Competition math isn't just about knowing formulas β it's about knowing how to attack a problem. The best problem solvers have a toolkit of strategies. When you get stuck, try one of these four techniques before you give up.
When a problem asks about a large number (like "the 100th term" or "100 people"), start small:
This works for sequences, tiling problems, handshake problems, and more.
When you know the final result and need the starting value, reverse every operation:
Key: reverse the order too β the last operation forward becomes the first operation backward.
If a problem feels abstract, draw it out:
Drawing turns words into something your visual brain can process. It's not "wasting time" β it saves time.
On multiple-choice tests (like AMC 8/10), you can often work from the answers:
This is called "backsolving" and it's a legitimate competition strategy. Use it when setting up the equation is hard.
Strategy: Find a pattern from small cases.
Step 1: List the first few odd numbers with their positions:
1st: 1, 2nd: 3, 3rd: 5, 4th: 7, 5th: 9, ...
Step 2: Find the pattern.
Each time, we add 2. The nth odd number = $ 2n - 1 $
\((Check: n=1 \\to 1, n=2 \\to 3, n=3 \\to 5.\) β)
Step 3: Apply to n = 50:
\[ 2(50) - 1 = 100 - 1 = 99 \]
Answer: 99
Strategy: Work backwards β reverse every operation, in reverse order.
\(Forward steps: start \\to add 5 \\to multiply by 2 \\to subtract 3 \\to 11\)
\(Reverse steps: 11 \\to add 3 \\to divide by 2 \\to subtract 5 \\to answer\)
Step 1: Reverse "subtract 3" \(\\to add 3 to 11\)
$ 11 + 3 = 14 $
Step 2: Reverse "multiply by 2" \(\\to divide by 2\)
$ 14 \div 2 = 7 $
Step 3: Reverse "add 5" \(\\to subtract 5\)
$ 7 - 5 = 2 $
Check forward: \(Start with 2. Add 5 \\to 7. Multiply by 2 \\to 14. Subtract 3 \\to 11.\) β
Answer: 2
Strategy: Draw it out / find the pattern.
Method 1 (Drawing): Draw 6 points (people). Connect every pair with a line (handshake). Count the lines.
Method 2 (Pattern / Formula):
\(Person 1 shakes hands with 5 others \\to 5 handshakes\)
Person 2 already shook with Person 1, so 4 new handshakes
Person 3: 3 new, Person 4: 2, Person 5: 1, Person 6: 0
\[ 5 + 4 + 3 + 2 + 1 = 15 \]
General formula: For $n$ people, handshakes = $ \frac{n(n-1)}{2} $
Check: $ \frac{6 \times 5}{2} = 15 $. β
Answer: 15