Welcome! Today we're tackling Building Numbers from LCM — where the LCM isn't the final answer, but just the starting point.

You've been computing HCF and LCM as endpoints.

Now the LCM becomes a stepping stone — the answer to the actual question is built FROM the LCM.

We'll see three ways to build the final answer:

Construction Type	Final Step
Remainder problems	Add the remainder
Overshoot problems	Subtract the overshoot
Constant-deficit problems	Subtract the deficit

These construction problems are some of the trickiest in the chapter.

The difference between adding and subtractingkey in the final step is where most marks are lost.

By the end of this section, you'll:

Handle all three types confidently
Know exactly when to add and when to subtract

1. Least number with a given remainder: why you ADD

Building Numbers from LCM 🔧

You've been computing HCF and LCM as endpoints. Now let's see how the LCM becomes a stepping stone — the answer to the actual question is built FROM the LCM.

The first construction type asks for the smallest number that leaves a specific remainder when divided by several divisors.

The LCM gives you the clean multiple — and the answer is a step beyond itkey.

📋 Given Info

Key Insight 💡

If a number leaves remainder 7 when divided by 35, then:

$(number - 7)$ is exactly divisible by 35

Similarly for the other divisors:

$(number - 7)$ is exactly divisible by 56
$(number - 7)$ is exactly divisible by 91

So $(number - 7)$ must be a common multiple of all the divisors.

✍️ Question

Your Challenge 🎯

Find the least number which when divided by 35, 56, and 91 leaves remainder 7 in each case.

Show why you ADD the remainder to the LCM rather than subtracting it.

(Work through the factorisation, find the LCMkey, and explain your reasoning for the final step)

Think about what the question says: the number leaves remainder 7Same remainder every time — that's the pattern when divided by 35, 56, and 91.

This means:

$number = 35 \times (some quotient) + 7$
$number = 56 \times (some quotient) + 7$
$number = 91 \times (some quotient) + 7$

✍️ MCQ

Choose one

If a number leaves remainder 7 when divided by 35, what can we say about

(\text{number} - 7)

?

✓

Rearranging: $(\text{number} - 7)$ When you subtract 7, it becomes exactly divisible is divisible by 35, 56, and 91. So $(number - 7)$ is a common multipleNumber minus 7 is a common multiple of all three of all three.

For the LEAST such number, we want the smallest common multiple: $(\text{number} - 7) = \text{LCM}(35, 56, 91)$ key formulaFor the LEAST number, we need the LCM.

✍️ MCQ

Choose one

If

(\text{number} - 7) = \text{LCM}(35, 56, 91)

, how do we find the number itself?

✓

Computing the LCM:

Let's find the prime factorisationBreak each number into its prime building blocks first of each divisor:

$35 = 5 \times 7$

$56 = 2^{3} \times 7$

$91 = 7 \times 13$

All primes involved: $2, 5, 7, 13$ .

Notice that 7 appears in ALL three numbers.

For LCM, we take the highest powerNot the lowest, not what's common — the highest of each prime:

Highest power of $2$ : $2^{3}$
Highest power of $5$ : $5^{1}$
Highest power of $7$ : $7^{1}$
Highest power of $13$ : $13^{1}$

✍️ MCQ

Choose one

Why do we take the highest power of each prime when finding the LCM?

✓

$\text{LCM} = 2^3 \times 5 \times 7 \times 13 = 8 \times 5 \times 7 \times 13 = 3640$

So the LCM is 3640.answer

✍️ FIB

Fill in the blank

If

\text{LCM} = 3640

and (number

- 7

)

= \text{LCM}

, what is the required number?

✓

So $number - 7 = 3640$ , meaning $number =$ $3640 + 7$ $=$ $\mathbf{3647}$ answer.

Why ADD? Because the number is the LCM plus the leftover. The LCM is the last 'clean' multipleIt sits beyond the nearest clean multiple before the number. The number sits 7 units beyond+7Add the remainder to the LCM, never subtract it.

✍️ MCQ

Choose one

If you divide

3647

by

35

, what remainder do you get?

✓

The number line shows exactly what 'remainder 7' means geometrically — the number sits 7 units beyond the clean LCM multiple. This visual makes clear why we ADD the remainder rather than subtract.

✍️ MCQ

Choose one

The number leaves remainder 7. Should it be to the LEFT or RIGHT of the LCM on this number line?

✓

Verify:

$3647 \div 35 = 104$ remainder $7$ ✓

$3647 \div 56 = 65$ remainder $7$ ✓

$3647 \div 91 = 40$ remainder $7$ ✓

All three divisions give remainder 7Verify by dividing and checking remainders match — our answer is correct!Every division must give the same remainder

✍️ MCQ

Choose one

If a number leaves remainder

5

when divided by

12

,

18

, and

24

, what would be the form of the answer?

✓

2. Greatest N-digit multiple: why you SUBTRACT

New Challenge: Going Down Instead of Up 📉

In the previous problem, we built UP from 0 by adding a remainder to a multiple of the LCM.

Now we reverse direction:

Instead of building up from 0, you start from the largest N-digit number
Come down to the nearest multiple

This time you subtract.

📋 Given Info

Given Information:

The greatest 4-digit number is 9999
You need the largest multiple of $LCM (15, 24, 36)$ that does not exceed 9999
If 9999 is not itself a multiple, you go DOWN to the nearest one

✍️ Question

Problem 🧮

Find the greatest 4-digit number exactly divisible by 15, 24, and 36.

Show:

The LCM computation (using prime factorization)
The division of 9999 by the LCM
Explain why you subtract the remainder from 9999

Step 1 — Find LCM(15, 24, 36)

First, we break each number into prime factors:

$15 = 3 \times 5$
$24 = 2^{3} \times 3$
$36 = 2^{2} \times 3^{2}$

All primes involved: 2, 3, 5. Take the highest power of eachTake the highest power of each prime factor:

$2^{3} = 8$
$3^{2} = 9$
$5^{1} = 5$

LCM = $8 \times 9 \times 5 = 360$ 360 is divisible by all three numbers

✍️ MCQ

Choose one

We found that

\text{LCM}(15, 24, 36) = 360

. What does this tell us about the number we're looking for?

✓

Step 2 — Divide 9999The number we're checking against the LCM by 360We divide by 360 to find how close we are to a multiple

$9999 \div 360 = 27$ Division shows the remainder is 279 with remainder 279How far 9999 overshoots the nearest multiple.

✍️ MCQ

Choose one

Why do we divide 9999 by 360 in this problem?

✓

This means: $360 \times 27 = 9720$ The largest 4-digit multiple of 360, and $9999 - 9720 = 279$ Subtract 279 to get a number divisible by all three.

✍️ MCQ

Choose one

If

9999 \div 360

gives remainder

279

, what is the greatest 4-digit number exactly divisible by 360?

✓

The number line shows exactly how 9999 overshoots the last clean multiple of 360. The remainder of 279 is the 'extra' we must subtract to land on a perfect multiple.

Step 3 — Subtract the remainder

9999 is 279 units PAST the last multiple of 360. Think of it as standing at 9999 and walking backward to the nearest multiple. You walk back 279 steps.

$9999 - 279 = 9720$

✍️ MCQ

Choose one

In the previous problem, we added the remainder to find the answer. In this problem, we subtracted the remainder. Why?

✓

Why SUBTRACT (not add)? Adding 279 would give $9999 + 279 = 10278$ — a 5-digit number, which overshoots. We want the greatest 4-digit number, so we go DOWN to 9720.

Verify: $9720 \div 15 = 648$ No remainder means exact divisibility, $9720 \div 24 = 405$ , $9720 \div 36 = 270$ . All exact. ✓

So 9720Answer! is the greatest 4-digit number exactly divisible by 15, 24, and 36.

✍️ MCQ

Choose one

To find the smallest 5-digit number exactly divisible by

360

, after dividing

10000

by

360

and finding the remainder, you would:

✓

⚠️ Common mistake: Adding the remainder instead of subtracting.

❌ Wrong approach	✅ Correct approach
$9999 + 279 = 10278$ Adding gives a 5-digit number, breaking the constraint	$9999 - 279 = 9720$ Subtracting keeps us in the valid range
Goes above 9999 (5 digits!)This breaks the 4-digit requirement	Stays within 4 digits ✓Answer stays within required range

✍️ MCQ

Choose one

A student calculates

9999 + 279 = 10278

as the answer. What is wrong with this approach?

✓

Remember — you're going DOWNGo down from the limit, not up from 9999, not up.

When finding the greatestWe want the largest valid number number in a range, we need the largest multiple that does not exceedMust not go over the upper bound the limit. Subtracting the remainderThis is the key operation takes us back to the nearest multiple below.

3. Constant-deficit problems: recognising equal gaps

The Trickiest Variant 🔍

The trickiest variant looks like it has different remainders — but a hidden pattern turns it into a clean subtraction problem.

The secret is in the gaps between the divisors and their remainders.

📋 Given Info

Here's the problem:

Find the smallest number which when divided by 28 gives remainder 8, and when divided by 32 gives remainder 12.

Notice that the remainders are different (8 and 12), so this doesn't look like the earlier 'same remainder' type we've seen.

✍️ Question

Your Task ✏️

Compute (divisor − remainder) for each condition:

$28 - 8 = ?$
$32 - 12 = ?$

What do you notice?

Use this observation to find the answer. Explain what 'constant deficit' means and why the answer is LCM minus this deficit.

At first glance, the remainders are 8 and 12 — different. But compute the 'gap'Divisor minus remainder gives you the gap between each divisor and its remainder:

$28 - 8 = 20$ $32 - 12 = 20$

Equal!Equal gaps signal a pattern This is the signal for a constant-deficit problemThis is a constant-deficit problem.

The "deficit" or "gap"The number falls short by the same amount is the same (20) for both conditions — this tells us the number falls short of a multipleShort of a multiple by the same amount each time of each divisor by exactly 20.

✍️ MCQ

Choose one

In a constant-deficit problem, the number is always less than a multiple of each divisor by the same amount. If the deficit is

20

and the LCM of the divisors is

L

, what would the answer be?

✓

Now let's fill in those calculations and see the equal deficit of 20 emerge from both conditions — this equality is the key that unlocks the problem.

✍️ MCQ

Choose one

If

\text{LCM}(28, 32) = 224

and the constant deficit is

20

, what is the smallest number satisfying both conditions?

✓

What does this mean? The number is 20 SHORTThe constant deficit for both divisors of the next multiple of each divisor.

Number line showing a number N positioned 8 units past one multiple of 28 and 20 units before the next multiple of 28, with similar visualization for 32 below it

Divided by 28: remainder 8 means the number is 8 past a multiple, but also 20 short of the NEXT multiple28 minus 8 equals 20 ( $28 - 8 = 20$ ).
Divided by 32: remainder 12 means the number is 12 past a multiple, but also 20 short of the NEXT multiple32 minus 12 equals 20 ( $32 - 12 = 20$ ).

In both cases, the number falls short by exactly 20.This tells you it's an LCM problem

✍️ MCQ

Choose one

If a number leaves remainder 8 when divided by 28, how many units is it SHORT of the next multiple of 28?

✓

So (number + 20)Number plus 20 is divisible by both is a multiple of both 28 and 32. The smallest such value is LCM(28, 32)The least common multiple of 28 and 32.

So our answer will be LCM minus 20.LCM minus 20 gives your answer

✍️ T/F

True or False?

In a constant-deficit problem where the deficit is

d

and the divisors are

a

and

b

, the answer is

\text{LCM}(a, b) - d

. True or False?

✓

Let's find the LCM:

$28 = 2^{2} \times 7$ $32 = 2^{5}$

\text{LCM} = 2^5 \times 7 = 32 \times 7 = 224

(The key rule for finding LCM)

✍️ MCQ

Choose one

We found that LCM

(28, 32) = 224

. To get our answer, we need to:

✓

Therefore: number + 20 = 224, so number = 224 - 20 = 204LCM minus 20 gives us 204.

Verify:Catches calculation mistakes before they cost you marks

$204 \div 28 = 7$ remainder $8$ ✓
$204 \div 32 = 6$ remainder $12$ ✓

Both correct!

✍️ FIB

Fill in the blank

In a constant-deficit problem, if the divisors are

18

and

24

, and the deficit is

6

, what is the smallest such number?

✓

Recognition trickThis is the essential technique to identify the problem type: Whenever you see different remainders, compute (divisor - remainder)Calculate this difference for every given condition for each. If they're all equal, it's a constant-deficit problemWhen differences match, use the shortcut formula.

✍️ MCQ

Choose one

A number leaves remainder 5 when divided by 20, and remainder 9 when divided by 24. Compute the deficits

(20 - 5)

and

(24 - 9)

. Is this a constant-deficit problem?

✓

\text{Answer} = \text{LCM} - \text{deficit}

(Subtract the deficit from the LCM to get your answer)

4. Distinguishing the three construction types

You've now seen all three types of LCM construction problems.

The critical skill is telling them apart — because each type has a different final step, and mixing them up gives a completely wrong answer.

Type	Final Operation
Remainder problem	LCM + remainder
N-digit problem	Limit − overshoot
Constant-deficit	LCM − deficit

Let's see if you can classify problems correctly.

✍️ Question

Classification Challenge 🎯

For each problem below, state which type it is and what the final operation is (add or subtract, and what). Do NOT solve — just classify.

(A) Find the least number which when divided by 12 and 18 leaves remainder 4 in each case.

(B) Find the greatest 3-digit number exactly divisible by 8, 12, and 15.

(C) Find the smallest number which when divided by 36 and 45 leaves remainders 11 and 20 respectively.

Let's classify each one.

(A) 'Least number... leaves remainder 4 in each case.'

Same remainder for all divisorsThis phrase signals Type 1 classification. This is Type 1Your classification trigger.

Setup: (number - 4) = LCM(12, 18). Answer = LCM + 4Add L C M and remainder for final answer.

The number sits ABOVENumber is above L C M by remainder amount the LCM by the remainder amount. ADD.keyAlways add when remainder is same for all

✍️ MCQ

Choose one

For Problem A, why do we ADD the remainder to the LCM instead of subtracting it?

✓

(B) 'Greatest 3-digit number exactly divisible by 8, 12, and 15.'

Greatest N-digit exact multiple. This is Type 2Exactly means a clean multiple, no remainder.

Setup: Find LCM(8, 12, 15)Start by finding the LCM of the divisors, then find the largest multiple of LCM that fits in 3 digits.

Divide 999 by LCMSee how much 999 goes past the LCM multiple, get remainder, subtract remainder from 999Subtract to get back to the exact multiple.

✍️ MCQ

Choose one

In Type 2 problems (greatest N-digit exact multiple), why do we subtract the remainder instead of adding it?

✓

The number sits BELOWType 2 answers fall below the digit limit 999. SUBTRACTSince 999 overshoots, subtract to come back down the overshoot.

(See the number line diagram — 999 overshoots the target, so we walk backward.)

(C) 'Divided by 36 leaves remainder 11, divided by 45 leaves remainder 20.'

Different remainders — check deficits: $36 - 11 = 25$ Calculate the deficit for each divisor, $45 - 20 = 25$ Same calculation for the second divisor. Equal!When deficits match, that's your signal This is Type 3Equal deficits means constant-deficit type.

✍️ MCQ

Choose one

In Problem C, the deficits are both

25

. What does this tell us about where the answer sits relative to the LCM?

✓

Setup: (number + 25) = LCM(36, 45). Answer = LCM - 25Answer equals LCM minus the deficit.

The number sits BELOWThe number is positioned below the LCM the LCM by the deficit amount. SUBTRACT.Key!Type 3 uses subtraction, opposite of Type 1

Quick summary:

Type 1Type 1 means you add the remainder (same remainder given): ANSWER = LCM + remainderSame remainder means add
Type 2These types require subtraction (greatest N-digit): ANSWER = limit - overshootSubtract for Types 2 and 3
Type 3Equal deficits also means subtract (equal deficits): ANSWER = LCM - deficitSubtract the deficit

Types 2 and 3 both subtractThis is where marks are lost. Type 1 addsOnly Type 1 uses addition. The common errorThe common mistake in exams is adding when you should subtract, or vice versa.

✍️ MCQ

Choose one

For a constant-deficit problem where the deficit is

25

and

\text{LCM} = 180

, the answer is:

✓

Building Numbers from LCM: Remainder, N-Digit, and Constant-Deficit Problems

1. Least number with a given remainder: why you ADD

2. Greatest N-digit multiple: why you SUBTRACT

3. Constant-deficit problems: recognising equal gaps

4. Distinguishing the three construction types