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AFCAT Numerical Ability · Formula sheet

LCM and HCF formulas

9 formulas, 1 reference table and 20 common traps for AFCAT Numerical Ability LCM and HCF, grouped by subtopic.

Full notes with worked examples

Finding the HCF and LCM

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HCF and LCM by prime factors

HCF and LCM from prime powers

HCF=∏pmin⁡(a,b),LCM=∏pmax⁡(a,b)\text{HCF} = \prod p^{\min(a, b)}, \qquad \text{LCM} = \prod p^{\max(a, b)}

HCF of large numbers by repeated division

Euclid's division step

HCF(a,b)=HCF(b,  a mod b)\text{HCF}(a, b) = \text{HCF}(b,\; a \bmod b)

HCF and LCM of fractions

Fractions

HCF=HCF of numeratorsLCM of denominators,LCM=LCM of numeratorsHCF of denominators\text{HCF} = \dfrac{\text{HCF of numerators}}{\text{LCM of denominators}}, \qquad \text{LCM} = \dfrac{\text{LCM of numerators}}{\text{HCF of denominators}}

Common traps

The HCF keeps only shared primes

For 23×52^3 \times 5 and 2×32 \times 3, the HCF is 2. The 5 and the 3 belong to one number each, so they go into the LCM (which is 120) and never into the HCF.

The LCM takes the highest power

The LCM of 22×32^2 \times 3 and 2×322 \times 3^2 is 22×32=362^2 \times 3^2 = 36. Taking 2×32 \times 3 from either number gives the HCF, 6, by mistake.

Divide by the remainder

Each step divides the last divisor by the last remainder. Dividing the first number again by the new remainder breaks the chain and gives a wrong HCF.

The answer is the last divisor

When the remainder reaches 0, the HCF is the number you divided by in that step, not the quotient and not the remainder before it.

The denominators swap jobs

For the HCF of fractions the denominators take their LCM, not their HCF. Check any answer by dividing: each fraction divided by the HCF must give a whole number.

Lowest terms first

24\tfrac 24 and 36\tfrac 36 are both 12\tfrac 12, so their HCF is 12\tfrac 12. Applying the rule to the unreduced fractions gives 112\tfrac{1}{12}, which is wrong.

HCF and LCM Together: the Product Rule and Coprime Parts

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Product of two numbers equals HCF times LCM

Product rule (two numbers only)

a×b=HCF(a,b)×LCM(a,b)a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)

Numbers as the HCF times coprime parts

Coprime parts

a′=Ha,  b′=Hb,  HCF(a,b)=1  ⇒  LCM=Haba' = Ha,\; b' = Hb,\; \text{HCF}(a, b) = 1 \;\Rightarrow\; \text{LCM} = Hab

What the HCF and LCM can and cannot be

FactExample
The LCM is a multiple of the HCFHCF 6 allows LCM 90, since 90=15×690 = 15 \times 6
HCF ≤ the smaller number, LCM ≥ the larger14 and 35: HCF 7, LCM 70
Coprime numbers: HCF 1, LCM is the product9 and 16: HCF 1, LCM 144
One number divides the other: HCF is the smaller, LCM the larger13 and 39: HCF 13, LCM 39
Two consecutive numbers are coprime20 and 21: HCF 1, LCM 420

Common traps

Two numbers only

For 3, 6 and 12 the product is 216, but HCF × LCM is 3×12=363 \times 12 = 36. The product rule holds for a pair of numbers, never for three or more.

Check the answer is a whole number

If HCF × LCM divided by the given number is not a whole number, or its HCF with that number is not the HCF given, recheck the arithmetic before picking an option.

Reduce the ratio first

Numbers in the ratio 6:86 : 8 with HCF 5 are 15 and 20, because 6:8=3:46 : 8 = 3 : 4. Writing 30 and 40 gives a pair whose HCF is 10.

The LCM of the ratio terms, not their product

Numbers 3k,4k,6k3k, 4k, 6k have LCM 12k12k, since LCM(3,4,6)=12\text{LCM}(3, 4, 6) = 12. Multiplying the terms to get 72k72k works only when the terms share no factor.

Ruling out is not proving

Any divisor of the LCM could be the HCF. A question asking what the HCF will be can only rule out the options that do not divide the LCM; the one left is the answer only because the others fail.

Greatest and Least Number Problems: Tiles, Bells and Remainders

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The greatest number that divides or measures: tiles, containers and remainders

Greatest divisor leaving remainders

HCF(a−r1,  b−r2,  c−r3)\text{HCF}(a - r_1,\; b - r_2,\; c - r_3)

Bells, lights and laps: when events coincide again

Next time together

next meeting=start+LCM(a,b,c)\text{next meeting} = \text{start} + \text{LCM}(a, b, c)

The least number leaving given remainders

Least number with a remainder

N=LCM(a,b,c)×m+r,m=1,2,3,…N = \text{LCM}(a, b, c) \times m + r, \quad m = 1, 2, 3, \ldots

Multiples of the LCM in a range or with a condition

Multiples of the LCM

multiple=L×m,L=LCM(a,b,c)\text{multiple} = L \times m, \quad L = \text{LCM}(a, b, c)

Common traps

Take the remainder off first

The HCF of the original numbers answers a different question. If each division must leave 7, the divisor goes exactly into each number minus 7.

One unit before the HCF

A floor 3 m 60 cm by 2 m 40 cm is 360 cm by 240 cm, so the largest tile is 120 cm. Mixing metres and centimetres gives a meaningless HCF.

Counting the start

If bells coincide every 15 minutes, in 2 hours they coincide 8 times after the start, and 9 times if the start counts. Read which one the question asks.

Together again is an LCM

The next time together is the smallest common multiple of the intervals. Adding the intervals, or taking their HCF, gives a time when they are not all ringing.

Different remainders, same gap

When the remainders differ, compare each with its divisor. Remainders 3, 8 and 13 on 7, 12 and 17 are each 4 short, so subtract 4 from the LCM; adding any of the remainders is wrong.

Adding the remainder can add a digit

For the greatest 3-digit number leaving remainder 7 on division by 333, 999 plus 7 has four digits. Step back one multiple: 666+7=673666 + 7 = 673.

Divisible by all is the LCM, not the product

A number divisible by 6 and 10 is a multiple of 30. Using the product, 60, misses 30 and 90.

Largest to subtract leaves one copy, not zero

Subtracting the whole number leaves 0, which every number divides, but the question means a positive result. Leave the smallest positive multiple, the LCM itself.

Read which number is bounded

'The greatest 5-digit number to be added' bounds the number added, not the sum. The sum can be as large as the given number plus 99,999.

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