, otherwise you lose a factor: 18 passes the 2-test and the 6-test yet is not divisible by 12. Same trap for 8 (use 8 directly, not 2 and 4).

A mod-9 condition often has TWO digit solutions, not one

If the required residue is

0

, both

P=0

and

P=9

satisfy it. A question asking "the value of

P

" when two exist is asking you to notice; one asking for a count is counting both. Always solve the congruence and then enumerate the range rather than stopping at the first hit.

With two hidden digits the condition fixes only their SUM

9 \mid N

pins

A+B

to a residue, never

A

and

B

individually — so the answer is a count of pairs. Remember that

A+B=0

is a legitimate total (both digits zero) and is divisible by 9, which is the case students drop.

The tail rule works only for divisors built from 2s and 5s

It fails the moment a factor of 3 or 7 appears, because

10^k

is then not a multiple of the divisor. There is no "last two digits" test for 12 or 24. If the divisor is

2^a 5^b m

with

m>1

, split it: handle

2^a5^b

by the tail and

m

by its own rule.

Reading the tail of a described number is where this goes wrong

When the number is described rather than printed — "write 1 to 100 in order" — the hard part is working out what the last four digits actually are. The string ends

\ldots 99\,100

, so the final four characters are

9100

, not

0099

or

9910

. Write out the tail explicitly before dividing.

Trial is the intended method here, so do not hunt for a rule

On a 13-divisibility question with one hidden digit, students lose two or three minutes trying to recall a test that does not exist. Ten divisions is faster and certain. The exam-craft point is recognising immediately that there is no rule to recall, which is why 7 and 13 have a row of their own in the table above.

Unit Digit & Cyclicity

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Reducing the exponent modulo 4

Exponent reduction

b^{n} \text{ ends in the } r\text{-th cycle entry}, \quad r = n \bmod 4 \;\;(r=0 \Rightarrow \text{4th entry})

Unit digits of sums, differences and products of powers

Unit digit of a combination

(A \pm B) \bmod 10 = \big[(A \bmod 10) \pm (B \bmod 10)\big] \bmod 10

A number that is odd and a multiple of 5 must end in 5

Odd multiple of five

5 \mid N \;\text{and}\; N \text{ odd} \;\Longrightarrow\; N \equiv 5 \pmod{10}

Counting how many unit digits an expression can produce

Number of cases to check

\text{cases} = \prod_{\text{moving terms}} (\text{period of that term})

The unit-digit cycle of each base

Last digit of base	Cycle of unit digits	Period
0	0	1
1	1	1
2	2, 4, 8, 6	4
3	3, 9, 7, 1	4
4	4, 6	2
5	5	1 Every positive power of a number ending in 5 ends in 5. There is no alternation.
6	6	1
7	7, 9, 3, 1	4
8	8, 4, 2, 6	4
9	9, 1	2

Read the cycle left to right starting at exponent 1. Every period divides 4, so exponent modulo 4 settles every case.

Common traps

The base's other digits are irrelevant, and the exponent's are not

67^{32}

,

7^{32}

and

1257^{32}

all end in the same digit — only the base's last digit counts. But you must use the whole exponent when reducing modulo 4: the exponent's last digit alone is not enough, since

14

and

34

end alike yet leave different remainders on division by 4.

A remainder of 0 sends you to the END of the cycle, not the start

This is the single commonest error in the whole subtopic.

2^{100}

has

100 \bmod 4 = 0

, and the answer is 6 — the fourth entry — not 2. Think of it as finishing a lap: remainder 0 means you are standing on the last step, and the amber node in the diagram above is exactly that position.

Reduce the exponent modulo 4, never modulo 10

The cycle length is 4, so 4 is the modulus for the exponent. Students who reduce the exponent modulo 10 — because the question is about the last digit — get a number between 0 and 9 that means nothing here. The 10 lives in the answer; the 4 lives in the exponent.

Factor a difference of powers before taking unit digits

For

3^{98}-3^{89}

, taking each term's unit digit gives

9-3

and the tempting answer 6 — which happens to be right here, but the method is unsafe: it breaks whenever the first unit digit is the smaller. Factoring to

3^{89}(3^9-1)

turns it into a product, which never needs borrowing.

A negative difference needs plus 10, not a minus sign

If the unit digits give

4-7=-3

, the last digit is

7

, not

-3

or

3

. Add 10 once. A digit is always in the range 0 to 9.

The word ODD in the question is what changes the answer from 0 to 5

"The product of all odd primes up to 110" ends in 5; "the product of all primes up to 110" ends in 0, because including 2 makes the product even. CDS sets both versions. The single word doing the work is odd, and it is easy to read past.

Collapse the period-1 terms before you start enumerating

In

5^{a}+7^{b}+11^{c}+13^{d}

two of the four terms never change:

5^a

ends in 5 and

11^c

ends in 1. Treating all four as variable means enumerating sixteen cases instead of the necessary sixteen over only the two that move — and, worse, invites you to imagine

5^a

cycling, which it does not.

The question asks for a COUNT, or sometimes for a SUM of the distinct values

CDS sets both phrasings, and they sit next to each other in the same paper. "How many distinct remainders" wants 5; "the sum of all distinct remainders" wants the total of that set. Re-read the last line of the stem before you answer.

Prime Numbers & Primality

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Testing a number for primality by trial division

Trial-division bound

n \text{ is prime} \iff p \nmid n \ \text{ for every prime } p \le \sqrt{n}

Forcing a prime to be 2 with a parity argument

The forcing rule

p+q \text{ odd (with } p,q \text{ prime)} \;\Longrightarrow\; \{p,q\} \ni 2

Coprimality and Euclid's lemma

Coprimality via the difference

\gcd(n,\,n+1)=1 \quad\text{for every integer } n

Breaking a number into its prime factors

Unique factorisation

N = p_1^{a_1} p_2^{a_2}\cdots p_k^{a_k} \quad\text{uniquely}

Primes, composites, and the numbers that are neither

Claim	Verdict	Why
1 is prime	False	It has one divisor, not two
1 is composite	False	It is neither
2 is prime	True	Divisors 1 and 2 only
Every prime is odd	False	2 is even This is the single most useful exception in the chapter — it is what lets you force one prime to be 2.
Number of primes below 100	25	2, 3, 5, ..., 89, 97
Number of primes below 50	15	So 10 lie between 50 and 100
Possible unit digits of a prime	1, 2, 3, 5, 7, 9	Six digits; 0, 4, 6, 8 give an even number above 2
Smallest odd composite	9	1 is neither; 3, 5, 7 are prime
A product of two composites can be coprime	True	4 and 9 share no prime factor

Nine rows CDS asks about directly. The 25-below-100 count and the six possible unit digits are pure recall.

Forms that look like they generate primes but do not

Form or claim	Always prime?	First failure
$6n-1$	No	$n=6$ gives $35=5\times 7$
$6n+1$	No	$n=4$ gives $25=5^2$
Every prime $>3$ is $6n\pm1$	True	This is the valid direction True one way, false the other. The question always tests the false direction.
$2^n-1$ (Mersenne)	No	$n=11$ gives $2047=23\times 89$
$n^2+n+41$	No	$n=40$ gives $1681=41^2$
Product of first $n$ primes, plus 1	No	$n=6$ gives $30031=59\times 509$ It is prime for $n=1$ to $5$ — 3, 7, 31, 211, 2311 — which is why the statement looks safe.
Prime triples spaced by 2	Only once	$3,5,7$ ; one of any such triple is a multiple of 3
Difference of two primes $>2$	Always even	Both are odd

Learn the failure, not the pattern. Each right-hand entry is a complete answer to a statement question.

Common traps

Coprime does not mean prime

4

and

9

are coprime and both composite;

3

and

4

are coprime with one of each. "Relatively prime" is a statement about the pair, not about either number, so every combination of prime and composite is possible — which is exactly what the 2019 and 2016 statement questions test.

Numbers near 400 or 1000 look prime and often are not

437 = 19\times 23

and

1073 = 29\times 37

both survive every easy test — odd, not a multiple of 3, not ending in 5 — and fail only at a two-digit prime. Push the trial division all the way to

\sqrt{n}

; stopping at 13 because "nothing small worked" is how both of those get called prime.

Having forced the 2, still check the survivor is prime

The parity step tells you one prime is 2; it does not tell you the rest works. For a sum of 45 you still verify 43 is prime. And when the answer options list values that never occur — as in the 2019 question whose three primes are 2, 31 and 67 while the options offer 17, 29 and 43 — "none of these" is the intended answer, not a sign you slipped.

Euclid's lemma needs the divisor to be PRIME

"If

p \mid qr

then

p\mid q

or

p \mid r

" is false for composite

p

:

4

divides

2\times 6 = 12

but divides neither 2 nor 6. CDS plants exactly this by dropping the word prime from the statement.

The converse of a true statement about primes is usually false

Every prime above 3 has the form

6n\pm1

, so it is tempting to accept "

6n-1

is always prime". It is not —

35

settles it. Whenever a statement about primes reads like a generating rule, look for the first small counterexample rather than checking a few cases that work.

The LCM of two distinct primes is their product

Given "the LCM of two primes is 2231", do not search: two distinct primes share no factor, so their LCM is

pq

. Factorising

2231 = 23\times 97

answers the question immediately. If the two primes were equal the LCM would be the prime itself, so a composite LCM guarantees they are distinct.

Factors, Divisor Counting & Trailing Zeros

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Canonical prime-power form

Canonical form

N = p_1^{a_1}\,p_2^{a_2}\cdots p_k^{a_k}

Counting the divisors of a number

Divisor count

d(N)=\prod_{i=1}^{k}(a_i+1)

$a_i$ the exponent of the i-th prime in N

Summing the divisors of a number

Divisor sum

\sigma(N)=\prod_{i=1}^{k}\frac{p_i^{a_i+1}-1}{p_i-1}

Odd divisors, divisors of a square, and working backwards

Divisors of a square

d(N^{2})=\prod_{i=1}^{k}(2a_i+1)

Counting the zeros at the end of a factorial

Zeros at the end of n factorial

Z(n!)=\sum_{i\ge 1}\left\lfloor \frac{n}{5^{i}}\right\rfloor

Trailing zeros of a general product: the scarcer prime wins

Zeros of a general product

n=\min\big(v_2(P),\,v_5(P)\big)

Common traps

Evaluate the expression before you factorise it

For

24^3-16^3-8^3

there is no shortcut identity — compute

13824-4096-512 = 9216

and factorise that as

2^{10}\cdot 3^2

. Students who try to factor term by term and subtract exponents get nonsense, because exponents do not subtract across a difference.

Read whether 1 and N are to be excluded

d(38808)=72

, but the question asks for divisors "exclusive of 1 and itself", so the answer is 70. The same trap runs on 1000:

d=16

, answer 14. The formula always counts both ends — the subtraction is yours to do.

The divisor sum is a product of sums, not a sum of products

For

2^2\cdot 3\cdot 5

the answer is

(1+2+4)(1+3)(1+5)

, not

(1+2+4)+(1+3)+(1+5)

. Multiplying the brackets is what generates each divisor once; adding them counts almost nothing correctly.

Working backwards from a divisor count usually leaves several shapes

d(N)=15

admits

p^{14}

and

p^{4}q^{2}

. What eliminates the first is the size clue: the smallest

p^{14}

is

2^{14}=16384

, which has five digits, so a four-digit

N

must be

p^4q^2

. Always list every factorisation of the count, then use the stated size to cut.

Counting only the multiples of 5 undercounts

For

25!

the multiples of 5 number 5, which tempts the answer

10^5

. But 25 itself carries two fives, so the true count is

5+1=6

. Every term

\lfloor n/25\rfloor

,

\lfloor n/125\rfloor

is a real contribution, not a refinement you can skip.

Outside a factorial, do not assume the fives are the scarce prime

In

n!

the twos always outnumber the fives, so counting fives suffices. In

2^{8}\cdot 5^{3}\cdot 7

it is the other way round, and in

6^{23}\cdot75^{9}\cdot105^{2}

the counts are 23 twos against 20 fives — close enough that guessing loses the mark. Count both, every time.

Check the parity of a SUM before counting any factors

For

P+Q

where

P

is a product of odd numbers and

Q

of even ones, the sum is odd, so it ends in no zero at all — and no amount of counting fives inside

P

and

Q

is relevant. The 2026 question is exactly this shape, and the factor-counting route wastes minutes before failing.

HCF & LCM — Laws and Fractions

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HCF and LCM from the prime factorisations

HCF and LCM by exponents

\mathrm{HCF}=\prod p_i^{\min(a_i,b_i)}, \qquad \mathrm{LCM}=\prod p_i^{\max(a_i,b_i)}

The product law for two numbers

Product law (two numbers)

\mathrm{HCF}(a,b)\cdot\mathrm{LCM}(a,b)=ab

Writing the pair as H times coprime parts

The Ha, Hb substitution

x=Ha,\; y=Hb,\; \gcd(a,b)=1 \;\Longrightarrow\; \frac{\mathrm{LCM}}{\mathrm{HCF}}=ab

The subtraction property and its consequences

Subtraction property

\mathrm{HCF}(n,\;n+k)=\mathrm{HCF}(n,\;k)

HCF and LCM of fractions and decimals

Fractions: the crossed recipes

\mathrm{LCM}\!\left(\frac{a_i}{b_i}\right)=\frac{\mathrm{LCM}(a_i)}{\mathrm{HCF}(b_i)}

The HCF of two numbers of the form a to the n, minus one

HCF of power-minus-one

\gcd\!\left(a^{m}-1,\;a^{n}-1\right)=a^{\gcd(m,n)}-1

Spotting HCF and LCM data that cannot exist

The ratio test

\frac{\mathrm{LCM}}{\mathrm{HCF}} = ab \in \mathbb{Z}^{+}

Common traps

An LCM that is not a multiple of the HCF is impossible

Given HCF 12, an LCM of 80 can never occur, because

12 \nmid 80

. This is the fastest elimination in the whole unit and CDS uses it directly. Check divisibility before doing any arithmetic.

The product law is a TWO-number law

For three numbers, HCF times LCM is not the product. The 2026 data-sufficiency question on three numbers with HCF 5 and LCM 30 has to be solved by listing the candidate triples, not by dividing — each number must be a multiple of 5 and a divisor of 30, which leaves only four triples to test.

List only the COPRIME factor pairs, and expect more than one to survive

With

ab=28

the pairs are

(1,28)

and

(4,7)

—

(2,14)

is excluded because 2 and 14 share a factor, which would make the HCF larger than stated. And when two coprime pairs both satisfy every stated condition, the question genuinely has two answers; the 2022 question with HCF 9 and LCM 126 is exactly that case.

The property transfers the HCF, it does not compute it

\mathrm{HCF}(n,\,n+10)=\mathrm{HCF}(n,10)

tells you the answer divides 10 — not that it is 10. The 2024 question requiring

\mathrm{HCF}(n,\,n+10)=10

therefore forces

10 \mid n

, which is an extra condition you must impose before counting the possible LCMs.

The two fraction recipes are crossed — do not use the same one twice

HCF of fractions uses HCF-over-LCM; LCM of fractions uses LCM-over-HCF. Applying the numerator rule to the denominators as well gives a value that is neither. A quick check catches it: the LCM must be divisible by each fraction, and the HCF must divide each.

Pull out the common constant before applying the identity

3^{29}-9

is not of the form

a^{n}-1

. Rewrite it as

9(3^{27}-1)

and likewise

3^{38}-9 = 9(3^{36}-1)

; then the identity gives

\gcd = 9\!\left(3^{9}-1\right) = 3^{11}-9

. Applying the identity to the exponents 29 and 38 directly gives a near-miss answer that appears in the option list.

Some CDS questions carry data that cannot exist, and that is deliberate

The 2021 question gives HCF 120 with one number 104 — but

120 \nmid 104

, so no such pair exists, and the 2023 question asks for an LCM-to-HCF ratio of

3:2

, which no pair can have. Run the three checks, state the conflict, and give the value the intended arithmetic yields. Do not assume you have miscalculated: on this corpus the data is sometimes the thing that is wrong.

HCF & LCM — Applications and Remainder Recipes

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When the answer is the HCF: the largest common measure

Tile count from the HCF

\text{tiles}=\frac{L}{\mathrm{HCF}}\times\frac{W}{\mathrm{HCF}}

When the answer is the LCM: things coinciding again

Coincidences within a window

\text{count}=\left\lfloor \frac{\text{window}}{\mathrm{LCM}}\right\rfloor

Largest, smallest and how many multiples in a range

Multiples in a range

\#\{\text{multiples of } L \le N\}=\left\lfloor \frac{N}{L}\right\rfloor

Recipe 1: same unknown remainder means take the HCF of the differences

Same-remainder recipe

N_{\max}=\mathrm{HCF}\big(b-a,\;c-b\big)

Recipe 2: the same remainder from every divisor means LCM times k, plus r

Common-remainder recipe

N = L\,k + r, \qquad L=\mathrm{LCM}(d_1,\dots,d_k)

Recipe 3: a constant shortfall means LCM times k, minus d

Constant-shortfall recipe

N = L\,k - d, \qquad d = d_i - r_i \ \text{(the same for every } i)

Layering an extra condition on a remainder recipe

Layered condition

m \mid (Lk+r) \;\Longrightarrow\; Lk \equiv -r \pmod m

Common traps

Largest tile and minimum number of tiles are the same question

A question asking for the minimum number of tiles still wants the largest tile, because bigger tiles means fewer of them. Both phrasings point at the HCF. Do not switch to the LCM because the word "minimum" appeared.

How many MORE times excludes the start

If they ring together at 9 am and you are asked how many more times in the next 72 hours, count the multiples of the LCM inside the window and do not count the 9 am ring itself. With an LCM of 1575 minutes in 4320 minutes, only 1575 and 3150 fit — the answer is 2, not 3.

An LCM multiple need not be a perfect square

For the least perfect square divisible by 3, 4, 5, 6 and 7, the LCM is

420 = 2^2\cdot3\cdot5\cdot7

— but 420 is not a square, because three of its exponents are odd. Raise each to the next even value:

2^2\cdot3^2\cdot5^2\cdot7^2 = 44100

. Stopping at the LCM, or at a multiple like 17640 that is divisible by everything but is not square, are both offered as options.

Unknown remainder means differences; known remainder means subtract it

These are two different recipes and using the wrong one is the commonest error here. "Leaves the same remainder" (unspecified) means take the HCF of the differences. "Leaves remainder 5 in each case" means subtract 5 from each number first and take the HCF of those. Taking differences when the remainder is given still works, but subtracting a remainder that was never given does not.

The stated remainder must be smaller than every divisor

"Remainder 7 on division by 18 as well as 11" is legitimate because

7<11

. If a question offered remainder 9 with a divisor of 8 it would be inconsistent. Check the smallest divisor against

r

before applying the recipe — it is a one-second sanity test that occasionally is the question.

Check the shortfall is really constant before reaching for this

Remainders 2, 8, 11, 20 for divisors 6, 12, 15, 24 look unrelated, but each is exactly 4 short — so

N = 120k-4

. Conversely, if the shortfalls come out unequal, neither this recipe nor recipe 2 applies and you must solve the congruences directly. Compute all the differences first; it takes seconds and decides the whole method.

Do not stop at the family — the extra condition is the question

N = 180k+5

is only the first half of the 2024 question; the answer must also be a multiple of 11, which forces

k \equiv 7 \pmod{11}

and gives

N = 1265

. An option list built around the family alone will contain several values satisfying the remainder conditions and failing the divisibility one.

Remainders by Congruence & Cyclicity

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Replacing a number by its remainder

Reduce the base first

a \equiv r \pmod n \;\Longrightarrow\; a^{k} \equiv r^{k} \pmod n

When the base is one less than the modulus

Alternating powers

a\equiv -1 \pmod n \;\Longrightarrow\; a^{k}\equiv(-1)^{k}\pmod n

Finding the cycle of powers modulo n

Cycle reduction

a^{t}\equiv 1 \pmod n \;\Longrightarrow\; a^{k}\equiv a^{\,k \bmod t} \pmod n

Fermat's little theorem

a^{p-1}\equiv 1 \pmod p \quad (p \text{ prime},\ p \nmid a)

Pairing terms that cancel modulo n

Cancelling pair

a+b\equiv 0 \pmod n,\ k \text{ odd} \;\Longrightarrow\; a^{k}+b^{k}\equiv 0

Remainders of sums, differences and products of given remainders

Combining remainders

m\equiv r_1,\ n\equiv r_2 \pmod d \;\Longrightarrow\; m\pm n\equiv r_1\pm r_2,\ \ mn\equiv r_1r_2

Common traps

A remainder must land in 0 to n minus 1

If your working produces

-1

modulo 18, the remainder is 17, not

-1

. And a remainder on division by 9 can never be 9 — CDS puts exactly that value in the option list for the

37^{1000}

question. Normalise before you answer.

Odd power of minus one is the modulus minus one, not minus one

65^{99} \equiv -1 \pmod{11}

means the remainder is 10. Writing

-1

or

1

are the two wrong answers offered. Similarly

7^{84}

modulo 344 uses

343 \equiv -1

with an even exponent, giving 1 — the parity of the exponent is doing all the work, so read it carefully.

Reduce the exponent modulo the CYCLE, not modulo the divisor

For

2^{50}

modulo 9 the cycle is 6, so reduce 50 modulo 6 — not modulo 9. Mixing the two moduli is the standard error and produces a plausible wrong answer. The divisor sets the arithmetic; the cycle length sets the exponent reduction.

The exponent p minus one gives 1; the exponent p gives the base back

2^{100} \bmod 101 = 1

but

2^{101} \bmod 101 = 2

. CDS has set both, one year apart, and the option lists overlap. Check whether the exponent is

p-1

or

p

before answering, and remember the theorem needs the modulus prime — which is exactly what the 2017 statement question tests.

Pair the bases before reducing them individually

In

13^5+14^5+15^5+16^5

modulo 29 the structure is two pairs:

13+16=29

and

14+15=29

. Reducing each base on its own gives four unhelpful residues and a long computation; noticing the pairing gives 0 immediately. Always add the outermost bases together first to check for this.

m greater than n does not mean its remainder is greater

With

m\equiv 4

and

n\equiv 6

modulo 12 and

m>n

, the difference is

4-6=-2\equiv 10

— the same as the sum's remainder, which is why the 2025 question can truthfully say both are 10. Students who assume

r_1>r_2

because

m>n

compute

6-4=2

and get it wrong.

Divisibility by Factorisation

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Pulling the smallest power out of a sum of like powers

Power-sum extraction

a^{m}+a^{m+1}+\cdots+a^{m+k}=a^{m}\!\left(1+a+\cdots+a^{k}\right)

The a-to-the-n minus b-to-the-n identity

Difference of like powers

a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)

The a-to-the-n plus b-to-the-n identity

Sum of like powers, odd exponent

a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2}b+\cdots+b^{n-1}\right) \quad (n \text{ odd})

Rewriting mixed bases as powers of one number

Rebasing

\left(a^{j}\right)^{k}=a^{jk}

The largest number that ALWAYS divides an expression

The cap from the smallest case

d \mid f(n)\ \forall n \;\Longrightarrow\; d \mid f(n_{\min})

When a variable divides a polynomial in itself

Constant-term criterion

x \mid \left(a_kx^{k}+\cdots+a_1x+c\right) \iff x \mid c

Common traps

A new prime can only come from the bracket

2^{122}+2^{124}+2^{126}+2^{128}+2^{130} = 2^{122}\times 341

, and

341 = 11\times 31

. The answer is 11 — the power of 2 cannot supply an odd factor, so checking the bracket alone is both necessary and sufficient. Students who try to test the whole number for divisibility by 11 get nowhere.

a plus b needs an EVEN exponent for a difference

97^{30}-14^{30}

is divisible by

97^2-14^2 = 9213 = 3\times37\times83

precisely because 30 is even — which is how 37 and 83 both appear. For an odd exponent only

a-b

is available. Check the parity before claiming

a+b

.

The parity conditions for a sum and a difference are opposite

a-b

divides a difference for every exponent;

a+b

divides a sum only for an odd exponent. Swapping these is the single commonest error in this unit. Write down which you have — sum or difference — then check the exponent's parity against the right rule.

Rebase before you sort, and sort before you factor

2^{122}+4^{62}+8^{42}+4^{64}+2^{130}

looks like five unrelated terms. Rebasing gives

2^{122},2^{124},2^{126},2^{128},2^{130}

— an evenly spaced run — and factoring out

2^{122}

leaves

1+4+16+64+256 = 341 = 11\times31

. Attempting to factor before rebasing gets nowhere.

Whole numbers include zero, and that can break the statement

5\cdot 8^{m}+2^{3m} = 6\cdot 8^{m}

, which is divisible by 48 for every

m \ge 1

— but at

m=0

it equals 6, and 6 is not divisible by 48. Since "whole numbers" include 0, the statement is false. Always test the smallest value the stated domain actually permits.

Count the divisors of the constant, not the values you happen to test

For

x^{3}+x^{2}+16

the condition is

x\mid 16

, giving

1,2,4,8,16

— five values. Testing

x=1,2,3,\ldots

by substitution finds the same answers much more slowly and risks stopping early. Reduce to the constant first, then count its divisors.

Perfect Squares, Cubes & Difference of Squares

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The nearest perfect square above or below

Distance to the neighbouring squares

k^{2}\le N<(k+1)^{2}: \quad \text{subtract } N-k^{2}, \quad \text{add } (k+1)^{2}-N

Difference of squares and the parity constraint on factor pairs

Difference of squares

m^{2}-n^{2}=(m-n)(m+n)

Completing the square to force a factorisation

Reduce to a constant difference of squares

n^{2}+bn+c=k^{2} \;\Longrightarrow\; (2k)^{2}-(2n+b)^{2}=4c-b^{2}

Expressions that are always perfect squares

Four consecutive integers plus one

n(n+1)(n+2)(n+3)+1=\left(n^{2}+3n+1\right)^{2}

Cubes, fourth powers and taxicab numbers

Taxicab identity

1729 = 1^{3}+12^{3} = 9^{3}+10^{3}

The last digit of a perfect square

Unit digit of n	Unit digit of n squared
0	0
1 or 9	1
2 or 8	4
3 or 7	9
4 or 6	6
5	5 So the possible endings are exactly 0, 1, 4, 5, 6, 9 — and 2, 3, 7, 8 never occur.

Six reachable endings out of ten. The four unreachable ones are the examinable content.

Common traps

The last-digit test only rules out, never rules in

Of 2222, 11664, 343343 and 220347, the endings 2, 3 and 7 eliminate three immediately — but the survivor 11664 still has to be checked, and it happens to be

108^{2}

. Treat the test as a filter that saves time, not as a proof of squareness.

The number itself may not be a square even when it looks round

The smallest four-digit number is 1000, and it is tempting to answer 1000 for "smallest four-digit perfect square". But 1000 is not a square:

31^2=961

has three digits and

32^2=1024

has four, so the answer is 1024. Bracket with actual squares rather than trusting the round number.

Mixed-parity factor pairs must be discarded

For

N=72

the pair

(8,9)

multiplies correctly but gives

m=8.5

— not an integer. Only

(2,36)

,

(4,18)

and

(6,12)

survive, so the answer is 3. Counting all factor pairs rather than the same-parity ones is the standard error and inflates every answer in this concept.

A prime target forces a unique pair

If

N

is prime the only factorisation is

1\times N

, so

m=\frac{N+1}{2}

and

n=\frac{N-1}{2}

uniquely. For

N=199

that is

m=100

,

n=99

, giving

mn = 9900

— no searching needed.

An odd middle coefficient needs the factor of 4

For

n^{2}+19n+92

, completing the square directly gives halves. Multiply by 4 first:

4(n^2+19n+92) = (2n+19)^{2}+7

, so

(2k)^{2}-(2n+19)^{2}=7

. Since 7 is prime the factors are

\pm1

and

\pm7

, giving

2n+19=\pm3

and hence

n=-8

or

n=-11

, summing to

-19

. Skipping the multiplication loses both solutions.

Pair the OUTER factors, not adjacent ones

The identity works because

n(n+3)

and

(n+1)(n+2)

differ by exactly 2. Pairing

n(n+1)

with

(n+2)(n+3)

instead gives two quadratics differing by

4n+6

, and the substitution collapses. Always multiply the first by the last.

m to the n has a trivial solution that the question does not intend

m^{n}=1331

is satisfied by

m=1331,\ n=1

as well as by

m=11,\ n=3

, and the trivial reading gives

1330^{0}=1

— which appears in the option list. The stem's intent is the genuine power, so read any restriction such as "different from 1" carefully, and prefer the non-trivial factorisation when both are admissible.

Rational & Irrational Numbers

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Converting a recurring decimal to a fraction

Purely recurring decimal

0.\overline{d_1d_2\cdots d_k}=\frac{d_1d_2\cdots d_k}{\underbrace{99\cdots9}_{k}}

Deciding irrationality of roots, sums and products

Root test

\sqrt{n} \in \mathbb{Q} \iff n \text{ is a perfect square}

What makes a number rational, and what the decimal expansion reveals

Number	Rational?	Reason
0.5	Rational	Terminates
0.333...	Rational	Recurs, equals one third
$\sqrt{75}$	Irrational	Equals $5\sqrt3$ , and 75 is not a perfect square
$\sqrt{59049}$	Rational	Equals 243, since $59049=3^{10}$
0.12112211122211112222...	Irrational	Blocks grow, so it never repeats A visible pattern is not a repeating block. Recurrence needs a fixed block repeated forever.
$\pi$	Irrational	Non-terminating, non-repeating
$4\pi r^{2}$ with $r$ rational	Irrational	A non-zero rational multiple of $\pi$
$\sqrt2 \times \sqrt{50}$	Rational	Equals 10 — a product of irrationals can be rational

The decimal expansion is the test. Note the two rows that go against first instinct: a square root can be rational, and a product of irrationals can be too.

Common traps

A square root is not automatically irrational

\sqrt{59049} = 243

because

59049 = 3^{10}

, so it is rational.

\sqrt{n}

is irrational exactly when

n

is not a perfect square. Check for squareness before calling a root irrational — CDS plants a large perfect square in the option list precisely to catch this.

A pattern is not the same as a recurring block

0.12112211122211112222\ldots

is clearly patterned, but the blocks lengthen, so no fixed block repeats and the number is irrational. Recurrence means one unchanging block forever, as in

0.\overline{45}

.

The recurring bar changes the value, and the question turns on it

0.\overline{9} = 1

exactly, so

0.\overline{9} - 0.9 = 0.1

, not

0.0999\ldots

. Likewise

0.\overline{459}

is

\frac{17}{37}

while the terminating

0.459459459

is a fraction over a power of 10 and has no 37 in its denominator. Read whether the bar is present before converting.

Simplify the surd before judging it

\sqrt{75}

looks irreducible but equals

5\sqrt3

;

\sqrt{59049}

looks irrational but equals 243. Pull out every square factor first. The 2019 statement question hinges on exactly this:

\sqrt{75}

being called rational is the planted falsehood.

Number System formulas

Division, Parity & Consecutive Integers

The division algorithm

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

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

Centring a run of consecutive integers on its middle term

Using the parity of a total to count the odd terms

Parity bookkeeping for sums and products

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

Common traps

Place Value & Digit Problems

Writing a number in expanded algebraic form

Reversing a two-digit number: the 11 and 9 identities

Reversing a three-digit number and swapping just two digits

The cyclic sum of a three-digit number is 111 times its digit sum

Numbers built by repeating a block of digits

Strings of repeated ones and nines

Only the last few digits decide the last few digits

Solving equations whose unknowns are single digits

Common traps

Divisibility Rules & Missing Digits

Recovering a hidden digit from a divisibility condition

Divisors for which only the tail of the number matters

What to do when the divisor has no usable test

The divisibility test table

Common traps

Unit Digit & Cyclicity

Reducing the exponent modulo 4

Unit digits of sums, differences and products of powers

A number that is odd and a multiple of 5 must end in 5

Counting how many unit digits an expression can produce

The unit-digit cycle of each base

Common traps

Prime Numbers & Primality

Testing a number for primality by trial division

Forcing a prime to be 2 with a parity argument

Coprimality and Euclid's lemma

Breaking a number into its prime factors

Primes, composites, and the numbers that are neither

Forms that look like they generate primes but do not

Common traps

Factors, Divisor Counting & Trailing Zeros

Canonical prime-power form

Counting the divisors of a number

Summing the divisors of a number

Odd divisors, divisors of a square, and working backwards

Counting the zeros at the end of a factorial

Trailing zeros of a general product: the scarcer prime wins

Common traps

HCF & LCM — Laws and Fractions

HCF and LCM from the prime factorisations

The product law for two numbers

Writing the pair as H times coprime parts

The subtraction property and its consequences

HCF and LCM of fractions and decimals

The HCF of two numbers of the form a to the n, minus one

Spotting HCF and LCM data that cannot exist

Common traps

HCF & LCM — Applications and Remainder Recipes

When the answer is the HCF: the largest common measure

When the answer is the LCM: things coinciding again

Largest, smallest and how many multiples in a range

Recipe 1: same unknown remainder means take the HCF of the differences

Recipe 2: the same remainder from every divisor means LCM times k, plus r

Recipe 3: a constant shortfall means LCM times k, minus d

Layering an extra condition on a remainder recipe

Common traps

Remainders by Congruence & Cyclicity

Replacing a number by its remainder

When the base is one less than the modulus

Finding the cycle of powers modulo n

Pairing terms that cancel modulo n

Remainders of sums, differences and products of given remainders

Common traps

Divisibility by Factorisation

Pulling the smallest power out of a sum of like powers

The a-to-the-n minus b-to-the-n identity

The a-to-the-n plus b-to-the-n identity

Rewriting mixed bases as powers of one number

The largest number that ALWAYS divides an expression