Unit 2 · Part 1 — Linear Systems with Integer Constraints0%
STEP 1 OF 24 · Lesson Opening
Today: Linear Systems with Integer Constraints
A 60–75 minute lesson that turns AIMO Q3–Q5 multi-variable problems into reliable marks.
📌 What you will learn today
Topic
Solving systems of linear equations where the unknowns are required to be integers — and the special trick of reconstructing integers from their pairwise sums.
Category
Algebra (ALG) — sub-topic Linear Systems.
Solves these AIMO problems
2023 Q32015 Q32014 Q42010 Q5
Four past-paper problems — every one solvable by elimination plus a small integer-trick.
AoPS Reference
Optional background: Introduction to Algebra by Richard Rusczyk, for background in equations, ratios and algebraic modelling. This lesson is self-contained. Publisher's book information.
Why this matters
Linear equations help turn relationships into quantities you can calculate and verify.
The selected official problems practise elimination, pair sums and integer constraints.
Topic coverage and marks vary between papers; this lesson does not promise a particular competition score.
Time required
About 60–75 minutes for the full lesson, plus 30 minutes drilling past papers afterwards.
How this lesson is structured
We start with a picture — two roads crossing on a map. Once you see why each equation is
a road, and the solution is the spot where they meet, the algebra writes itself. Then we
add the integer constraint — the secret ingredient that makes Olympiad problems
different from school problems.
Phase 3 (Steps 8–11): Four worked Olympiad examples, fully explained.
Phase 4 (Step 12): Three practice problems with hints and full solutions.
Phase 5 (Steps 13–16): Four real AIMO past papers in exam format with progressive hints.
Step 17: Lesson quiz (3 problems, auto-graded).
Step 18: Cheat sheet + self-assessment.
Steps 19–24: Atomic-skill checks, a fifth worked example, two synthesis tasks, extra practice and a guided official-problem reference.
Pedagogical note for the student: Linear systems are deceptively simple. The
danger isn't in the algebra — it's in missing the integer constraint or
not noticing that the question asks for a combination, not a single variable.
Read every Olympiad statement twice. Underline what's actually asked.
STEP 2 OF 24 · Phase 1 · Visual
What is a system of equations?
Forget formulas. Start with a map.
Imagine two straight roads on a map. Each road is described by an equation. The roads
only cross at one point. That point is the solution: the only
pair (x, y) that lies on both roads at the same time.
Two lines crossing at exactly one point: the unique solution (x, y) = (2, 3).
Why this matters
A system of equations is just two (or more) equations that must all
hold at the same time. Geometrically, each equation is a line, and the solution is the
intersection. Algebraically, we need a recipe for finding it without drawing.
A 2-variable linear system
x + y = 5
2x + 3y = 13
Solution
x = 2, y = 3 — the unique point on both lines.
🔑 The big idea: For linear equations in real variables, n unknowns are uniquely determined by n independent equations. Independent means that none of these equations can be obtained by combining the others. Any additional equations must be consistent with that solution. Integer or other extra constraints can reduce the possibilities further. This equation-count rule does not apply to arbitrary non-linear systems.
STEP 3 OF 24 · Phase 1 · Visual
Why integer constraints matter
Olympiad problems live on the grid, not the smooth plane.
In school, the solution to x + y = 5 is a whole continuous line — infinitely many
real points. If variables count whole coins or blocks, they must be non-negative integers
(positive if zero is excluded). Money does not automatically mean an integer number of dollars:
it may include cents. Use the units and integer restrictions stated in the question.
The line x + y = 5 passes through infinitely many points, but only 6 lattice points (gold dots) have non-negative integer coordinates: (0,5), (1,4), (2,3), (3,2), (4,1), (5,0).
🔑 An integer restriction is an additional constraint, not another equation. A single non-degenerate linear equation in two real unknowns has infinitely many solutions. But adding
"and the answers must be positive integers" sometimes pins it down to a unique solution
(or a small finite list to check).
What this looks like in AIMO problems
You'll see the integer constraint hiding in plain English:
"How many red blocks…" — count must be a non-negative integer.
"integer weights in kg" — weight must be a positive integer.
"price in whole dollars" — value must be a positive integer.
"distinct integers" — strict ordering, no repeats.
⚠ Common student error: solving a 2-variable system with only one equation
and producing a fractional answer. If you only have one equation, you must use the
integer constraint to finish.
STEP 4 OF 24 · Phase 1 · Visual
Pairwise sums — a hidden system
When the equations are disguised as a list of sums.
Some Olympiad problems give you a list of pair-sums (or triple-sums) and ask you to
reconstruct the original integers. This is a linear system — just one with a lot of
equations and a clean trick to solve them.
Let's say you have 5 distinct integers, ordered a < b < c < d < e. You can form
C(5, 2) = 10 pairwise sums. Visualise them as the entries of a triangular grid:
Each integer (a, b, c, d, e) participates in exactly 4 pairs. So the sum of all 10 pair-sums = 4(a + b + c + d + e).
🔑 The core trick: the sum of all pair-sums (or triple-sums) is a clean
multiple of the total. Always start there. For n integers, each appears in
C(n−1, k−1) of the k-subset sums, so:
Sum of all pair-sums = (n − 1) × (total).
Sum of all triple-sums = C(n − 1, 2) × (total).
💡 The smallest pair-sum is a + b (the two smallest). The largest is d + e
(the two largest). The second smallest is a + c, the second largest is c + e.
These four observations — plus the total — let you reconstruct everything.
STEP 5 OF 24 · Phase 2 · Derivation
Derivation 1 — Concrete elimination
Two equations, two unknowns. The bread-and-butter technique.
Solve the system:
2x + 3y = 13
x + y = 5
Step 1 — Choose a variable to eliminate
The second equation is the simplest, so we'll eliminate x. Multiply the second by 2:
Elimination is faster when the coefficients are easy to scale (like multiplying by 2 or 3).
Substitution is faster when one variable is already isolated (e.g. y = …).
Choose elimination when coefficients align; substitution is also effective when one variable is easy to isolate.
🔑 The recipe: scale → subtract → solve one variable → back-substitute → verify.
Five lines on the page, every time.
STEP 6 OF 24 · Phase 2 · Derivation
Derivation 2 — Three variables, two equations
An AIMO classic: not enough equations… or is there?
This is a preview of AIMO 2015 Q3 (we'll re-do it as a worked example later). Suppose:
3w + 7d + t = 329
4w + 10d + t = 441
There are three unknowns (w, d, t) but only two equations. With no integer
constraint, infinitely many solutions exist. But notice what the question is
likely to ask: not "find w", but "find w + d + t" — a clever
combination.
\(w + d + t = w + d + (329 - 3w - 7d)\)\( = 329 - 2w - 6d\)\( = 329 - 2(w + 3d)\) factor!\( = 329 - 2(112)\) use ⓐ\( = 329 - 224 = 105\)
🔑 The combination trick: when you have fewer equations than unknowns, look
for a linear combination the question is really asking for. Often it's a
multiple of one of the equations you've already derived.
⚠ Don't try to solve for w, d, t individually here — you can't, and you don't
need to. Olympiad time is precious; only compute what's asked.
STEP 7 OF 24 · Phase 2 · Derivation
Derivation 3 — Reconstructing integers from pair sums
A 4-step algorithm you'll use over and over.
Suppose 4 integers a ≤ b ≤ c ≤ d have pairwise sums (in sorted order):
5, 8, 9, 11, 12, 15. Find a, b, c, d.
Step 1 — Sum of all pair sums
\(5 + 8 + 9 + 11 + 12 + 15 = 60\)For 4 integers, each appears in (n − 1) = 3 pairs.\(3(a + b + c + d) = 60 \Rightarrow a + b + c + d = 20\)
Step 2 — Smallest and largest pair
The smallest pair-sum is a + b (5). The largest is c + d (15).
Together they sum to 20 — consistent ✓.
Step 3 — Second smallest pair
The second smallest is a + c = 8 (because a + c ≤ a + d ≤ b + d and
a + c ≤ b + c ≤ b + d). So:
\(a + b = 5, a + c = 8, c + d = 15\)From a + b + c + d = 20 and a + b = 5: \(c + d = 15 ✓\)
Step 4 — Solve for individual integers
\(b = 5 - a, c = 8 - a, d = 15 - c = 15 - (8 - a) = 7 + a.\)The two middle pair-sums must be {a + d, b + c} = {2a + 7, 13 − 2a}.Both assignments must be checked: 2a + 7 = 9 gives a = 1, while 2a + 7 = 11 gives a = 2.Thus (a,b,c,d) = (1,4,7,8) or (2,3,6,9). Both satisfy the ordering and every pair-sum.
① Total = (sum of all pair-sums) ÷ (n − 1)
② Smallest sum = a + b. Largest sum = c(n) + c(n−1).
③ For four sorted values, the second-smallest pair-sum is a + c (counting multiplicities).
④ Check both assignments of the middle sums; more than one valid reconstruction may exist.
STEP 8 OF 24 · Phase 3 · Worked Example 1
Worked Example 1 — Plain 2-variable system
A clean warm-up. Two equations, two unknowns, integer answers.
Solve the system: 2x + 5y = 23 and 3x + 2y = 18.
Find x and y on paper; enter only x in the answer box. Both are positive integers.
Type your answer (single integer):
💡 Hints — open as needed
Two equations, two unknowns with integer coefficients. The question asks for x (and y); the coefficients suggest elimination will be clean.
Eliminate x by scaling. Multiply eq 1 by 3 and eq 2 by 2 to make the x coefficients match, then subtract.
Answer: \(x = 4, y = 3\)
Multiply eq 1 (2x + 5y = 23) by 3 and eq 2 (3x + 2y = 18) by 2: 6x + 15y = 69 and 6x + 4y = 36. Subtract: 11y = 33 → y = 3. Back-substitute into eq 2: 3x + 6 = 18 → x = 4. Verify: 2(4) + 5(3) = 23 ✓ and 3(4) + 2(3) = 18 ✓.
Tried first?
STEP 9 OF 24 · Phase 3 · Worked Example 2
Worked Example 2 — Find a combination, not the variables
Two equations, three unknowns — and a beautiful combination trick.
A selection of 3 whatsits, 7 doovers and 1 thingy costs $329. A selection of 4 whatsits,
10 doovers and 1 thingy costs $441. What is the total cost of 1 whatsit, 1 doover and 1 thingy?
Type your answer (single integer):
💡 Hints — open as needed
Three unknowns (w, d, t) but only two equations — you cannot find them individually. The question asks for the combination w + d + t.
Subtract eq 1 from eq 2 to get w + 3d = 112; express t from eq 1 and substitute into w + d + t to get a clean number.
Answer: \(w + d + t = $105\)
eq 2 − eq 1: (4w + 10d + t) − (3w + 7d + t) = 441 − 329 → w + 3d = 112. From eq 1: t = 329 − 3w − 7d. So w + d + t = 329 − 2w − 6d = 329 − 2(w + 3d) = 329 − 224 = $105.
Tried first?
STEP 10 OF 24 · Phase 3 · Worked Example 3
Worked Example 3 — Pair sums of 4 integers
The reconstruction algorithm in action.
Four integers a ≤ b ≤ c ≤ d have pairwise sums (in sorted order)
4, 5, 7, 8, 10, 11. Find the four integers on paper, then enter only their sum in the answer box.
Type your answer (single integer):
💡 Hints — open as needed
Six pair-sums from four integers a ≤ b ≤ c ≤ d are given sorted. Each integer appears in exactly 3 of the 6 sums.
Total of all 6 sums = 3(a+b+c+d). The smallest sum is a+b, the second-smallest a+c; solve linearly.
Answer: Sum of the four integers = 15 (a, b, c, d) = (1, 3, 4, 7)
Sum of all 6 pair-sums = 4+5+7+8+10+11 = 45 = 3S → S = 15. Smallest a+b = 4, largest c+d = 11, second-smallest a+c = 5. Then b = 4−a, c = 5−a, d = 6+a; the middle sums {2a+6, 9−2a} = {7,8} give a = 1/2 or a = 1. Integrality rejects a = 1/2. So (a,b,c,d) = (1,3,4,7); all six sums check: 4,5,8,7,10,11 ✓.
Tried first?
STEP 11 OF 24 · Phase 3 · Worked Example 4
Worked Example 4 — Coloured blocks (AIMO-style)
Several equal-weight collections plus an integer constraint.
Joel has blocks. Each colour has a fixed positive integer weight in kg, and different colours have
different weights. Three collections all weigh the same:
(i) 5 red + 3 blue + 5 green; (ii) 4 red + 5 blue + 4 green; (iii) 7 red + 4 blue + n green.
The shared weight w satisfies 30 < w < 50. Find n, then find the weight of
6 red + 7 blue + 3 green. Enter only that final weight in kg in the answer box.
Type your answer (single integer):
💡 Hints — open as needed
Three block-collections share the same weight w, with r, b, g distinct positive integers and 30 < w < 50. The question asks for a new combination, not the individual weights.
From (i)=(ii), obtain r + g = 2b, and substitute this into (i) to express w using b alone. The bound on w fixes b. List the remaining distinct positive r,g pairs and test which makes the green-block count n a non-negative integer.
Answer: Weight = 42 kg
From (i)=(ii), r+g=2b. Then w=5(r+g)+3b=13b, so 30<w<50 forces b=3 and w=39. Therefore r+g=6. Distinct positive weights leave (r,g)=(1,5),(2,4),(4,2),(5,1). The third collection gives 7r+ng=27, so these pairs respectively require n=4,13/4,−1/2,−8. Only n=4 is a valid block count. Thus r=1,b=3,g=5, and the requested weight is 6+21+15=42 kg.
Tried first?
STEP 12 OF 24 · Phase 4 · Practice
Practice — three problems with hints
Try first. Click for a hint, then the answer, then the full solution.
EASY
P1. Solve 2x + y = 7 and x + 2y = 8.
Multiply the second equation by 2 to match x-coefficients, then subtract.
x = 2, y = 3.
Multiply eq 2 by 2: 2x + 4y = 16. Subtract eq 1: 3y = 9 ⇒ y = 3. Back-sub into eq 1: 2x + 3 = 7 ⇒ x = 2. Verify: 2 + 6 = 8 ✓.
MEDIUM
P2. Three numbers satisfy a + b + c = 20, a + 2b + 3c = 35, and a + 3b + 6c = 55. Find c.
Subtract consecutive equations to peel off a, then again to peel off b.
c = 5 (and b = 5, a = 10).
eq2 − eq1: b + 2c = 15. eq3 − eq2: b + 3c = 20. Subtract: c = 5. Then b = 15 − 10 = 5, and a = 20 − 5 − 5 = 10. Verify all three: ✓.
MEDIUM
P3. Four positive integers a ≤ b ≤ c ≤ d have pairwise sums (sorted) 5, 8, 9, 11, 12, 15. Find all possible values of the largest integer d.
Total = 60 ÷ 3 = 20. Smallest sum is a + b, largest is c + d. Then second-smallest is a + c.
d = 8 or d = 9.
a+b+c+d = \(\frac{60}{3}\) = 20. a+b=5, c+d=15, a+c=8 ⇒ b = 5−a, c = 8−a, d = 7+a. Middle sums {a+d, b+c} = {2a+7, 13−2a} = {9,11}. The two assignments give a = 1 or a = 2, hence (1,4,7,8) or (2,3,6,9). Both have precisely the six given pair-sums, so the possible largest integers are 8 and 9.
STEP 13 OF 24 · Phase 5 · AIMO Exam
AIMO 2023 Q3 — Five integers from pair sums
2 marks · exam format · type the answer, then submit.
AIMO 2023 · Q3 · 2 marks
The ten pairwise (two-at-a-time) sums of five distinct integers are
0, 1, 2, 4, 7, 8, 9, 10, 11, 12. Find the sum of the five integers.
Your answer (a single integer):
Stuck? Open hints in order:
Hint 1 — How many pairs from 5 integers?
C(5, 2) = 10. Yes, all 10 are listed.
Hint 2 — How often does each integer appear?
Each of the 5 integers is paired with the other 4. So each appears in 4 of the 10 sums.
Hint 3 — Sum of all pair-sums?
Total = 4(a + b + c + d + e). The total of the 10 given sums is 64.
Solution
Sum of all 10 pair-sums = 0+1+2+4+7+8+9+10+11+12 = 64.Each integer appears in 4 pairs: \(4S = 64 \Rightarrow S = 16.\)
🔁 This is the same trick as Worked Example 3 — sum the pair-sums, divide by (n − 1).
STEP 14 OF 24 · Phase 5 · AIMO Exam
AIMO 2015 Q3 — Whatsits, doovers and a thingy
3 marks · exam format.
AIMO 2015 · Q3 · 3 marks
A selection of 3 whatsits, 7 doovers and 1 thingy cost a total of $329. A selection of 4
whatsits, 10 doovers and 1 thingy cost a total of $441. What is the total cost, in dollars,
of 1 whatsit, 1 doover and 1 thingy?
Your answer (a single integer in dollars):
Stuck? Open hints in order:
Hint 1 — Set up the equations
Let w, d, t be the prices. Then 3w + 7d + t = 329 and 4w + 10d + t = 441.
Hint 2 — Subtract the equations
eq2 − eq1 gives w + 3d = 112.
Hint 3 — Express what's asked
Compute w + d + t directly. From eq 1: t = 329 − 3w − 7d. So w + d + t = 329 − 2w − 6d = 329 − 2(w + 3d).
🔁 Identical to Worked Example 2 — fewer equations than unknowns, but the
question asks for a clean linear combination.
STEP 15 OF 24 · Phase 5 · AIMO Exam
AIMO 2014 Q4 — Coloured blocks
3 marks · exam format.
AIMO 2014 · Q4 · 3 marks
Joel has blocks, each with a positive integer weight in kg. All blocks of one colour have the
same weight, and different colours have different weights. Three collections share the same
total weight w kg: (i) 5 red + 3 blue + 5 green; (ii) 4 red + 5 blue + 4 green; (iii) 7 red +
4 blue + some green. If 30 < w < 50, what is the total weight in kilograms of 6 red + 7 blue
+ 3 green blocks?
Your answer (a single integer in kg):
Stuck? Open hints in order:
Hint 1 — Equate (i) and (ii)
5r + 3b + 5g = 4r + 5b + 4g gives r + g = 2b, so b = \(\frac{(r + g)}{2}\).
Hint 2 — Equate (i) and (iii)
5r + 3b + 5g = 7r + 4b + ng. Substitute b = \(\frac{(r + g)}{2}\) and clear fractions to get 5r = (9 − 2n)g.
Hint 3 — Test small n
For positive integer r, g, you need 9 − 2n > 0, so n ≤ 4. Try n = 4 first.
Solution
From (i)=(ii): \(b = \frac{(r + g)}{2}.\)From (i)=(iii): \(5r = (9 - 2n)g.\)Using r+g=2b in the first collection gives w=13b. Thus b=3,w=39 and r+g=6. Distinct positive weights give (r,g)=(1,5),(2,4),(4,2),(5,1). In 7r+ng=27 these require n=4,13/4,−1/2,−8 respectively. Only n=4 is a non-negative integer. Hence r=1,g=5,b=3 uniquely.w = 5(1) + 3(3) + 5(5) = 39, and 30 < 39 < 50 ✓.\(6r + 7b + 3g = 6 + 21 + 15 = 42\text{ kg}.\)
🔁 Identical to Worked Example 4. The complete enumeration in the solution is essential;
the integer constraint plus the range of w pins everything down.
STEP 16 OF 24 · Phase 5 · AIMO Exam
AIMO 2010 Q5 — Five bank balances from triple sums
3 marks · exam format.
AIMO 2010 · Q5 · 3 marks
Jess has five bank accounts. If three account balances at a time were added, the following
ten amounts would result: 94, 97, 99, 100, 101, 103, 104, 106, 107, 109. What is the sum of
the lowest and highest balances?
Your answer (a single integer):
Stuck? Open hints in order:
Hint 1 — How many triples and how often does each balance appear?
C(5, 3) = 10 triples. Each balance appears in C(4, 2) = 6 of them.
Hint 2 — Total of the balances
Sum of all 10 triple-sums = 1020. So 6S = 1020 ⇒ S = 170.
Hint 3 — Smallest and largest triples
Sort the balances a ≤ b ≤ c ≤ d ≤ e. The smallest triple-sum is a+b+c = 94, so d+e = 170 − 94 = 76. The largest triple-sum is c+d+e = 109, so a+b = 170 − 109 = 61.
Hint 4 — Pin down a and e
Second-smallest triple = a+b+d = 97, so d = 97 − 61 = 36 ⇒ e = 76 − 36 = 40. Second-largest = b+d+e = 107, so b = 107 − 76 = 31 ⇒ a = 61 − 31 = 30. Lowest + highest = 30 + 40.
Solution
Sort balances a ≤ b ≤ c ≤ d ≤ e. Sum of all 10 triple-sums = 1020.Each balance appears in C(4, 2) = 6 triples ⇒ 6S = 1020 ⇒ S = 170.Smallest triple a+b+c = 94 ⇒ d + e = 76.Largest triple c+d+e = 109 ⇒ a + b = 61.Second-smallest = a+b+d = 97 ⇒ d = 36, e = 40.Second-largest = b+d+e = 107 ⇒ b = 31, a = 30.Middle: c = 170 − (a+b+d+e) = 170 − 137 = 33. Verify all 10 triples ✓.Lowest + highest = 30 + 40 = 70.
🔁 Same engine as the pair-sum algorithm. For triple sums, each appears in C(n−1, 2) of them.
STEP 17 OF 24 · Lesson Quiz
Lesson quiz — three problems, 8 marks total
Exam conditions. No hints. Submit at the end.
📝 Linear Systems Lesson Quiz — 8 marks · target time 12 min
Type your numerical answer in each box. Click Grade at the bottom.
Q1 · 2 marks
Three positive integers a, b, c satisfy a + b + c = 15, a + b = 8, and
b + c = 12. Find b.
Q2 · 3 marks
Find z, given x + y + z = 24, x + 2y + 3z = 41, and x + 3y + 6z = 64.
Enter z in the answer box.
Q3 · 3 marks
Four positive integers have pairwise sums (sorted) 4, 5, 7, 8, 10, 11. Find the sum of the
four integers.
STEP 18 OF 24 · Summary
Cheat sheet — Linear Systems with Integer Constraints
Print this page and keep it in your AIMO folder.
① The 5-line elimination recipe
Scale one equation so a coefficient matches.
Subtract to eliminate a variable.
Solve the new (smaller) system.
Back-substitute.
Verify in both originals.
⚠ Pick the variable whose LCM is smallest — fewer arithmetic mistakes.
② Fewer equations than unknowns? Look at what's asked.
If the question asks for a linear combination (e.g. w + d + t), don't try to
solve for individual variables. Express the combination, then plug in any auxiliary
equation you've derived.
Example: w + d + t = 329 − 2(w + 3d).
If w + 3d = 112, answer = 329 − 224 = 105.
③ Pair-sum reconstruction (n integers)
Total of all pair-sums = (n − 1) × S.
Smallest pair-sum = a₁ + a₂. Largest = aₙ₋₁ + aₙ.
For sorted values and the complete multiset of pair-sums, the second-smallest is a₁ + a₃.
Solve linearly. Verify the middle pair-sums match.
For triple sums: total = C(n − 1, 2) × S.
④ Integer constraints restrict the candidate set
If the explicit algebra leaves you with one equation in two unknowns, the integer
constraint may leave zero, one or several possibilities. Check every candidate against
any range condition (like 30 < w < 50).
⚠ Reject a candidate only when it violates a stated condition, then verify the remaining candidates.
⑤ Sanity checks before you write the answer
Substitute back into every original equation.
Check distinctness, positivity, and the stated range.
Re-read the question — is the answer a single variable, a sum, or a count?
Common pitfalls
Trying to solve for individual variables when the question only needs a combination.
Forgetting that "positive integer weight" or "whole dollars" is a real constraint.
Confusing repeated values with distinct pairs: keep multiplicities when sorting pair-sums.
Arithmetic slips when scaling — always re-multiply once, not from memory.
Assuming every Olympiad problem asks for an integer. Follow the actual response and domain requirements.
When to use this technique
If a problem mentions any of:
"n things, each with the same/integer weight (or price)" — system with integer constraint.
"the pairwise sums are…" or "the three-at-a-time sums are…" — reconstruction algorithm.
"find the total cost of one of each" — combination trick (don't solve individuals).
"the largest balance" or "the smallest weight" — extract a₁ or aₙ from a sorted list.
… then this is a linear-system problem with an integer constraint. Apply the recipe.
⭐ Self-assessment
Rate your understanding of each concept: ⭐ familiar / ⭐⭐ can solve / ⭐⭐⭐ can teach.
① I can solve a 2-variable linear system by elimination in under 2 minutes.
② When fewer equations than unknowns, I look for a linear combination the question asks for.
③ I can reconstruct n integers from their pairwise sums using the (n − 1) trick.
④ I use integer and range constraints to test all remaining possibilities, without assuming uniqueness.
⑤ I have walked through the four AIMO past papers in this lesson and could re-solve them next week.
⭐ \(\frac{0}{15}\) — click stars to record your mastery
🎉 Core recap complete — continue through Steps 19–24 to finish this lesson.
Tomorrow we'll do Part 2 — Non-Linear Systems (symmetric identities and how
they collapse two-variable problems into one-line answers).
Tonight: re-solve AIMO 2015 Q3 and 2010 Q5 with
pencil and paper. Aim for under 5 minutes each. The brain consolidates better when you re-derive on paper.
📅 Unit tests: Take Past-Paper-Test.html first for verified
AIMO questions, then Mock-Test.html for the 10-question unit test reusing verified past-paper questions.
Answers and review unlock only after submission on each page.
A5·110r + 13b = w with positive integers r, b and 30<w<50. If r=1, find the largest such b.
A5·2Same equation 10r + 13b = w, 30<w<50. If r=2, find the largest such b.
Visual: equation as a balance
Each side of an equation must stay balanced. Subtraction (elimination) shifts both pans equally.
STEP 20 · v3 PACK · Worked Example 5
Worked Example 5 — Five integers from all 10 pair-sums
⭐⭐⭐⭐ · Pass-1 ceiling. The same pair-total technique as 2023 Q3, with a new complete list.
Worked Example 5
Five distinct positive integers have all 10 pairwise sums equal to:
3, 4, 5, 6, 7, 8, 9, 10, 11, 13.
Find the sum of the five integers.
Your answer (single integer):
Hints
Observe
10 sums from 5 distinct integers — each integer appears in exactly 4 sums. Sum of all 10 sums = 4·S.
Strategy
Add all 10 pair-sums, divide by 4. The "distinct positive" wording is for context — it doesn't change the trick. Why this technique: any time pairwise data is given symmetrically, the (n−1) multiplier collapses it instantly.
Solution
Sum of all ten pairwise sums = 76.Each of the 5 integers appears in 4 pairs, so 4S = 76 → S = 19.(The integers are 1, 2, 3, 5, 8.)
\(19\)
🔁 Same skill as 2023 Q3 — sum the pairwise data, divide by (n−1) = 4.
Four distinct positive integers a<b<c<d satisfy: the smallest pairwise sum is 3, the second-smallest pairwise sum is 5, the largest pairwise sum is 11, and the six pairwise sums total 42. Find d.
Your answer:
Toolbox switching:
Step 1 — apply A3 (Σ pair-sums = 3·S, since each of 4 numbers is in 3 sums): S = \(\frac{42}{3}\) = 14.
Step 2 — apply A4: a+b=3 (smallest pair), c+d=11 (largest pair). Verify 3+11=14 ✓.
Step 3 — apply A5 with the second-smallest sum: a+b=3 gives (a,b)=(1,2); a+c=5 gives c=4; c+d=11 gives d=7. Check: the six sums are 3, 5, 6, 8, 9, 11 — total 42 ✓. So d = 7.
Phase-5 5-Step Observe Reference (covers all 4 AIMO problems above)
Apply this template before starting any AIMO problem. Refer back to AIMO steps 13–16 with this lens.
2023 Q3 — 5-step Observe
Keyword identification: "ten pairwise sums of five integers" → pair-sum trick (A3).
Known quantities: 10 specific integers (the sums) totalling 64.
Unknown quantities: sum S = a+b+c+d+e (only the total — not the individuals).
Intermediate variable: none needed; the (n −1) trick is direct.
Hidden constraints: The five integers are distinct. Different pairs may still have equal sums; count each pair once, retaining multiplicities.
Strategy: Σ pair-sums = (n −1)·S = 4S. Divide by 4. Why this technique applies in general: any pairwise data given symmetrically reduces to a single linear equation in the total.
2015 Q3 — 5-step Observe
Keyword: "what is the total cost of 1 + 1 + 1" → linear combination (A2), not individual values.
Known: two equations 3w+7d+t=329 and 4w+10d+t=441.
Unknown: w+d+t (one combination, not 3 values).
Intermediate: α, β such that αEq1 + βEq2 has all coefficients 1.
Hidden constraint: prices are real (not necessarily integer); under-determined system is fine.
Strategy: solve coefficient system for α, β (here α=3, β=−2). Why: any time the question asks for a specific linear combination instead of individuals, search for that combination directly.
2014 Q4 — 5-step Observe
Keyword: "blocks of integer weight" + range "30 < w < 50" → integer constraint pinning (A5).
Known: three weight equations (one with unknown green count).
Unknown: 6r + 7b + 3g, plus the green-count parameter.
Intermediate: g = 2b − r from (i)=(ii), then w = 5(r+g)+3b = 13b.
Hidden constraint: r,b,g are positive integers (not just real); 30<w< 50 narrows the search.
Strategy: reduce to 2 unknowns + scan integer pairs in the 30<w< 50 window. Why: integer + bounded constraint problems always end in an enumeration step.
2010 Q5 — 5-step Observe
Keyword: "ten three-at-a-time sums of five balances" → count each balance’s appearances.
Known: the ten triple-sums, with total 1020.
Unknown: the sum of the smallest and largest account balances.
Intermediate: total balance S; the smallest/largest and second-smallest/second-largest triple-sums.
Hidden constraint: Order the five balances a≤b≤c≤d≤e and preserve all ten triple-sums.
Strategy: Each balance occurs in six triples, so S=1020/6=170. The extreme triples determine a+b and d+e; the next extremes determine b and d, then a and e.
💡 Stuck? Open this for guiding questions (no spoilers)
Ask yourself, in order:
How many independent equations do I really have? Subtract one pair to reduce.
What does the question literally ask for — a single variable, or a combination?
If pair-sums are involved, did I use the rule total = (n − 1) × S?
Is there an integer constraint I haven't used yet?
Did I check my answer against every original equation, not just one?