Two-step LCM/HCF cases (extra condition, N-digit bounds) — notes, question types & practice · HCF & LCM

01

Overview

The harder questions in this topic add one extra condition to a plain LCM or HCF setup. The master frame is $N = L \times k + r$ , where $L$ is the LCM of the divisors. For divisors 5 and 6 with remainder 2: $L = 30$ , so $N = 30k + 2$ , giving 32, 62, 92, and so on.

02

The master frame

$k = 1$ gives the least positive value; larger k walks up the same family. Any extra condition, such as divisibility by one more number or a digit bound, simply selects k.

Rule: Write $N = L \times k + r$ first. Then fit the extra condition onto k.

03

An extra divisibility condition

Test $k = 1, 2, 3, \ldots$ until $Lk + r$ satisfies the condition. For 'N leaves remainder 2 with 5 and 6, and N is divisible by 7': $N = 30k + 2$ . Trying k gives 32, 62, 122, 152, 182, and $182 = 7 \times 26$ works. So $N = 182$ .

04

Least and greatest n-digit numbers

Multiples of $L$ are the candidates. Least four-digit multiple of 60: $1000 = 16 \times 60 + 40$ , so add $60 - 40 = 20$ to reach 1020. Greatest works from the top: divide 99,999 by 60, subtract the remainder, and land on the greatest multiple under the bound.

Tip: Step up from 10, 100, 1000 for the least; step down from 9, 99, 999 for the greatest. With a remainder r, fit $Lk + r$ inside the range instead of plain multiples.

05

Rebuilding numbers from HCF and LCM

Numbers with HCF $h$ are $ha$ and $hb$ with co-prime parts, and $ab = \dfrac{\text{LCM}}{h}$ . With HCF 4 and LCM 48: $ab = 12$ , whose co-prime pairs are $(1, 12)$ and $(3, 4)$ . A given sum picks the pair: sum 28 needs $a + b = 7$ , so the parts are 3 and 4 and the numbers are 12 and 16.

Watch: List only co-prime factor pairs. The pair $(2, 6)$ for $ab = 12$ is impossible, because the numbers would share an extra factor 2.

06

Counting possible pairs

Count the co-prime factor pairs of $ab$ , and remember the pair $(1, ab)$ . Product 1764 with HCF 14 gives $ab = \dfrac{1764}{14^2} = 9$ , and 9 has the single co-prime pair $(1, 9)$ : the numbers must be 14 and 126. Count each unordered pair once, with the smaller part first, so $(1, 9)$ and $(9, 1)$ are the same pair.

07

Sanity checks that catch slips

Two quick filters catch most slips in this subtopic. First, the HCF must divide the LCM, both numbers, and their sum and difference. Second, every rebuilt pair must multiply back to the product and carry the stated HCF. Run both checks before marking the answer.

08

The same unknown remainder

When no remainder is given but it is the same for all numbers, the divisor divides every pairwise difference. For 51, 123 and 171: the differences are 72, 48 and 120, and $\gcd(72, 48, 120) = 24$ . The common remainder is $51 \bmod 24 = 3$ , the same for all three.

Example: Check: $123 = 5 \times 24 + 3$ and $171 = 7 \times 24 + 3$ . Both leave remainder 3, so 24 is right.

09

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Least or greatest n-digit number divisible by a set

How to spot it:

The question asks for the least or greatest number of a given digit count divisible by several divisors.

\text{least: next multiple of } L \text{ at or above } 10^{n-1}; \quad \text{greatest: } 10^n - 1 - \left((10^n - 1) \bmod L\right)

Method

Take $L$ as the LCM of the divisors.
Least: divide the smallest n-digit number by $L$ and step up to the next multiple.
Greatest: divide the largest n-digit number by $L$ and subtract the remainder.
With a remainder r in the question, fit $Lk + r$ inside the range instead.

Why it works:

Multiples of the LCM are exactly the numbers divisible by every divisor; the digit bounds only pick where to stop.

Try this

Find the greatest five-digit number exactly divisible by 12, 15 and 20.

Show solution

$L = \text{LCM}(12, 15, 20) = 60$ .
$99999 = 1666 \times 60 + 39$ .
Greatest $= 99999 - 39 = 99960$ .

Answer

99960

Type 2common2 practice Q

LCM + r with an extra condition

How to spot it:

An LCM plus remainder question carries one more filter, such as divisibility by another number or a digit bound.

N = Lk + r, \qquad \text{smallest } k \text{ meeting the condition wins}

Method

Write $N = L \times k + r$ with $L$ the LCM of the divisors.
Test $k = 1, 2, 3, \ldots$ against the extra condition.
With an n-digit bound, start k just above the bound minus r, divided by L.

Why it works:

N minus r is a multiple of L by construction; the extra condition only chooses which multiple.

Try this

Find the least number which leaves remainder 3 with 6, 7 and 8, and is divisible by 5.

Show solution

$L = \text{LCM}(6, 7, 8) = 168$ , so $N = 168k + 3$ .
Test k: $171, 339, 507, 675$ ; and $675 = 5 \times 135$ .
$N = 675$ .

Answer

675

Type 3common4 practice Q

Rebuild numbers from HCF and LCM; count pairs

How to spot it:

HCF and LCM are given with a sum, difference or bound; find the numbers, or count the possible pairs.

\text{numbers} = ha,\ hb; \quad \gcd(a, b) = 1, \quad ab = \dfrac{\text{LCM}}{h}

Method

Compute $ab = \text{LCM} \div h$ (or product $\div h^2$ ).
List factor pairs of $ab$ and keep only co-prime pairs.
Use the given sum or difference to pick the pair; the numbers are $ha$ and $hb$ .
To count pairs, count the surviving co-prime pairs.

Why it works:

Two numbers with HCF h are h times two co-prime parts, and those parts must multiply to LCM divided by h.

Try this

The HCF and LCM of two numbers are 6 and 36. How many pairs of numbers fit?

Show solution

$ab = \dfrac{36}{6} = 6$ .
Co-prime pairs of 6: $(1, 6)$ and $(2, 3)$ .
Pairs: 6 with 36, and 12 with 18. That is 2 pairs.

Answer

2

Type 4common2 practice Q

Greatest number dividing with the same unknown remainder

How to spot it:

The question asks for the greatest number dividing a, b, c and leaving the same remainder in each case, with no remainder given.

\text{answer} = \gcd(a - b,\ b - c,\ a - c)

Method

Take the pairwise differences of the numbers.
The HCF of the differences is the greatest such divisor.
The common remainder is any number mod that HCF; confirm it is the same for all.

Why it works:

If two numbers leave the same remainder, their difference is exactly divisible by the divisor, so it divides every difference.

Try this

Find the greatest number which divides 61, 109 and 133 leaving the same remainder in each case.

Show solution

Differences: $109 - 61 = 48$ , $133 - 109 = 24$ , $133 - 61 = 72$ .
$\gcd(48, 24, 72) = 24$ .
Remainder: $61 \bmod 24 = 13$ , and $109 \bmod 24 = 133 \bmod 24 = 13$ .

Answer

24

Type 5common

LCM + r inside a digit bound

How to spot it:

The question asks for the least (or greatest) number of a given digit count leaving remainder r with several divisors.

N = L \times k + r, \quad 10^{n-1} \le N < 10^n

Method

Write $N = L \times k + r$ with $L$ the LCM of the divisors.
Least n-digit: take the smallest k with $Lk + r$ at or above the smallest n-digit number.
Greatest n-digit: take the largest k with $Lk + r$ at or below the largest n-digit number.
Check the remainder with every divisor.

Why it works:

The family of valid numbers is exactly LCM multiples plus r; the digit bound selects the first or last member.

Try this

Find the least four-digit number which leaves remainder 3 when divided by 6, 7 and 8.

Show solution

$L = \text{LCM}(6, 7, 8) = 168$ , so $N = 168k + 3$ .
Need $168k + 3 \ge 1000$ , so $k \ge 5.97$ , giving $k = 6$ .
$N = 6 \times 168 + 3 = 1011$ . Check: $1011 \bmod 6 = 1011 \bmod 7 = 1011 \bmod 8 = 3$ .

Answer

1011

10

Formula sheet

Extra divisibility condition

N = Lk + r, \quad N \equiv 0 \pmod{p} \ \Rightarrow\ Lk \equiv -r \pmod{p}

Reconstruction from HCF and LCM

ab = \dfrac{\text{LCM}}{h}, \quad \gcd(a, b) = 1, \quad \text{numbers} = ha,\ hb

Pair count

\#\{(a, b): ab = M,\ \gcd(a, b) = 1,\ a \le b\}

Same unknown remainder

\text{answer} = \gcd(a - b,\ b - c,\ a - c)

11

Shortcuts that save time

⚡ Test k until it clicks

Once N = LCM x k + r is written, only k remains. Test k = 1, 2, 3 against the extra condition; small values click fast.

Example

Find the least number which leaves remainder 1 with 3, 5 and 7, and is divisible by 8.

Show solution

$L = \text{LCM}(3, 5, 7) = 105$ , so $N = 105k + 1$ .
Test: $106, 211, 316, 421, 526, 631, 736$ ; $736 = 8 \times 92$ .
$N = 736$ .

Answer

736

⚡ n-digit multiple scan

For the least n-digit multiple, divide the smallest n-digit number by the LCM and step up to the next multiple. For the greatest, subtract the remainder from the largest n-digit number.

Example

Find the least three-digit number exactly divisible by 9 and 12.

Show solution

$\text{LCM}(9, 12) = 36$ .
$100 = 2 \times 36 + 28$ , so step up by $8$ .
$108 = 3 \times 36$ .

Answer

108

⚡ Sum or difference picks the pair

With HCF, LCM and a sum or difference given, list co-prime pairs of LCM / HCF and match h times the parts to the sum.

Example

Two numbers have HCF 6, LCM 180 and sum 66. Find the numbers.

Show solution

$ab = \dfrac{180}{6} = 30$ ; co-prime pairs $(1, 30)$ , $(2, 15)$ , $(3, 10)$ , $(5, 6)$ .
Parts sum: $6(a + b) = 66$ gives $a + b = 11$ .
Pair $(5, 6)$ : numbers 30 and 36.

Answer

30 and 36

12

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Answering LCM + r when an extra condition forces a larger multiple.

Fit N = LCM x k + r through the extra condition and take the smallest k that works.

Mistake 02

Listing factor pairs of LCM / HCF without checking co-primality.

Only co-prime pairs are possible; drop the rest before matching the sum.

Mistake 03

Forgetting the pair (1, M) when counting possibilities.

That pair is always co-prime and always counts.

Mistake 04

Counting (a, b) and (b, a) as two pairs.

Order does not matter; count each pair once with a at most b.

Mistake 05

Skipping the check that the HCF divides the sum or difference.

The HCF must divide both numbers, so it divides their sum and difference too.

13

Quick revision

Read this the night before the exam.

Master frame: $N = L \times k + r$ ; extra conditions only choose $k$ .
Least n-digit multiple: step up from $10 \ldots 0$ ; greatest: step down from $99 \ldots 9$ .
Numbers from HCF and LCM: $ab = \text{LCM} \div h$ , co-prime parts only.
Pair count: co-prime factor pairs of LCM $\div h$ (or product $\div h^2$ ).
Same unknown remainder: HCF of pairwise differences.
Common remainder check: any number mod the answer, same value for all.

14

Practice: 14 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 11 min · wrong answers go to your mistake notebook automatically.