How to Solve It
This is a classic HCF remainder problem — one of the most common question types in number theory exams. This guide walks you through exactly what the question is asking, the core formula you need, how to approach every step, and where students go wrong. No answer given — just the method, clearly explained.
🔢 Stuck on number theory or HCF problems? Our maths specialists explain it step by step.
Get Maths Help →What the Question Is Really Asking — Before You Touch Any Numbers
Find the largest number that divides 398, 436, and 542 and leaves remainders of 7, 11, and 15 respectively.
That word respectively matters. It means: when the unknown number divides 398, the remainder is 7. When it divides 436, the remainder is 11. When it divides 542, the remainder is 15. Each divisor–remainder pair is linked. You are not looking for a number that leaves any combination of those remainders — the pairing is fixed.
The question is asking for the largest such number. That word is the second clue. When you want the largest common divisor of something in mathematics, you are working with the Highest Common Factor — HCF, also called the Greatest Common Divisor (GCD). So this is an HCF problem. The work is figuring out exactly what to take the HCF of.
Read the Question Twice Before Starting
The most common mistake on this type of problem happens before any calculation — students jump straight to finding HCF of 398, 436, and 542 directly. That gives the wrong answer. The numbers you feed into the HCF calculation are not the original numbers. You need to transform them first. The formula below explains exactly what that transformation is.
The Core Idea — Why HCF Solves This
Call the unknown number d. When d divides 398 and leaves remainder 7, that means d divides (398 − 7) exactly — no remainder. Why? Because if 398 = d × q + 7 for some quotient q, then 398 − 7 = d × q, which is perfectly divisible by d.
The same logic applies to all three numbers. So:
398 ÷ d leaves remainder 7
So d divides (398 − 7) exactly.
d divides 391 exactly436 ÷ d leaves remainder 11
So d divides (436 − 11) exactly.
d divides 425 exactly542 ÷ d leaves remainder 15
So d divides (542 − 15) exactly.
d divides 527 exactlyNow you have three numbers — 391, 425, and 527 — that d must divide exactly. You want the largest such d. The largest number that divides all three exactly is their HCF. That is the connection. That is why this is an HCF problem.
The insight is simple once you see it: subtracting each remainder from its paired number removes the leftover, leaving something the unknown divisor fits into perfectly.
— Core principle of HCF-remainder problems in number theoryThe Key Formula — Write This Down
If a number d divides N₁, N₂, N₃ leaving remainders r₁, r₂, r₃ respectively, then:
d = HCF of (N₁ − r₁), (N₂ − r₂), (N₃ − r₃)
Provided that d must be greater than each remainder (since a remainder must always be less than the divisor — this is a constraint you need to verify at the end).
Applied directly to this question:
398 − 7 = ?
Subtract the first remainder from the first number. You get a new number that d must divide exactly.
Work this out yourself436 − 11 = ?
Subtract the second remainder from the second number. Again, d must divide this exactly.
Work this out yourself542 − 15 = ?
Subtract the third remainder from the third number. D must divide this exactly too.
Work this out yourselfOnce you have those three results, your answer is the HCF of all three. That single HCF is the largest number d that satisfies all three divisibility conditions simultaneously.
An Alternative Route — Differences Method
Some textbooks and exam solutions use a slightly different but equivalent approach: instead of subtracting remainders from the original numbers, they find the HCF of the pairwise differences of the adjusted numbers. The two methods give the same answer. The subtract-then-HCF method shown above is generally cleaner for three numbers. Use whichever your course or textbook presents — both are valid, both earn marks.
How to Approach Each Step of Your Working
Exam markers and tutors want to see clear, logical working — not just a final number. Here is exactly how to lay out your solution for this problem.
Solution Framework — Largest Number Leaving Remainders 7, 11, 15
How to structure your working for full marks — without giving the answer away
Write: “Let the required number be d. We need d such that when 398, 436, and 542 are divided by d, the remainders are 7, 11, and 15 respectively.” One sentence setting up the problem. This shows the examiner you have read the question correctly and are not just applying a formula blindly.
Write: “Since d leaves remainder 7 when dividing 398, d must divide (398 − 7) exactly.”
Repeat the same statement for the other two pairs.
398 − 7 = ___ 436 − 11 = ___ 542 − 15 = ___ Do the three subtractions and write the results. These three values are what you will find the HCF of.
Write: “The largest such d is the HCF of [your three results from Step 2].”
This is the logical link between the problem setup and the HCF calculation — make it explicit. Do not skip this statement. It shows you understand why you are computing an HCF, not just that you know you should.
Use either the prime factorisation method or the Euclidean algorithm (division method) to find the HCF of your three numbers. Show all working — do not just state the HCF. The working is worth marks even if you make an arithmetic error.
HCF(a, b, c) = HCF(HCF(a, b), c) See Section 5 below for how to execute both methods clearly.
Your answer d must be larger than all three remainders (7, 11, and 15). This is because in any division, the remainder is always strictly less than the divisor. Check this condition and state it explicitly.
d > 15? ← check this If your HCF turns out to be 17, say — that is greater than 15, so the constraint is satisfied. If it were, say, 5 (less than some remainders), the problem would have no valid solution as stated.
Write: “Therefore, the largest number that divides 398, 436, and 542 leaving remainders 7, 11, and 15 respectively is [your HCF].”
✓ Box or underline your final answer — exam markers scan for it One sentence. Mirrors the language of the question. Do not just write a number alone.
Two Methods for Computing HCF — Which One to Use
Once you have your three adjusted numbers, you need to find their HCF. There are two standard methods. Both work. Pick whichever your course uses.
Break Each Number into Prime Factors, Then Find Common Factors
Write each of your three adjusted numbers as a product of prime factors. The HCF is the product of every prime factor that appears in all three factorisations, each taken to its lowest power. This method is clean and visual — great for smaller numbers where the prime factors are easy to find.
HCF = product of shared prime factors at lowest powersRepeated Division Until Remainder Is Zero
Take two of your numbers. Divide the larger by the smaller. Take the remainder. Divide the previous divisor by that remainder. Repeat until the remainder is zero. The last non-zero remainder is the HCF of those two numbers. Then find HCF(that result, third number) the same way. This works for any number size and is faster for larger numbers.
HCF(a, b): divide until remainder = 0How to Show Your HCF Working Clearly
Correct vs. incomplete working — for the Euclidean algorithm approach
A Reliable Check: HCF of Three Numbers = HCF(HCF(First Two), Third)
You do not have to find HCF of all three numbers at once. Find HCF of the first two, then find HCF of that result with the third number. The final result is the same. This is useful with the Euclidean algorithm — you can apply it twice, each time working with just two numbers, which is easier to manage without errors.
How to Check Your Answer Before You Write It Down
Once you have a candidate answer for d, do not write it down immediately. Take 30 seconds to verify. This is the difference between a confident correct answer and an answer you are not sure about.
🔍 Three Verification Checks — Do All Three Before Finalising
Divide 398 by your answer d. Does the remainder equal 7 exactly? If not, either your subtraction in Step 2 went wrong or your HCF calculation has an error. Trace back.
Divide 436 by d — remainder should be 11. Divide 542 by d — remainder should be 15. All three must pass. If only two pass, the HCF calculation likely has an error.
Confirm d is greater than 15 (the largest remainder). A divisor must always be greater than its remainder. If d ≤ 15, something has gone wrong earlier in your working.
These checks are also a useful exam technique if you finish with time to spare. Work backwards from your answer to verify it satisfies all the original conditions. A wrong answer you have verified is still wrong — but running the checks gives you confidence to move on or tells you exactly where to look for an error.
Variations of This Problem Type You Will See in Exams
The same underlying method handles a family of related questions. Know how each variation differs so you are not thrown when the numbers or framing change.
| Variation | How the Question Is Phrased | What Changes in the Method | Key Watch-Out |
|---|---|---|---|
| Same remainder for all | “Find the largest number that divides 245, 1029, and 1345 leaving the same remainder in each case.” | You do not know the remainder — but it cancels out. Take pairwise differences of the original numbers and find HCF of those differences. | Do not subtract a specific remainder. Instead: HCF(1029−245, 1345−1029, 1345−245) — or HCF of any two differences. |
| Zero remainder (exact divisibility) | “Find the largest number that exactly divides 270, 315, and 405.” | All remainders are zero. HCF of the original numbers directly — no subtraction needed. | Straightforward HCF — make sure you recognise it as a special case of the remainder formula with r=0. |
| Different remainders (this question) | “Leaves remainders 7, 11, and 15 respectively.” | Subtract each specific remainder from its paired number. Find HCF of the three results. | The pairing is fixed — remainder 7 goes with 398, not with any other number. Do not mix them up. |
| Two numbers only | “Find the largest number that divides 616 and 830 leaving remainders 6 and 8.” | Same method — just two adjusted numbers instead of three. HCF of (616−6) and (830−8). | Easier computation — but the same logical setup applies. Always subtract before computing HCF. |
| Remainder given as a fraction of divisor | Some competitive exam questions give the remainder as a fraction of the number. Algebraically different — set up an equation. | This is a harder variant that requires algebraic manipulation before applying HCF. | Only appears in competitive entrance exams (CAT, GMAT). Not the standard school or university format. |
The Pattern That Links All Variations
Every variation of this problem type reduces to the same question: what number divides a set of quantities exactly? The strategy is always to transform the problem so you have a set of numbers that your unknown divisor divides perfectly — then find the HCF of that set. The variation changes how you arrive at that set, not what you do with it once you have it.
Mistakes That Cost Marks on This Type of Question
| # | ❌ Mistake | Why It Happens | ✓ How to Avoid It |
|---|---|---|---|
| 1 | Finding HCF of the original numbers (398, 436, 542) directly | The most common error. Students know the answer involves HCF so they jump straight to HCF of the given numbers without stopping to apply the formula. | Always subtract the remainders first. Write the formula at the top of your working before you do any arithmetic: “d = HCF(N₁−r₁, N₂−r₂, N₃−r₃).” That forces you to complete the subtraction step. |
| 2 | Mixing up which remainder belongs to which number | The question says “respectively” — this means the pairing is fixed. Students occasionally subtract 11 from 398 or 7 from 436, which gives completely wrong adjusted numbers. | Before calculating, write three paired statements: 398→r=7, 436→r=11, 542→r=15. Keep these visible throughout your working and check each subtraction against the correct pair. |
| 3 | Arithmetic error in the subtraction step | Simple calculation slips — especially under exam pressure — produce wrong adjusted numbers, which then produce a wrong HCF even if the method is perfect. | Do each subtraction explicitly and check it. 398 − 7: write 398, subtract 7, get 391. Check: 391 + 7 = 398. ✓ Takes five seconds and catches errors before they cascade. |
| 4 | Stopping at HCF of two numbers instead of all three | Students find HCF of the first two adjusted numbers and forget the third. This gives a number that divides two of the adjusted values but not necessarily the third. | After finding HCF of the first two adjusted numbers, explicitly compute HCF(that result, third adjusted number). Write it out: “HCF(a, b) = x. Now HCF(x, c) = ?” |
| 5 | Not verifying that d > largest remainder | If the HCF is less than or equal to the largest remainder (15 in this case), the problem is logically inconsistent — you cannot have a remainder larger than the divisor. Students miss this constraint check entirely. | After finding the HCF, write one check line: “d = [answer] > 15 (largest remainder). ✓ Constraint satisfied.” If d were ≤ 15, you would need to recheck your working. |
| 6 | No working shown — just a final number | In competitive or school exams, writing only the final answer gives no partial credit if it is wrong, and raises doubt about method even if it is right. | Show every step: the three subtractions, the HCF working (factor tree or Euclidean steps), and the constraint check. Even if you make an arithmetic slip, method marks are available for correct logical steps. |
| 7 | Confusing HCF with LCM | Some students use LCM methods for problems that ask for the largest divisor, perhaps from muscle memory of LCM-type problems on previous topics. | The trigger word is “largest number that divides” — that points to HCF. “Smallest number divisible by” or “smallest multiple” points to LCM. Recognise the trigger phrase and make a conscious method choice at the start. |
The Mathematical Background — Where This Method Comes From
The approach used here is grounded in the Division Algorithm, which is one of the fundamental theorems of number theory. It states that for any integers a and b (with b > 0), there exist unique integers q (quotient) and r (remainder) such that a = bq + r, where 0 ≤ r < b.
This theorem guarantees that when we write 398 = d × q + 7, the relationship is exact and unique — no ambiguity. Subtracting the remainder from both sides gives 398 − 7 = d × q, which means d divides (398 − 7) exactly. That is the step the entire HCF method rests on.
Division Algorithm (Number Theory)
The formal statement that any integer can be expressed uniquely as quotient × divisor + remainder, with the remainder between 0 and the divisor. This is the theorem your method applies.
Reference: Hardy & Wright, An Introduction to the Theory of Numbers · OpenStax PrecalculusEuclidean Algorithm
Dating to Euclid’s Elements (~300 BCE), this is one of the oldest algorithms in mathematics. The Khan Academy article on the Euclidean algorithm covers it step by step with worked examples directly relevant to this problem type.
khanacademy.org → search “Euclidean algorithm” — verified free resourceGCD/HCF in Competitive Mathematics
The NCERT Class 10 Mathematics textbook (India) and AQA/Edexcel GCSE Mathematics specifications both include HCF remainder problems as a core topic. Your course notes or textbook will have the same method presented in your syllabus’s preferred notation.
ncert.nic.in · aqa.org.uk · edexcel.comVerified External Resource: Khan Academy on GCD and the Euclidean Algorithm
Khan Academy’s free article and video on the Greatest Common Divisor and the Euclidean Algorithm — available at khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/the-euclidean-algorithm — provides a clear, worked explanation of the Euclidean algorithm with step-by-step division tables. It is one of the most reliable free mathematics resources available and directly covers the computational method you need for Step 4 of this problem. Use it to check your HCF computation technique, not the specific numbers in this question.
HCF Remainder Problem FAQs
The Method in One Paragraph
This problem type trips students up not because the mathematics is difficult, but because the setup step — subtracting the remainders before computing HCF — is easy to miss when you are under exam pressure. Once you have the three adjusted numbers, the rest is mechanical: compute HCF using whichever method your course teaches, verify the constraint, write the answer clearly.
The skill being tested here is recognising the problem type from the language of the question — “largest number,” “divides,” “leaving remainders respectively” — and knowing immediately that this calls for subtract-then-HCF. That recognition is what separates students who get these right reliably from those who occasionally stumble on the setup.
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