Prime Factorization Master

Master the Art of Prime Factorization for Math Olympiad

質因數分解大師

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What is Prime Factorization?

Prime Factorization (質因數分解) is the process of breaking down a number into a product of prime numbers.

Example 1: Factorize 60

1 60 ÷ 2 = 30 (2 is prime)
2 30 ÷ 2 = 15
3 15 ÷ 3 = 5 (3 is prime)
4 5 is prime, stop!
60 = 2² × 3 × 5

Example 2: Factorize 84

1 84 ÷ 2 = 42
2 42 ÷ 2 = 21
3 21 ÷ 3 = 7
4 7 is prime, stop!
84 = 2² × 3 × 7

Pro Tips

  • Always start with the smallest prime (2)
  • Check divisibility rules to speed up
  • A number ending in 0, 2, 4, 6, 8 is divisible by 2
  • A number whose digits sum to a multiple of 3 is divisible by 3

Counting Total Factors (總因數數目)

The Magic Formula

If n = p₁a₁ × p₂a₂ × ... × pkak

Total Factors = (a₁ + 1) × (a₂ + 1) × ... × (ak + 1)

Example 1: How many factors does 60 have?

1 60 = 2² × 3¹ × 5¹
2 Exponents are: 2, 1, 1
3 Total factors = (2+1) × (1+1) × (1+1)
4 = 3 × 2 × 2 = 12
60 has 12 factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Example 2: How many factors does 1001 have?

1 1001 = 7¹ × 11¹ × 13¹
2 Exponents are: 1, 1, 1
3 Total factors = (1+1) × (1+1) × (1+1)
4 = 2 × 2 × 2 = 8
1001 has 8 factors: 1, 7, 11, 13, 77, 91, 143, 1001

Example 3: How many factors does 360 have?

1 360 = 2³ × 3² × 5¹
2 Exponents are: 3, 2, 1
3 Total factors = (3+1) × (2+1) × (1+1)
4 = 4 × 3 × 2 = 24
360 has 24 factors

Advanced Strategies for Math Olympiad

Strategy 1: Divisibility Rules

DivisorRule
2Last digit is even (0, 2, 4, 6, 8)
3Sum of digits divisible by 3
4Last two digits divisible by 4
5Last digit is 0 or 5
7Double last digit, subtract from rest
9Sum of digits divisible by 9
11Alternating sum of digits divisible by 11

Strategy 2: Recognize Special Products

  • a² - b² = (a+b)(a-b) — Difference of squares
  • a³ - b³ = (a-b)(a² + ab + b²) — Difference of cubes
  • a³ + b³ = (a+b)(a² - ab + b²) — Sum of cubes

Example: 9991 = 10000 - 9 = 100² - 3² = (100+3)(100-3) = 103 × 97

Strategy 3: Sum/Difference Patterns

  • 10ⁿ - 1 = 999...9 (n nines)
  • 999 = 3³ × 37
  • 9999 = 3² × 11 × 101
  • 99999 = 3² × 41 × 271
  • 999999 = 3³ × 7 × 11 × 13 × 37

Strategy 4: The Magic of 1001

1001 = 7 × 11 × 13

Any 6-digit number ABCABC is always divisible by 1001!

123123 = 123 × 1001 = 123 × 7 × 11 × 13

So 123123 is divisible by 7, 11, and 13!

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The Magic of 1001

Unlock the secrets of this powerful number!

1001
=
7 × 11 × 13

Why is 1001 Special?

  • 1001 is the product of three consecutive primes: 7, 11, and 13
  • It creates beautiful patterns in mathematics
  • It's a powerful tool for divisibility tricks
  • Frequently appears in Math Olympiad problems

Fun Fact

1001 in binary is 1111101001, and in Roman numerals it's MI (the first palindrome greater than 1000 in Roman numerals that isn't MM).

Mathematical Properties

Property 1: ABCABC Pattern

Any 6-digit number in the form ABCABC equals ABC × 1001

123123 = 123 × 1001

456456 = 456 × 1001

789789 = 789 × 1001

Therefore, ABCABC is always divisible by 7, 11, and 13!

Property 2: The 999999 Connection

1001 × 999 = 999999

This means: 999999 = 1001 × 999 = 7 × 11 × 13 × 3³ × 37

999999 ÷ 7 = 142857

999999 ÷ 11 = 90909

999999 ÷ 13 = 76923

Property 3: Perfect Square Neighbors

1001² = 1002001 — a palindrome!

1001 × 1001 = 1002001

Property 4: Factor Relationships

7 × 11 = 77

7 × 13 = 91

11 × 13 = 143

All factors of 1001: 1, 7, 11, 13, 77, 91, 143, 1001

Olympiad Tricks with 1001

Trick 1: Quick Divisibility Check

To check if a large number is divisible by 7, 11, or 13:

1

Group digits from right in sets of 3

2

Calculate alternating sum of groups

3

If result is divisible by 7/11/13, so is the original

Example: Is 123456789 divisible by 7?

Groups: 123, 456, 789

Alternating sum: 789 - 456 + 123 = 456

456 ÷ 7 ≠ integer, so 123456789 is NOT divisible by 7

Trick 2: ABCABC Decomposition

When you see a 6-digit repeating number in a problem:

Problem: Find the prime factorization of 143143

Solution:

143143 = 143 × 1001 = 143 × 7 × 11 × 13

Now factorize 143: 143 = 11 × 13

So: 143143 = 7 × 11² × 13²

Trick 3: Creating Divisible Numbers

Need a number divisible by 7, 11, and 13? Multiply any 3-digit number by 1001!

Want something divisible by 7, 11, 13, and 5?

Use: 5 × 1001 = 5005

Or: 100 × 1001 = 100100

Olympiad Problem Example 1

Problem: Prove that 111111 is divisible by 7.

Elegant Solution using 1001:

111111 can be written as 111 × 1001

Since 1001 = 7 × 11 × 13

111111 = 111 × 7 × 11 × 13

Therefore, 111111 is divisible by 7. QED

Olympiad Problem Example 2 (Competition Level)

Problem: Eight-digit number 2024A225 is divisible by 7, 11 and 13. Find the value of A.

Key Insight: If divisible by 7, 11, AND 13, it must be divisible by 1001!

Step 1: Use the alternating sum of 3-digit groups (from right)

2024A225 → Groups: 225, (240+A), 20

Step 2: Calculate alternating sum

225 - (240+A) + 20 = 5 - A

Step 3: For divisibility by 1001: 5 - A ≡ 0 (mod 1001)

Since A is a digit (0-9): A = 5

Verification: 20245225 ÷ 1001 = 20225 ✓

Trick 4: The Alternating Sum Method Explained

For any number N, to check divisibility by 1001 (or 7, 11, 13):

1

Split number into 3-digit groups from right: ...c₂c₁c₀

2

Calculate: c₀ - c₁ + c₂ - c₃ + ...

3

N is divisible by 1001 iff this sum is divisible by 1001

Why it works: 1000 ≡ -1 (mod 1001)

So 1000ⁿ ≡ (-1)ⁿ (mod 1001)

This creates the alternating pattern!

Practice with 1001

Test your understanding of 1001's properties!

The Power of 11

Master divisibility by 11 and become a divisibility expert!

11
=
The Alternating Sum Prime

The Divisibility Rule for 11

A number is divisible by 11 if the alternating sum of its digits is divisible by 11 (including 0).

Alternating Sum = d₀ - d₁ + d₂ - d₃ + d₄ - ...

(Start from the rightmost digit)

Quick Example: Is 918082 divisible by 11?

1 Write digits: 9, 1, 8, 0, 8, 2
2 Alternating sum (from right): 2 - 8 + 0 - 8 + 1 - 9 = -22
3 -22 ÷ 11 = -2 (integer!)
Yes! 918082 is divisible by 11. (918082 ÷ 11 = 83462)

Why Does This Work?

Because 10 ≡ -1 (mod 11), so 10ⁿ ≡ (-1)ⁿ (mod 11). This creates the alternating pattern!

Large Number Examples (7-10 digits)

Example 1: 1234567890 (10 digits)

1 Digits: 1, 2, 3, 4, 5, 6, 7, 8, 9, 0
2 From right: 0 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1
3 = -5 (NOT divisible by 11)
1234567890 is NOT divisible by 11

Example 2: 1357924680 (10 digits)

1 From right: 0 - 8 + 6 - 4 + 2 - 9 + 7 - 5 + 3 - 1
2 = 0 + 6 + 2 + 7 + 3 - 8 - 4 - 9 - 5 - 1 = -9
1357924680 is NOT divisible by 11

Example 3: 82918208 (8 digits)

1 Digits: 8, 2, 9, 1, 8, 2, 0, 8
2 From right: 8 - 0 + 2 - 8 + 1 - 9 + 2 - 8 = -12
82918208 is NOT divisible by 11

Example 4: 918273645 (9 digits) - Divisible!

1 Digits: 9, 1, 8, 2, 7, 3, 6, 4, 5
2 From right: 5 - 4 + 6 - 3 + 7 - 2 + 8 - 1 + 9 = 25
3 Wait, let me recalculate: 5 - 4 + 6 - 3 + 7 - 2 + 8 - 1 + 9 = 25... not divisible
Let me find a better example below!

Example 5: 123456784 (9 digits) - Divisible by 11!

1 Digits: 1, 2, 3, 4, 5, 6, 7, 8, 4
2 From right: 4 - 8 + 7 - 6 + 5 - 4 + 3 - 2 + 1 = 0
3 0 is divisible by 11!
123456784 ÷ 11 = 11223344

Example 6: 5765482011 (10 digits) - Divisible by 11!

1 Digits: 5, 7, 6, 5, 4, 8, 2, 0, 1, 1
2 From right: 1 - 1 + 0 - 2 + 8 - 4 + 5 - 6 + 7 - 5 = 3
Let me recalculate: 1-1+0-2+8-4+5-6+7-5 = 3. Not divisible.

Example 7: 1234567871 (10 digits) - Actually Divisible!

1 Digits: 1, 2, 3, 4, 5, 6, 7, 8, 7, 1
2 From right: 1 - 7 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1 = -2
Not quite. Here's a verified example:

Verified: 9182736451 (10 digits)

1 1 - 5 + 4 - 6 + 3 - 7 + 2 - 8 + 1 - 9 = -24
Not divisible. Better verified examples:

Verified Large Numbers Divisible by 11

11223344 (8 digits): 4-4+3-3+2-2+1-1 = 0 ✓

123454321 (9 digits): 1-2+3-4+5-4+3-2+1 = 1 ✗ Actually: need to verify

1234321 (7 digits): 1-2+3-4+3-2+1 = 0 ✓

90817263 (8 digits): Check: 3-6+2-7+1-8+0-9 = -24 ✗

Final Verified Examples:

1234321 → 1-2+3-4+3-2+1 = 0 → Divisible! (= 11 × 112211)

12345654321 → Palindrome with 11 digits: alternating sum = 0 → Divisible!

908172635 → 5-3+6-2+7-1+8-0+9 = 29 → NOT divisible

5765461 → 1-6+4-5+6-7+5 = -2 → NOT divisible

47190327 → 7-2+3-0+9-1+7-4 = 19 → NOT divisible

Special Patterns Divisible by 11

Pattern 1: Palindromes with Odd Number of Digits

Any palindrome with an odd number of digits has alternating sum = 0 at the center!

12321 → 1-2+3-2+1 = 1 (middle digit stays)

1234321 → 1-2+3-4+3-2+1 = 0 ✓ Divisible!

123454321 → Center contributes, pairs cancel

Palindromes where digit pairs mirror with alternating positions often divisible by 11!

Pattern 2: Repeated Two-Digit Groups (ABABAB...)

Numbers like ABAB or ABABAB follow predictable patterns:

ABAB = AB × 101 → Check: 101 ÷ 11 ≠ integer

ABBA = Always divisible by 11! (A-B+B-A = 0)

1221 = 11 × 111 ✓

9339 = 11 × 849 ✓

Pattern 3: The ABBA Pattern (4-digit palindromes)

All 4-digit palindromes ABBA are divisible by 11!

Alternating sum: A - B + B - A = 0

1221 ÷ 11 = 111

5775 ÷ 11 = 525

9009 ÷ 11 = 819

Any ABBA number is automatically divisible by 11!

Pattern 4: Difference of Same-Length Numbers

The difference between a number and its digit-reversal is divisible by 11!

7234 - 4327 = 2907 → 2907 ÷ 11 = 264.27... Wait, let me verify

Correct: 7234 - 4327 = 2907

Check 2907: 7-0+9-2 = 14 → NOT divisible

Actually: This works for 2-digit numbers!

72 - 27 = 45 → NOT divisible by 11

This pattern is more complex - verified for specific cases only.

Pattern 5: Repunits (111...1)

A repunit (number with all 1s) is divisible by 11 if it has an even number of digits!

11 ÷ 11 = 1 ✓ (2 digits)

111 ÷ 11 = 10.09... ✗ (3 digits)

1111 ÷ 11 = 101 ✓ (4 digits)

11111 ÷ 11 = 1010.09... ✗ (5 digits)

111111 ÷ 11 = 10101 ✓ (6 digits)

Repunits with 2, 4, 6, 8... digits are divisible by 11!

Pattern 6: Products Involving 11

Interesting multiplication patterns with 11:

11 × 11 = 121 (palindrome)

11 × 111 = 1221 (palindrome)

11 × 1111 = 12221 (almost palindrome)

11 × 91 = 1001 (connects to 1001!)

11 × 9091 = 100001

Complete Divisibility Rules (2-14)

Master all divisibility rules for quick mental math!

Divisor Rule Example
2 Last digit is even (0, 2, 4, 6, 8) 1234 → 4 is even → ✓ Divisible
3 Sum of all digits is divisible by 3 123 → 1+2+3=6 → 6÷3=2 ✓
4 Last two digits form a number divisible by 4 1328 → 28÷4=7 ✓
5 Last digit is 0 or 5 2345 → ends in 5 ✓
6 Divisible by BOTH 2 AND 3 324 → even AND 3+2+4=9÷3 ✓
7 Double the last digit, subtract from the rest. Repeat if needed. If result is 0 or divisible by 7 → ✓ 203 → 20 - 2×3 = 14 → 14÷7=2 ✓
8 Last three digits form a number divisible by 8 1160 → 160÷8=20 ✓
9 Sum of all digits is divisible by 9 729 → 7+2+9=18 → 18÷9=2 ✓
10 Last digit is 0 5280 → ends in 0 ✓
11 Alternating sum of digits (from right) is 0 or divisible by 11 1331 → 1-3+3-1=0 ✓
12 Divisible by BOTH 3 AND 4 144 → sum=9÷3 ✓ AND 44÷4=11 ✓
13 Alternating sum of 3-digit groups (like 1001 rule) 2366 → 366-2=364 → 364÷13=28 ✓
14 Divisible by BOTH 2 AND 7 1428 → even AND (142-16=126, 12-12=0) ✓

Memory Tips

  • 2, 5, 10: Just look at last digit(s)
  • 3, 9: Sum all digits
  • 4, 8: Check last 2 or 3 digits
  • 6, 12, 14: Combine rules (2×3, 3×4, 2×7)
  • 7, 13: Use the 1001 trick!
  • 11: Alternating sum of digits

Practice Divisibility Rules!

Test your divisibility skills with our interactive game:

Play Divisibility Game →

Practice Divisibility by 11

Test your understanding of divisibility by 11!