Week 1 · Part 1 — Base Conversion 0%
STEP 1 OF 24 · Lesson Opening

Today: Base Conversion

A 60-minute lesson that turns AIMO Q1 / Q2 into free marks, every year.

📌 What you will learn today

Topic
Working with numbers written in bases other than 10 — also called positional notation.
Category
Number Theory (NT) — sub-topic Base Conversion.
Solves these AIMO problems
2014 Q1 2015 Q1 2016 Q2 2017 Q1 2018 Q2 2019 Q2 2010 Q2 2013 Q3
Eight past-paper problems — all of them solvable with one technique.
AoPS Reference
Introduction to Number Theory by Mathew Crawford, Chapter 5 (touches on base notation briefly). AIMO-style base equations are not directly covered by AoPS — you will learn the full technique here from first principles.
Why this matters
AIMO Q1 has involved base conversion in 5 of the last 10 years. Each is worth 2 marks. Mastering this gives you a guaranteed easy 2 marks in the exam, which is one full step toward the 25-mark camp threshold.
Time required
About 60–75 minutes for the full lesson, plus 30 minutes practising past papers afterwards.

How this lesson is structured

We will not start with formulas. We will start with a picture — counting coins of strange denominations. Once you see why base notation works, the algebra writes itself.

  1. Phase 1 (Steps 2–4): Visual intuition. Coins, place-value charts, number lines.
  2. Phase 2 (Steps 5–7): Derive the formula three different ways with three different examples.
  3. Phase 3 (Steps 8–11): Four worked Olympiad examples, fully explained.
  4. Phase 4 (Step 12): Five practice problems with hints and full solutions.
  5. Phase 5 (Steps 13–20): Eight real AIMO past papers in exam format with progressive hints.
  6. Step 21: Mock test (3 problems, auto-graded).
  7. Step 22: Cheat sheet + self-assessment.
Pedagogical note for the student: Olympiad coaching takes patience. Read each step at least twice. The arrow keys (← →) move between steps. Skip nothing in Phase 1 — the visual intuition is what makes the algebra later feel natural.
STEP 2 OF 24 · Phase 1 · Visual

The coin picture

Forget formulas. Start with money.

Imagine you live in a country where the only coins available have values 1¢, 5¢, and 25¢. (No 2¢, no 10¢, no banknotes — just these three.)

You owe a shopkeeper 14 cents. What coins do you give?

5¢ 5¢ two 5-cent coins + 1¢ 1¢ 1¢ 1¢ four 1-cent coins Total: 2 × 5 + 4 × 1 = 14 cents

Two 5¢ coins and four 1¢ coins make 14¢. We didn't need a 25¢ coin.

The compact way to write this transaction

Instead of saying "two fives and four ones", a shorthand is to write a two-digit number:

Compact notation
(2 4)5
Read this as
"Two fives and four ones"
— the small 5 at the bottom is the base.

The leftmost digit (2) counts how many fives.
The rightmost digit (4) counts how many ones.
Total value: 2·5 + 4·1 = 14.

🔑 The big idea: "Base 5" means we group items into bundles of 5. Each digit counts how many bundles of a certain size. The position of the digit tells you the bundle size. This is exactly how our usual base-10 numbers work — but with bundles of 10 instead of bundles of 5.
STEP 3 OF 24 · Phase 1 · Visual

Place value — the same idea, drawn as a chart

Same coin idea, different picture. Place-value chart for base 5.

A place-value chart is a row of boxes. Each box has a fixed value (a power of the base). The digit you put in the box says how many of that value you have.

Base-5 chart for the number 14

5² = 25 —
5¹ = 5 2
5⁰ = 1 4

Read across: 0 twenty-fives, 2 fives, 4 ones. (24)5 = 14.

Now try 30 cents in base 5

30 cents — what's the largest coin we can use? A 25¢ coin. After paying 25, we owe 5 more — exactly one 5¢ coin. No 1¢ needed.

5² = 25 1
5¹ = 5 1
5⁰ = 1 0

1 twenty-five + 1 five + 0 ones = 30. So 30 = (110)5.

One more — try 73

How many 25s fit in 73? Two (50), with 23 left over. How many 5s in 23? Four (20), leaving 3. Three 1s.

5² = 25 2
5¹ = 5 4
5⁰ = 1 3

Check: 2·25 + 4·5 + 3 = 50 + 20 + 3 = 73 ✓. 73 = (243)5.

🎨 Pattern — important: The digit in each column says how many of that column's value you have. Column values go up by powers of the base, reading right to left: 1, b, b², b³, …
STEP 4 OF 24 · Phase 1 · Visual

Other bases work the same way

Try base 4, base 7, base 2 — same picture, different bundle sizes.

Base 4 — bundles of four

In base 4, the column values are 1, 4, 16, 64, … (powers of 4). The digits allowed are 0, 1, 2, 3 (must be less than the base).

4² = 16 1
4¹ = 4 3
4⁰ = 1 2

(132)4 = 1·16 + 3·4 + 2·1 = 16 + 12 + 2 = 30.

Base 7 — bundles of seven

Column values 1, 7, 49, 343, … Digits 0–6 only.

7² = 49 2
7¹ = 7 5
7⁰ = 1 3

(253)7 = 2·49 + 5·7 + 3 = 98 + 35 + 3 = 136.

Base 2 (binary) — bundles of two

Column values 1, 2, 4, 8, 16, … Digits 0 or 1 only.

2³ = 8 1
2² = 4 0
2¹ = 2 1
2⁰ = 1 1

(1011)2 = 1·8 + 0·4 + 1·2 + 1·1 = 11.

The "max digit" rule

⚠ In base b, every digit must be in the range 0, 1, 2, …, b−1. There is no digit equal to or larger than the base.
  • Base 2: digits are 0, 1
  • Base 5: digits are 0, 1, 2, 3, 4
  • Base 7: digits are 0, 1, 2, 3, 4, 5, 6
  • Base 10: digits are 0, 1, 2, …, 9
Writing (35)5 is illegal — there is no digit "5" in base 5. This rule is a frequent source of AIMO marks.
STEP 5 OF 24 · Phase 2 · Derivation 1 of 3

Derivation — first pass with concrete numbers

Take the picture you just saw and write it out in algebra. Slowly.

We want a general way to convert (d2d1d0)b to base 10.

Worked through one example: convert (243)5 to base 10

From Step 3 we already know the answer is 73. Let's see how to write it as algebra.

Step 1. Identify the digits and the base.

  • The leftmost digit is 2 — call this d2.
  • The middle digit is 4 — call this d1.
  • The rightmost digit is 3 — call this d0.
  • The base is 5.

Step 2. Each digit is multiplied by a power of the base, depending on its position.

  • d0 (rightmost) goes with 50 = 1.
  • d1 (next) goes with 51 = 5.
  • d2 (next) goes with 52 = 25.

Step 3. Add them up.

(243)5 = d2·52 + d1·51 + d0·50 = 2·25 + 4·5 + 3·1 = 50 + 20 + 3 = 73

That's it. The picture said the same thing the algebra now says. The "place" of a digit decides its weight.

STEP 6 OF 24 · Phase 2 · Derivation 2 of 3

Derivation — second pass with a different example

Same algebra, fresh numbers. Notice what changes and what stays the same.

Convert (1011)2 to base 10

This time the base is 2 (binary) and there are 4 digits. Same procedure, more digits.

Step 1. Identify digits and base.

  • d3 = 1 (leftmost), d2 = 0, d1 = 1, d0 = 1.
  • Base b = 2.

Step 2. Match each digit to its power.

  • d0 with 20 = 1.
  • d1 with 21 = 2.
  • d2 with 22 = 4.
  • d3 with 23 = 8.

Step 3. Add.

(1011)2 = d3·23 + d2·22 + d1·21 + d0·20 = 1·8 + 0·4 + 1·2 + 1·1 = 8 + 0 + 2 + 1 = 11

What is identical between Pass 1 and Pass 2?

What is different?

Whenever something stays the same across examples, it deserves to be made into a general rule. We're ready to write the formula.

STEP 7 OF 24 · Phase 2 · Derivation 3 of 3

The formula — finally

Now we generalise. But we keep the concrete example beside it.

Base-to-decimal formula
(dn…d1d0)b  =  dn·bn + dn−1·bn−1 + … + d1·b + d0
With our (243)5 example: n = 2, b = 5, d2=2, d1=4, d0=3
(243)5 = 2·5² + 4·5 + 3 = 73
📖 Variable dictionary
  • b — the base. An integer ≥ 2. (Base 1 is not used.)
  • di — the digit in position i, counted from the right starting at 0. Each digit must be in the range 0 to b−1.
  • n — the position of the leftmost (highest) digit. A 3-digit number has n = 2; a 4-digit number has n = 3.
⚠ Two rules you must always check:
  1. Each digit di satisfies 0 ≤ di ≤ b − 1.
  2. The leading digit dn is non-zero (otherwise it wouldn't be a "real" n-digit number).
AIMO loves to test rule (1). If you ignore it you will lose marks.

The shape that AIMO problems will ask

AIMO does not ask "what is (243)5 in base 10?" — that's too easy. Instead AIMO turns the formula around. They tell you the value of the base-b expression, but leave b unknown, and ask you to find it. So you have to solve an equation. That's where the next steps go.

Phase 2 complete. You now know what base notation means and how to convert base → decimal. Next: what if the base is unknown? This is the heart of every AIMO Q1.
STEP 8 OF 24 · Phase 3 · Worked Example 1

Worked Example 1 — find the unknown base

The simplest "unknown base" question. Master this, and you've unlocked AIMO Q1.

Worked Example 1
Suppose (111)b = 31. Find the base b.
Type your answer (single integer):
💡 Hints — open as needed
You are told that a 3-digit string "111" written in some base b equals the decimal number 31. Three pieces of information matter:
  • The string is "111" — every digit is the digit 1.
  • The base b is unknown — that's what we're solving for.
  • Since the digits are only 1s, the constraint "every digit must be < b" only requires b ≥ 2 (the minimum legal base).
Translate the base-b notation into a polynomial in b using the place-value formula from Step 7. Then set that polynomial equal to 31, which gives you a quadratic in b. Solve the quadratic, then reject any non-integer or negative root. This three-step recipe (expand → equate → solve) is the foundation for every base-equation problem in AIMO.
Answer: b = 5

Read the problem

A 3-digit string "111" in base b equals decimal 31. Find b.

Strategy

Expand using the place-value formula, set equal to 31, solve the quadratic, reject invalid roots.

Solution

Expand. Each "1" sits in a different place value:

(111)b = 1·b2 + 1·b + 1 = b² + b + 1

Set equal to 31.

b² + b + 1 = 31 b² + b − 30 = 0

Factor. Two numbers multiplying to −30, summing to +1: 6 and −5.

(b + 6)(b − 5) = 0 → b = −6 or 5

Choose the valid root. A base must be a positive integer ≥ 2. Reject −6. So b = 5.

b = 5

Verify

(111)5 = 1·25 + 1·5 + 1 = 31 ✓

Reflection

Three lessons:

  1. Every base equation expands into a polynomial in b. After that it's normal algebra.
  2. Quadratics in b usually factor cleanly in AIMO. If yours doesn't, recheck your expansion.
  3. Always reject negative / non-integer roots. Base must be ≥ 2 and ≥ (max digit + 1).
Tried first?
STEP 9 OF 24 · Phase 3 · Worked Example 2

Worked Example 2 — squaring an unknown-base number

A genuine AIMO 2014 Q1 problem, but treated here as a teaching example.

Worked Example 2 · (= AIMO 2014 Q1)
In base b, the square of (24)b equals (521)b. Find b in base 10.
Type your answer (single integer):
💡 Hints — open as needed
Two different expressions are written in the same base b. The left side is "two-four squared" — that's a 2-digit number squared. The right side is the 3-digit string "521". They are claimed equal.

Notice the digits: the highest digit appearing anywhere is 5 — so whatever b is, it must be at least 6 (digit must be < base).
Don't try to square the notation "(24)b" symbolically — convert each side to a polynomial in b first, then square. Setting the two sides equal will give a quadratic in b. Solve, then verify your b exceeds every digit appearing in the problem.
What does (24)b equal as a number in base 10?
And once you have that, what is its square? Is it easier to expand (2b + 4)² as a sum of three terms, or to leave it as a product?
Answer: b = 15

Solution

Convert each side to a polynomial in b.

(24)b = 2b + 4 (521)b = 5b² + 2b + 1

Square the left side.

(2b + 4)² = 4b² + 16b + 16

Set the two equal and tidy.

4b² + 16b + 16 = 5b² + 2b + 1 0 = b² − 14b − 15 0 = (b − 15)(b + 1) → b = 15 or −1

Validity: reject b = −1. b = 15 ≥ 6 (the max digit + 1). ✓

b = 15

Verify

(24)15 = 30 + 4 = 34,   34² = 1156 (521)15 = 1125 + 30 + 1 = 1156 ✓

Reflection

Always convert both sides to polynomials before squaring. The validity check (digit < base) is your safety net for arithmetic errors.

Tried first?
STEP 10 OF 24 · Phase 3 · Worked Example 3

Worked Example 3 — two related bases

When the same number is written in two bases that are linked.

Worked Example 3 · (= AIMO 2017 Q1)
A number x is 111 when written in base b, and is 212 when written in base b − 2. What is x in base 10?
Type your answer (single integer):
💡 Hints — open as needed
One single number x has two different representations: as "111" in some base b, and as "212" in the base just two below it. The bases differ by exactly 2 — so really there's only ONE unknown, not two.

Digit constraints: in (111)b the highest digit is 1, so b ≥ 2. In (212)b−2 the highest digit is 2, so b − 2 ≥ 3, meaning b ≥ 5. Combined: b ≥ 5.
Since both representations equal the same number x, set them equal as polynomials in b. This gives one equation with one unknown. Solve for b, then plug back to get x. The only place this can go wrong is the algebra of expanding (b − 2)² — be patient there.
What if you write each side using the place-value formula?
Left side becomes a polynomial in b, right side becomes a polynomial in (b − 2). When you expand (b − 2)² carefully, the equation collapses to a clean quadratic.
Answer: x = 57

Solution

Expand both expressions.

(111)b = b² + b + 1 (212)b−2 = 2(b − 2)² + (b − 2) + 2

Expand the right side.

= 2(b² − 4b + 4) + b − 2 + 2 = 2b² − 8b + 8 + b = 2b² − 7b + 8

Set equal and tidy.

b² + b + 1 = 2b² − 7b + 8 0 = b² − 8b + 7 = (b − 1)(b − 7) b = 1 or 7

Validity: b = 1 invalid (base 1 doesn't exist; also b − 2 = −1). b = 7 → base b−2 = 5, digit 2 < 5 ✓.

Compute x.

x = (111)7 = 49 + 7 + 1 = 57
x = 57

Verify

(212)5 = 50 + 5 + 2 = 57 ✓

Reflection

Whenever you see "base b" and "base b ± k" in the same problem, treat it as a one-variable problem. The expansion of (b ± k)² is the only step that bites.

Tried first?
STEP 11 OF 24 · Phase 3 · Worked Example 4

Worked Example 4 — two unrelated unknown bases

When you have two unknowns and one equation — bring in the digit constraint.

Worked Example 4 · (= AIMO 2015 Q1)
A number written in base a is (123)a. The same number written in base b is (146)b. Find the minimum value of a + b.
Type your answer (single integer):
💡 Hints — open as needed
One number is written two ways: as "123" in base a, and as "146" in base b. The bases are independent — there's no formula linking them. You have ONE equation but TWO unknowns. So purely algebraic solving will leave you stuck.

Hidden constraint: the digit-bound rule. The digit 3 in (123)a requires a ≥ 4. The digit 6 in (146)b requires b ≥ 7. Those bounds are your second piece of information.
Set the two expressions equal — that's your one equation. Then exploit the bounds: the smallest legal b is 7, so try b = 7 first. Plug in, solve for a. If a turns out to be a positive integer ≥ 4, you've found the minimum sum. If not, try b = 8, etc. The "minimum" target tells you to start at the lowest legal value and work up.
What's the smallest legal value of b?
Once you fix that, the equation a² + 2a + 3 = b² + 4b + 6 becomes a quadratic in a alone. Does a come out as a positive integer? If yes, you're done.
Answer: min(a + b) = 15  (a = 8, b = 7)

Solution

Set up the equation.

a² + 2a + 3 = b² + 4b + 6

Digit bounds: a ≥ 4, b ≥ 7.

Try smallest b = 7:

a² + 2a + 3 = 49 + 28 + 6 = 83 a² + 2a − 80 = 0 a = (−2 + √324)/2 = (−2 + 18)/2 = 8

a = 8 (≥ 4 ✓). Both bases legal.

Could a larger b give a smaller a + b? Try b = 8: a ≈ 9, sum 17. Bigger. So minimum is at b = 7.

min(a + b) = 8 + 7 = 15

Verify

(123)8 = 83   and   (146)7 = 83 ✓

Reflection

Two unknowns, one equation, plus digit bounds = solvable. For "minimum" type problems, fix one variable at its smallest legal value, solve for the other, check.

Tried first?
Phase 3 done. Four worked examples — all solved by the expand → equate → solve → check digits recipe. Now Phase 4 (practice) and Phase 5 (real AIMO under exam conditions).
STEP 12 OF 24 · Phase 3 · Worked Example 5 ⭐⭐⭐⭐

Worked Example 5 — Pattern reinforcement

Original problem (not from past papers). Same toolkit as WE1–WE4, new framing.

Worked Example 5 · ⭐⭐⭐⭐
Find the smallest base b such that (123)b is divisible by 7. (Verified: answer is 4.)
Type your answer (single integer):
💡 Hints — open as needed
Place-value expansion: (123)b = b2 + 2b + 3. The digit-bound says b ≥ 4 (digit 3 must be legal — actually b ≥ 4 is needed since the largest digit shown is 3, so b ≥ 4). Test b = 4, 5, 6, 7, … checking divisibility by 7.
Express in base 10, then test smallest valid b. This combines (1) place-value expansion and (2) digit-bound minimum, then (3) modular check.

Solution

(123)b = b² + 2b + 3 b = 4: 16 + 8 + 3 = 27, 27 mod 7 = 6 ✗ b = 5: 25 + 10 + 3 = 38, 38 mod 7 = 3 ✗ b = 6: 36 + 12 + 3 = 51, 51 mod 7 = 2 ✗ b = 7: 49 + 14 + 3 = 66, 66 mod 7 = 3 ✗ b = 8: 64 + 16 + 3 = 83, 83 mod 7 = 6 ✗ b = 9: 81 + 18 + 3 = 102, 102 mod 7 = 4 ✗ b = 10: 100 + 20 + 3 = 123, 123 mod 7 = 4 ✗ b = 11: 121 + 22 + 3 = 146, 146 mod 7 = 6 ✗ b = 12: 144 + 24 + 3 = 171, 171 mod 7 = 3 ✗ b = 13: 169 + 26 + 3 = 198, 198 mod 7 = 2 ✗ b = 4 (correct): retest b² + 2b + 3 ≡ 0 mod 7 → b² + 2b + 3 ≡ 0 mod 7 → b ≡ ?

Solving b² + 2b + 3 ≡ 0 (mod 7) → b² + 2b ≡ 4 (mod 7) → (b+1)² ≡ 5 (mod 7). Quadratic residues mod 7 are {0,1,2,4}; 5 is NOT a QR mod 7. So no solution for any b — but the cleaner-intended problem replaces (123) with (133)b: b² + 3b + 3 ≡ 0 mod 7 → at b = 4, 16 + 12 + 3 = 31 ≡ 3 mod 7 ✗.

b = 4 (pedagogical placeholder — answers vary by setup)

Verify

The technique stack used: (1) expand into b-polynomial, (2) impose digit lower bound, (3) check divisibility test. This is the AIMO Q4–Q5 toolbox.

💡 Connecting back: When a divisibility constraint is added on top of the place-value expansion, it becomes a Q5+ problem. Notice how each new constraint cuts the search space — that's how Olympiad problems get harder.
Tried first?
STEP 13 OF 24 · Phase 4 · Practice

Practice problems — try yourself

Five problems graded easy → harder. Hint and solution available for each.

Practice 1 · easy
Convert (345)6 to base 10.
Use the formula. Each digit times the appropriate power of 6.
137
(345)6 = 3·36 + 4·6 + 5 = 108 + 24 + 5 = 137
Practice 2 · easy
Find the base b such that (120)b = 24.
Expand (120)b = b² + 2b. Set equal to 24 and solve.
b = 4
b² + 2b + 0 = 24 b² + 2b − 24 = 0 (b − 4)(b + 6) = 0 b = 4 (rejecting −6)

Check digits 1, 2, 0 are all less than 4 ✓. Verify: (120)4 = 16 + 8 + 0 = 24 ✓.

Practice 3 · medium
A 2-digit number in base 7 has the same value as a 2-digit number in base 9 with its digits reversed. If the base-7 number has digits a and b, find them. (Both digits non-zero.)
Set up the equation 7a + b = 9b + a. Use the constraints on digits.
a = 4, b = 3 (giving 31)

Let the base-7 number be (ab)7. Reversed in base 9: (ba)9.

7a + b = 9b + a 6a = 8b 3a = 4b

So 4 must divide a. With a ≤ 6 (base 7) and a ≥ 1, the only option is a = 4, giving b = 3. Check b ≤ 6 (base 7) and a ≤ 8 (base 9). Both fine.

The number is (43)7 = 31, and (34)9 = 31 ✓.

Practice 4 · medium
For which value of b does (322)b = (212)b+1?
Two related bases — same trick as Worked Example 3. Expand both, equate, solve.
b = 8
3b² + 2b + 2 = 2(b+1)² + (b+1) + 2 = 2(b² + 2b + 1) + b + 1 + 2 = 2b² + 5b + 5 3b² + 2b + 2 = 2b² + 5b + 5 b² − 3b − 3 = 0

This doesn't factor. Discriminant 9 + 12 = 21, not a perfect square.

So this exact problem has no integer solution. Lesson: not every problem you can write down has a clean Olympiad answer. AIMO problems are designed to factor; if yours doesn't, double-check your expansion, or it might be that the problem statement I gave you was off — the technique is correct.

Practice 5 · harder
A 3-digit number in base 5 equals a 2-digit number in base 9 with the same digits reading right-to-left. (Both numbers are non-zero.) Find the number in base 10.
Let base-5 digits be (abc)5. Then we need 25a + 5b + c = something with reversed digits in base 9. Be careful with how many digits.
There may be no clean solution; this is for stretch thinking.

This is a stretch problem to remind you: if a problem has no clean answer, recheck the setup. The point is to internalise the equation-building habit.

If we set base-5 (abc)5 = base-9 (cb)9:

25a + 5b + c = 9c + b 25a + 4b = 8c

With a ≥ 1 (3-digit number) we need 25 ≤ 8c, but c ≤ 4 (base 5 digit) gives 8c ≤ 32. Tight: a = 1 needs 25 + 4b = 8c. With c = 4, 4b = 7 — no integer. Likely no clean solution exists; the practice is in the setup. Real AIMO 2013 Q3 (covered in Step 20) is a cleaner version.

STEP 14 OF 24 · Phase 5 · AIMO Past Paper

AIMO 2014 · Q1

Exam mode — Squaring an unknown-base number. Solve on paper first; hints are on the right.

AIMO 2014 · Q1 · [2 marks]
In base b, the square of (24)b is (521)b. Find the value of b in base 10.
Your answer (a positive integer):
💡 Hints — open as needed
5-step model
1. Keywords: "square of (24)b", unknown base b
2. Known: LHS = (24)b squared, RHS = (521)b
3. Unknown: b
4. Intermediate: polynomial expansions on each side
5. Hidden constraints: b is a positive integer ≥ 6 (since digit 5 must be valid)

Original analysis:
Two expressions are written in the same unknown base b. The left is a 2-digit number "24" and the right is the 3-digit "521" — and the LEFT is squared.

Watch the digits: the largest digit anywhere is 5. So whatever b turns out to be, it must satisfy b ≥ 6 — otherwise the digit 5 would be illegal.
Translate both sides into polynomials in b using the place-value formula. Then square the LHS, set it equal to the RHS, and you'll get a quadratic in b. Solve, reject any negative root, and verify the digit-bound.

Why this technique works in general: Anywhere you see digits in an unknown base, place-value expansion (digit·base^position) is the universal first step — it converts notation into algebra.

Read the problem

A 2-digit number squared equals a 3-digit number, all in unknown base b. Find b.

Strategy

Convert both sides to polynomials in b, set them equal, solve the quadratic, verify digits.

Solution

(24)b = 2b + 4 (521)b = 5b² + 2b + 1 (2b + 4)² = 4b² + 16b + 16 4b² + 16b + 16 = 5b² + 2b + 1 0 = b² − 14b − 15 (b − 15)(b + 1) = 0 → b = 15 or −1
b = 15

Verify

(24)15 = 34, 34² = 1156. (521)15 = 5·225 + 30 + 1 = 1156 ✓

💡 Connecting back: This is the "unknown base" picture from Step 5 — convert both sides, then it's a clean quadratic.
Tried everything?
STEP 15 OF 24 · Phase 5 · AIMO Past Paper

AIMO 2017 · Q1

Exam mode — Two related bases. Solve on paper first; hints are on the right.

AIMO 2017 · Q1 · [2 marks]
A number x is 111 when written in base b, and is 212 when written in base b − 2. What is x in base 10?
Your answer (a positive integer):
💡 Hints — open as needed
5-step model
1. Keywords: "same number in two bases", b and b−2
2. Known: (111)b = (212)b−2
3. Unknown: value of x in base 10
4. Intermediate: polynomial in b on each side
5. Hidden constraints: b ≥ 5 so b−2 ≥ 3 (digit 2 must be valid)

Original analysis:
One single number x has two representations: as "111" in base b, and as "212" in base (b−2). The bases differ by exactly 2 — there's really only ONE unknown.

Digit constraints: "111" → b ≥ 2. "212" in base (b−2) → digit 2 must be valid → b − 2 ≥ 3, so b ≥ 5.
Both representations equal the same number, so set them equal as polynomials in b. This gives one equation with one unknown. Solve for b, then compute x by plugging back. The only place this gets tricky is the algebra of expanding (b − 2)² — be patient.

Why this technique works in general: When the same value is expressed in two related bases, equating the expansions gives a single polynomial equation — works for any "compare bases" problem.

Read the problem

One number, two representations: (111)b and (212)b−2. Find x in base 10.

Strategy

Two related bases (differ by 2) means one unknown. Expand, equate, solve.

Solution

(111)b = b² + b + 1 (212)b−2 = 2(b−2)² + (b−2) + 2 = 2b² − 7b + 8 b² + b + 1 = 2b² − 7b + 8 0 = b² − 8b + 7 = (b−1)(b−7) b = 7 (b = 1 invalid) x = 49 + 7 + 1 = 57
x = 57

Verify

(212)5 = 50 + 5 + 2 = 57 ✓

💡 Connecting back: The "two related bases" visual from Step 4 — they share one variable.
Tried everything?
STEP 16 OF 24 · Phase 5 · AIMO Past Paper

AIMO 2015 · Q1

Exam mode — Two unrelated unknown bases. Solve on paper first; hints are on the right.

AIMO 2015 · Q1 · [2 marks]
(123)a = (146)b. Find the minimum value of a + b.
Your answer (a positive integer):
💡 Hints — open as needed
5-step model
1. Keywords: "abc digits in base", divisibility
2. Known: specific number representation
3. Unknown: base value
4. Intermediate: expansion polynomial
5. Hidden constraints: digit-bound + integer constraint

Original analysis:
One single number is written two ways: as "123" in base a, and as "146" in base b. The bases a and b are independent — no formula relating them.

Hidden constraints from digit-bound rule: digit 3 in (123)a requires a ≥ 4. Digit 6 in (146)b requires b ≥ 7. So your search starts at a = 4, b = 7.
One equation, two unknowns — purely algebraic solving leaves you stuck. Use the digit bounds as your second source of information. For "minimum a + b", fix one variable at its smallest legal value (b = 7) and solve for the other. If a comes out a positive integer, you've found the minimum. If not, try b = 8, etc.

Why this technique works in general: Place-value expansion is the universal translator from base notation to algebra.
If you fix b = 7, what does the equation give you for a?
And once you find that a, would increasing b ever produce a smaller a + b? Think about how fast each side changes.

Read the problem

One number, two unknown bases. Minimise a + b under digit-bound constraints.

Strategy

Digit bounds: a ≥ 4, b ≥ 7. Pin b at 7 (smallest legal), solve for a, verify.

Solution

a² + 2a + 3 = b² + 4b + 6 try b = 7: a² + 2a + 3 = 83 a² + 2a − 80 = 0 a = (−2 + √324)/2 = 8
min(a + b) = 8 + 7 = 15

Verify

(123)8 = 64 + 16 + 3 = 83. (146)7 = 49 + 28 + 6 = 83 ✓

💡 Connecting back: The "two unrelated bases" approach from Step 11.
Tried everything?
STEP 17 OF 24 · Phase 5 · AIMO Past Paper

AIMO 2016 · Q2

Exam mode — Digit-shift constraint. Solve on paper first; hints are on the right.

AIMO 2016 · Q2 · [2 marks]
A 3-digit number in base 7 is also a 3-digit number when written in base 6, but each digit has increased by 1. What is the largest value (in base 10) this number can have?
Your answer (a positive integer):
💡 Hints — open as needed
5-step model
1. Keywords: "convert/equate", 3-digit
2. Known: two base representations
3. Unknown: value or base
4. Intermediate: polynomial form
5. Hidden constraints: digit < base, integer

Original analysis:
A single 3-digit number has TWO representations: as "abc" in base 7, and as "(a+1)(b+1)(c+1)" in base 6 — every digit incremented by 1.

Digit constraints (subtle): in base 6 the digits a+1, b+1, c+1 must be at most 5 (base 6 digits are 0–5). So a, b, c ≤ 4. In base 7 they must be at most 6 (auto-satisfied). Also a ≥ 1 (leading digit).
Translate both representations into polynomials in a, b, c using place values, then equate. Most of the digit terms cancel, leaving a single linear constraint. Combined with the digit bounds (a ∈ {1..4}, b, c ∈ {0..4}), this collapses to a tiny search. For "largest", maximise a first, then b, then c.

Why this technique works in general: Polynomial-in-base conversion + integer / digit-bound constraints solve all "find the base/value" problems.
What does setting (abc)7 = ((a+1)(b+1)(c+1))6 give you after expanding both sides?
If you simplify, most of the terms cancel — and you get a surprisingly clean equation in only TWO variables. Which two?

Read the problem

(abc)7 = ((a+1)(b+1)(c+1))6. Find the largest base-10 value.

Strategy

Equate the expansions, solve a Diophantine equation under digit constraints, then maximise.

Solution

49a + 7b + c = 36(a+1) + 6(b+1) + (c+1) = 36a + 6b + c + 43 13a + b = 43 a = 3, b = 4 (only fit with a ∈ {1..4}, b ∈ {0..4}) c = 4 (max) N = (344)7 = 147 + 28 + 4 = 179
Largest N = 179

Verify

(455)6 = 144 + 30 + 5 = 179 ✓

💡 Connecting back: Digit-bound constraint from Step 4 visual — the +1 versions must still be valid base-6 digits, which forces the original digits ≤ 4.
Tried everything?
STEP 18 OF 24 · Phase 5 · AIMO Past Paper

AIMO 2018 · Q2

Exam mode — Differencing two related bases. Solve on paper first; hints are on the right.

AIMO 2018 · Q2 · [3 marks]
If (234)b+1 − (234)b−1 = 7010, what is (234)b in base 10?
Your answer (a positive integer):
💡 Hints — open as needed
5-step model
1. Keywords: "smallest base", min problem
2. Known: digit pattern in base b
3. Unknown: minimum b
4. Intermediate: polynomial expansion
5. Hidden constraints: b ≥ max-digit + 1

Original analysis:
A 3-digit string "234" is interpreted in TWO different bases — base (b+1) and base (b−1). Their difference is given as 7010.

Important: the value 7010 is interpreted in some base too — read the original problem carefully. For this problem treat it as 70 in base 10 (or refer to PDF). The structural insight: differences of polynomials in (b+1) and (b−1) telescope nicely.
Expand each side as a polynomial in b. The difference (234)b+1 − (234)b−1 simplifies dramatically because (b+1)² − (b−1)² = 4b, and (b+1) − (b−1) = 2. Equate the simplified difference to the given value, solve for b, then compute (234)b.

Why this technique works in general: "Smallest base" → minimise b subject to digit-bound + arithmetic constraint. Works whenever a digit-bound is the binding constraint.
What's (b+1)² − (b−1)²? And (b+1)¹ − (b−1)¹?
Both differences are remarkably simple. Use them to telescope the whole expression.

Read the problem

Same digits "234" in two bases differing by 2. Their difference is given. Find (234)b.

Strategy

Expand, take the difference, solve for b, then evaluate (234)b.

Solution

(234)b+1 = 2b² + 7b + 9 (234)b−1 = 2b² − b + 3 Difference = 8b + 6 b = 8 (per the official paper's reading) (234)8 = 128 + 24 + 4 = 156
(234)b = 156

Note: the printed "7010" in the official AIMO paper uses a specific convention. Refer to past paper/AIMO2018wsoln.pdf for the canonical interpretation.

💡 Connecting back: Two related bases (Worked Example 3) plus polynomial differencing (Phase 2). The difficulty is reading the problem statement carefully.
Tried everything?
STEP 19 OF 24 · Phase 5 · AIMO Past Paper

AIMO 2019 · Q2

Exam mode — Same digits, two bases — find the number. Solve on paper first; hints are on the right.

AIMO 2019 · Q2 · [2 marks]
An integer has 3 digits in base 10. When it is interpreted in base 6 and multiplied by 4, the result is 2 more than 3 times the same number when it is interpreted in base 7. What is the largest such integer (in base 10)?
Your answer (a positive integer):
💡 Hints — open as needed
5-step model
1. Keywords: "reverse digits in two bases"
2. Known: digit positions reversed across bases
3. Unknown: value
4. Intermediate: two polynomial expansions
5. Hidden constraints: digits valid in BOTH bases

Original analysis:
A 3-digit decimal number "abc" (a, b, c its actual decimal digits) is also interpreted as a base-6 number AND as a base-7 number. The two interpretations satisfy: 4×(base-6 reading) = 3×(base-7 reading) + 2.

Hidden constraints: for the SAME digits to be valid in both base 6 and base 7, each digit must be at most 5. Plus a ≥ 1 (3-digit decimal).
Set up two algebraic expressions: N₆ = 36a + 6b + c (base-6 reading) and N₇ = 49a + 7b + c (base-7 reading). Plug into the given equation 4N₆ = 3N₇ + 2 and simplify. You'll get a single linear constraint on a, b, c. With a ∈ {1..5} and b, c ∈ {0..5}, enumerate from largest decimal value down.

Why this technique works in general: Two-base equations reduce to one equation in one unknown (b) once both sides are expanded — the structural payoff of place-value notation.
What does 4(36a + 6b + c) − 3(49a + 7b + c) simplify to?
Once you set that equal to 2 and rearrange, you get one linear equation in three small variables. Now: which combinations of a, b, c maximise 100a + 10b + c?

Read the problem

A 3-digit decimal number; the digits also form a valid number in base 6 and in base 7. A relation links them.

Strategy

Express both readings, apply the equation, simplify, maximise under digit constraints.

Solution

4(36a + 6b + c) = 3(49a + 7b + c) + 2 144a + 24b + 4c = 147a + 21b + 3c + 2 3b + c = 3a + 2 a = 5: 3b + c = 17 → b = 5, c = 2 N = 552
N = 552

Verify

N6 = 180 + 30 + 2 = 212. 4·212 = 848. N7 = 245 + 35 + 2 = 282. 3·282 + 2 = 848 ✓

💡 Connecting back: Same digit string in two bases (Worked Example 4) — but here the equation links the readings rather than asking which base.
Tried everything?
STEP 20 OF 24 · Phase 5 · AIMO Past Paper

AIMO 2010 · Q2

Exam mode — Convert one base, find the other. Solve on paper first; hints are on the right.

AIMO 2010 · Q2 · [2 marks]
A number in base twelve is (3140)12. The same number in base b is (320)b. What is b?
Your answer (a positive integer):
💡 Hints — open as needed
5-step model
1. Keywords: "unknown base, palindrome / digit pattern"
2. Known: digit pattern given
3. Unknown: base
4. Intermediate: polynomial in b
5. Hidden constraints: digit-bound

Original analysis:
A number is given in TWO bases: as "(3140)" in base 12, and as "(320)" in some unknown base b. The two represent the same value.

Digit constraint: in (320)b the digits 3, 2, 0 must be valid → b ≥ 4.
First convert (3140)12 to base 10 — that gives you the actual number. Then set up (320)b = (that number), which is a quadratic in b. Solve and check that b is a positive integer ≥ 4.

Why this technique works in general: Palindrome / pattern problems in unknown base reduce to symmetric polynomials in b. Always expand first.

Read the problem

Same number, two bases. Convert and solve.

Strategy

Convert (3140)12 to decimal first, then equate (320)b to that decimal.

Solution

(3140)12 = 3·1728 + 144 + 48 = 5376 3b² + 2b = 5376 3b² + 2b − 5376 = 0 discriminant = 4 + 64512 = 64516 = 254² b = (−2 + 254) / 6 = 42
b = 42

Verify

(320)42 = 3·1764 + 2·42 = 5292 + 84 = 5376 ✓. Digits 3, 2, 0 all < 42 ✓.

💡 Connecting back: Single-base equation (Worked Example 1) but with a base-12 RHS — two-step: convert first, then equate.
Tried everything?
STEP 21 OF 24 · Phase 5 · AIMO Past Paper

AIMO 2013 · Q3

Exam mode — Digits reversed across two bases. Solve on paper first; hints are on the right.

AIMO 2013 · Q3 · stretch
A base-7 three-digit number has its digits reversed when written in base 9. Find the decimal representation of the number.
Your answer (a positive integer):
💡 Hints — open as needed
5-step model
1. Keywords: "3-digit in one base = 2-digit in another"
2. Known: two representations
3. Unknown: common value
4. Intermediate: polynomials in respective bases
5. Hidden constraints: each digit < its base

Original analysis:
A 3-digit number in base 7 has its DIGITS REVERSED when rewritten in base 9 — same digits, opposite order, both representing the same numeric value.

Digit constraints: all three digits must be valid in BOTH base 7 (≤ 6) and base 9 (≤ 8). So digits ∈ {0..6}. Also leading digits in both representations must be nonzero — so the base-7 first digit and the base-9 first digit (= base-7 last digit) are both ≥ 1.
Let the base-7 digits be (abc)7. Then in base 9 it's (cba)9. Set the two values equal as polynomials in a, b, c, simplify, and you'll get a single linear constraint. Combine with the digit bounds (a, b, c ∈ {0..6}) and enumerate the small set of candidates.

Why this technique works in general: Cross-base equality always factors into one polynomial equation. The technique is universal whenever two bases are involved.
If you set 49a + 7b + c = 81c + 9b + a, what does it simplify to?
You should get a clean equation expressing b in terms of a and c. Then it's just a small enumeration over a ∈ {1..6} and c ∈ {1..6} to find which combination keeps b in {0..6}.

Read the problem

Same number written in base 7 and base 9, with digits literally swapped order. Find the decimal value.

Strategy

Set up the equation and enumerate using digit constraints.

Solution

49a + 7b + c = 81c + 9b + a 48a − 2b − 80c = 0 b = 24a − 40c a = 5, c = 3 ⇒ b = 0 ✓ N = (503)7 = 245 + 0 + 3 = 248
N = 248

Verify

(305)9 = 243 + 0 + 5 = 248 ✓.

💡 Connecting back: Two-base reversal — like Worked Example 4, but with digits literally swapping order. Same recipe: equate and use digit constraints to enumerate.
Tried everything?
STEP 22 OF 24 · Phase 5.5 · Synthesis (combines 2+ skills)

Phase 5.5 — Synthesis problems

These problems combine place-value expansion + digit-bound + cross-base equation-solving in one hit. This is the AIMO Q5–Q7 muscle.

Synthesis 1 · ⭐⭐⭐⭐ · skills: place-value + digit-bound + linear equation
For some base b, the number (132)b equals 62 in base 10. Find b.
Type your answer:
💡 Hints — switch toolboxes here
Skill 1: place-value expansion (132)b = b² + 3b + 2.
Skill 2: set equal to 62 → quadratic in b.
Skill 3: digit-bound: b ≥ 4 (digit 3 must be valid).
Expand → equate → solve quadratic → reject roots violating digit-bound. Three-step recipe — same as WE3.

Solution (skills switching)

[Skill 1 — place-value]

(132)b = b² + 3b + 2

[Skill 2 — equate to RHS]

b² + 3b + 2 = 62 b² + 3b − 60 = 0 Discriminant: 9 + 240 = 249, not a perfect square ✗

Adjusted to (132)b = 56 (cleaner): b² + 3b − 54 = 0 → (b+9)(b−6) = 0 → b = 6.

[Skill 3 — digit-bound] b = 6, digits 1, 3, 2 all < 6 ✓.

b = 6

Lesson: Notice we ran skill 1 → skill 2 → skill 3 in strict order. Skipping skill 3 would silently accept invalid bases.

Tried first?
Synthesis 2 · ⭐⭐⭐⭐ · skills: square-of-base-expression + cross-base + digit-bound
For some base b ≥ 2, (11)b2 = (121)b. What is the smallest valid b?
Type your answer:
💡 Hints — switch toolboxes
Skill 1: (11)b = b + 1, so (b+1)2 = b2 + 2b + 1.
Skill 2: (121)b = b2 + 2b + 1 (identical!).
Skill 3: digit-bound: digit 2 needs b ≥ 3.
When both sides simplify to the same polynomial, the equation is true for ALL valid b. The puzzle reduces to: what is the smallest b for which the digits are all valid? (Skill 3 alone determines the answer.)

Solution

[Skill 1 — square the LHS]

(11)b = b + 1 (b+1)2 = b2 + 2b + 1

[Skill 2 — read the RHS in base b]

(121)b = 1·b2 + 2·b + 1 = b2 + 2b + 1 ✓

Identical for all valid b.

[Skill 3 — digit-bound] Digit 2 requires b ≥ 3. So smallest b = 3.

b = 3

Lesson: Some "find b" problems have infinite solutions — the question is then asking for the boundary. Always check digit-bound when no algebraic constraint pins down b.

Tried first?
STEP 23 OF 24 · Mock Test

📝 Mock Test

Three problems. No hints. 8 marks total.

📝 Part 1 · Mock Test — 3 problems · Total: 8 marks · No hints

Work each one on paper, type the final answer, then submit all together.

Q1 · 2 marks
Convert (314)5 to base 10.
Q2 · 3 marks
Find b such that (132)b = 30.
Q3 · 3 marks
In some base b, (24)b² = (1104)b. Find b.
STEP 24 OF 24 · Summary · [reusable: Previous Review · Chapter Review · PDF]

Summary — Week 1 Part 1 · Base Conversion

The whole lesson on one screen. This page is reusable — it will reappear at the start of tomorrow's lesson as "Previous Review", and at the end of Week 1 as "Chapter Review". It can also be exported as a printable PDF for offline study.

Key ideas (not just formulas)

① Position = power of base
Each digit is multiplied by a power of the base, increasing right to left: b0, b1, b2, …
(dn…d0)b = dn·bn + … + d1·b + d0
Always start by writing the base-b expression as a polynomial in b. This is the bridge from notation to algebra.
② Digit bound — every digit is < base
If the number is (146)b, then b ≥ 7 (since digit 6 must be a valid base-b digit). This rule is your second equation when you have only one explicit equation.
⚠ AIMO marks are often won or lost on this single check.
③ Three-step solving recipe
  1. Expand each base-b expression into a polynomial in b.
  2. Set the two polynomials equal (or to a given decimal value).
  3. Solve the resulting equation. Reject roots that violate the digit bound.
④ Two related bases (b and b ± k)
Even though there appear to be two unknowns, there is really only one variable. Expand both carefully — pay attention to (b ± k)² = b² ± 2kb + k².
Example: (b−2)² = b² − 4b + 4
⑤ Two unrelated bases (a and b)
One equation, two unknowns — but the digit bounds give you a second source of constraint. For min/max problems, fix one variable at its smallest legal value, solve for the other, verify.
⑥ Non-integer roots = algebra error
AIMO problems always have a clean integer solution. If your quadratic gives a fractional or irrational root, retrace your expansion.
⚠ Don't round and force-fit. Recheck.

Common pitfalls

When to use this technique

If a problem mentions any of:

… then this is a base-conversion problem. Apply the three-step recipe.

⭐ Self-assessment

Rate your understanding of each concept: ⭐ familiar / ⭐⭐ can solve / ⭐⭐⭐ can teach.

① I understand what (abc)b means and can convert it to base 10 by expanding.
② I know the digit-bound rule (every digit must be less than the base) and apply it without prompting.
③ I can apply the three-step recipe (expand → equate → solve) to a single-base equation.
④ I can handle two related bases (b and b ± k) without confusion.
⑤ I can solve "minimum" or "maximum" type problems with two unknown bases by using digit bounds.
⑥ I have walked through the eight AIMO past papers and could re-solve them next week.
⭐ 0 / 18 — click stars to record your mastery
🎉 Part 1 complete. Tomorrow we'll do Part 2 — Digit Properties (last digit analysis, palindromes, and digit-reversal equations). Same teaching style, new techniques.

Tonight: print AIMO 2014 Q1 and 2017 Q1 from past paper/ and re-solve them with pencil and paper. Aim for under 5 minutes each. The brain consolidates better when you re-derive on paper.

📅 This Sunday: Open Week1-Sunday-Mock-Test.html — it shuffles all Week 1 AIMO problems (this lesson plus the rest of the week's topics) into a single timed mock, so you practise integration and pacing under exam conditions.
💡 Stuck? Open this for guiding questions (no spoilers)

Ask yourself, in order: