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CMS Math Olympiad · 6th Grade

Olympiad Mix

Problems about cleverness, non-standard thinking, and different types of reasoning. These are what make olympiads fun!

Motion problems

Formula triangle

Distance = speed × time: \(S = v \times t\)
Speed = distance ÷ time: \(v = \dfrac{S}{t}\)
Time = distance ÷ speed: \(t = \dfrac{S}{v}\)

Two motion types

Towards each other: they close in at speed \(v_1 + v_2\).
In the same direction: they catch up at speed \(v_1 - v_2\) (if \(v_1 > v_2\)).

Example — meeting

Two cyclists ride towards each other. The first goes 15 km/h, the second 10 km/h. The distance between them is 50 km. After how many hours will they meet?

Closing speed: \(15 + 10 = 25\) km/h.

Time: \(50 \div 25 = 2\) hours.

CMS 2021 — Q23

A bus leaves stop A at 08:10. There are 8 intermediate stops. Between stops it travels 9 minutes, and at each stop it waits 1 minute. At what time does it return to A?

From A back to A through 8 stops makes 9 travel segments.

Careful: the trip A→1→2→...→8→A has 9 segments and 8 stops.

Time: \(9 \times 9 + 8 \times 1 = 81 + 8 = 89\) min.

08:10 + 89 min = 09:39.

Try it

A train travels at 80 km/h. How many minutes does it take to travel 20 km?

· · ·

New operations

On CMS they often invent a new operation with its own symbol. Do not panic — just substitute the numbers into the rule.

Example

Define: \(a \oplus b = 2a + 3b\). Find \(3 \oplus 4\).

Substitute \(a = 3\), \(b = 4\):

\(3 \oplus 4 = 2 \times 3 + 3 \times 4 = 6 + 12 = 18\).

Strategy

1. Read the definition of the operation.
2. Substitute the numbers for the letters.
3. Compute carefully — wrong answers are built from order-of-operations mistakes.

Try it

Define \(a \star b = a^2 - b\). The value of \(5 \star 3\):

Try it

Define \(a \diamond b = \dfrac{a + b}{a - b}\). The value of \(7 \diamond 3\):

· · ·

Equation systems with pictures

CMS often uses fruits, stars, or other symbols instead of variables. The method is the same as for ordinary equations.

Example

If \(\triangle + \triangle + \square = 17\) and \(\triangle + \square = 9\), find \(\triangle\).

From the second equation: \(\square = 9 - \triangle\).

Substitute into the first: \(\triangle + \triangle + (9 - \triangle) = 17\).

\(\triangle + 9 = 17\), so \(\triangle = 8\). Check: \(\square = 1\), \(8+8+1=17\) ✓.

Trick: subtract the equations

Another method: subtract the second equation from the first:
\((\triangle + \triangle + \square) - (\triangle + \square) = 17 - 9\)
\(\triangle = 8\). Instantly!

Try it

\(\bigcirc + \bigcirc + \bigcirc = 18\) and \(\bigcirc + \star = 11\). The value of \(\star\) is:

CMS 2024 — Q5

\(\Gamma \times \Delta = 14\), \(\Delta \times E = 10\), \(\Gamma \times E = 35\). Find \(\Gamma + \Delta + E\).

Multiply all three: \((\Gamma \Delta E)^2 = 14 \times 10 \times 35 = 4900\).

\(\Gamma \Delta E = 70\).

\(\Gamma = \dfrac{70}{10} = 7\), \(\Delta = \dfrac{70}{35} = 2\), \(E = \dfrac{70}{14} = 5\).

Sum: \(7 + 2 + 5 = 14\).

· · ·

Sequences and patterns

To find the next number or the missing term, you need to spot the rule.

Types of patterns

Arithmetic: constant difference. 3, 7, 11, 15, ... (each time +4).
Geometric: constant multiplier. 2, 6, 18, 54, ... (each time ×3).
Squares: 1, 4, 9, 16, 25, ... (\(n^2\)).
Mixed: alternation, pairs, nested rules.

Example — CMS 2021 Q22

Row 1: 1 circle, Row 2: 3 circles, Row 3: 5 circles, Row 4: 7 circles... How many are in row 2021?

Each row has 2 more circles than the previous row.

Row \(n\): \(2n - 1\) circles.

Row 2021: \(2 \times 2021 - 1 = 4041\).

Try it

Sequence: 2, 5, 10, 17, 26, ... The next number is:

Differences: 3, 5, 7, 9, ... (growing by 2).
Next difference: 11. So: \(26 + 11 = 37\).

Or: this is \(n^2 + 1\): \(1+1, 4+1, 9+1, 16+1, 25+1, 36+1 = 37\).

· · ·

Age problems

Main rule

The age difference never changes!
If a father is 25 years older than his son now, then in 10 years he still will still be older by 25 years.

Example

A mother is 36 years old and her daughter is 12. In how many years will the mother be twice as old as the daughter?

Difference: \(36 - 12 = 24\) years (always!).

The mother will be twice as old when the daughter’s age equals the difference: 24 years.

The daughter is now 12, so we need \(24 - 12 = 12\) more years.

Check: in 12 years the mother is 48 and the daughter 24. \(48 = 2 \times 24\) ✓.

Try it

The father is 40 and the son is 10. In how many years will the father be exactly three times as old as the son?

Age difference: \(40 - 10 = 30\) (always!).
In \(x\) years: father \(40+x\), son \(10+x\).
\(40+x = 3(10+x)\) → \(40+x = 30+3x\) → \(10 = 2x\) → \(x = 5\).
Check: father 45, son 15. \(45 = 3 \times 15\) ✓.

· · ·

Working backwards

When the final result is known but you need the starting number, go backwards and use opposite operations.

CMS 2024 — Q1

Number → ×8 → −14 → ÷2 → +4 → gives 9. What was the starting number?

Work backwards from the end:

\(9 \xrightarrow{-4} 5 \xrightarrow{\times 2} 10 \xrightarrow{+14} 24 \xrightarrow{\div 8} 3\).

Check: \(3 \times 8 = 24\), \(24 - 14 = 10\), \(10 \div 2 = 5\), \(5 + 4 = 9\) ✓.

Inverse operations

\(+\) ↔ \(-\), \(\times\) ↔ \(\div\). Start from the result and work back to the beginning.

Try it

Number → ×3 → +7 → ÷5 → result 4. The starting number is:

Working backwards: \(4 \xrightarrow{\times 5} 20 \xrightarrow{-7} 13 \xrightarrow{\div 3} \dfrac{13}{3}\).
Check: \(\dfrac{13}{3} \times 3 = 13\), \(13 + 7 = 20\), \(20 \div 5 = 4\) ✓.

· · ·

Olympiad practice

Problem 1 · Easy

Define \(a \# b = a \times b + a + b\). The value of \(3 \# 4\):

Problem 2 · Easy

Seven consecutive integers. The sum of the three smallest is 33. The sum of the three largest is:

The three smallest are \(n, n+1, n+2\). Their sum is \(3n+3 = 33\) → \(n = 10\).
The numbers are 10, 11, 12, 13, 14, 15, 16.
The three largest are \(14+15+16 = 45\).

Or faster: each of the three largest numbers is 4 more than the corresponding smallest one → difference = \(3 \times 4 = 12\). Answer: \(33+12=45\).
(This is CMS 2024 Q7!)

Problem 3 · Medium

A box with 30 chocolates weighs 1100 g. If 12 chocolates are removed, it weighs 680 g. The weight of the empty box is:

12 chocolates weigh: \(1100 - 680 = 420\) g.
One chocolate weighs: \(420 \div 12 = 35\) g.
30 chocolates weigh: \(35 \times 30 = 1050\) g.
The box weighs: \(1100 - 1050 = 50\) g.
(This is CMS 2024 Q9!)

Problem 4 · Medium

Five different integers from 1 to 30 have sum 30. The greatest possible value of the largest number is:

To make the largest as big as possible, the rest must be as small as possible:
\(1 + 2 + 3 + 4 + x = 30\) → \(x = 20\).
Check: 1, 2, 3, 4, 20 — all different, all from 1 to 30 ✓.
(This is CMS 2021 Q21!)

Problem 5 · Medium

How many handshakes occur if 8 people meet and each shakes hands with every other person?

The first person shakes hands with 7 others, the second with 6 new people, the third with 5...
\(7 + 6 + 5 + 4 + 3 + 2 + 1 = 28\).

Or: \(\dfrac{8 \times 7}{2} = 28\).
(This is CMS 2021 Q25!)

Problem 6 · Medium+

Number → ×4 → −12 → ÷2 → +8 → result 18. The starting number is:

Working backwards: \(18 \xrightarrow{-8} 10 \xrightarrow{\times 2} 20 \xrightarrow{+12} 32 \xrightarrow{\div 4} 8\).

Check: \(8 \times 4 = 32\), \(32 - 12 = 20\), \(20 \div 2 = 10\), \(10 + 8 = 18\) ✓.

✦

Cheat sheet — Olympiad Mix

Motion: \(S = v \times t\). Towards each other: \(v_1 + v_2\). Catching up: \(v_1 - v_2\)
New operation: just substitute the numbers into the definition
System of equations: subtract one equation from the other
Sequences: look for differences, multipliers, or a formula for the \(n\)-th term
Age problems: the age difference never changes over time
Working backwards: start from the result and use opposite operations
Handshakes: \(\dfrac{n(n-1)}{2}\)
Largest possible number: make the rest as small as possible

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