Factors & Multiples

Study Sheet

Factors & Multiples

Everything about factors, multiples, primes, GCF, and LCM

Factors of a Number

Concept
What a Factor Is

A factor of a number divides into it evenly, leaving no remainder. If a×b=na \times b = n, then both aa and bb are factors of nn.

To find all the factors of a number, work in pairs. Start at 11 and count upward: for each number that divides evenly, write down both partners. Stop once the pairs start to repeat.

Every whole number greater than 11 factors into primes in exactly one way — the fundamental theorem of arithmetic — and that factorization answers most questions about it. The greatest common factor takes the smallest power of each shared prime; the least common multiple takes the largest power of every prime that appears.

Why can you stop searching once the pairs meet? Factors come in pairs whose product is the number, and in each pair one factor is at most the square root and the other is at least the square root. For 3636, 36=6\sqrt{36} = 6, so every pair has a partner of 66 or less; after testing 11 through 66 you have found every factor, both small and large.

Example
Finding every factor of 36
1234691218366 × 6
Each arc joins a factor pair; 6×66 \times 6 pairs with itself.

Test each number in order and record the factor pairs:

1×36,2×18,3×12,4×9,6×6.\begin{aligned} 1 \times 36, \quad 2 \times 18, \quad 3 \times 12, \quad 4 \times 9, \quad 6 \times 6. \end{aligned}

(Note 55 does not divide 3636, so it is skipped.) Collecting both partners in order:

1, 2, 3, 4, 6, 9, 12, 18, 36.1,\ 2,\ 3,\ 4,\ 6,\ 9,\ 12,\ 18,\ 36.

So 3636 has 99 factors.

Side note
Where it goes wrong

Stopping at the small factors: "the factors of 3636 are 1,2,3,4,61, 2, 3, 4, 6." Each of those has a partner: 36,18,12,936, 18, 12, 9 (and 66 itself). A factor list is not complete until every pair is written down.

Tip

Every whole number has 11 and itself as factors. When a factor pair is a number times itself (like 6×6=366 \times 6 = 36), that number is listed only once — this happens for perfect squares.

Try it
Try it: all the factors

List every factor of 6060. How many are there?

Show the answer

Test 11 to 77 (since 72=49<60<64=827^2 = 49 < 60 < 64 = 8^2) and record pairs: 1×601 \times 60, 2×302 \times 30, 3×203 \times 20, 4×154 \times 15, 5×125 \times 12, 6×106 \times 10 (77 does not divide 6060). The factors are 1,2,3,4,5,6,10,12,15,20,30,601, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60: twelve of them.

Multiples of a Number

Concept
What a Multiple Is

A multiple of a number is what you get by multiplying it by 1,2,3,4,…1, 2, 3, 4, \dots In other words, it is any number in that number's “times table.” Multiples go on forever, while factors are a short, finite list.

Multiples of 6:6, 12, 18, 24, 30, 36, …\text{Multiples of } 6:\quad 6,\ 12,\ 18,\ 24,\ 30,\ 36,\ \dots

The multiples of 88 are the landing spots of equal jumps of 88 starting at 00. That picture makes "is 5454 a multiple of 88?" a question about landing: 66 jumps reach 4848 and 77 reach 5656, so 5454 is jumped over. Division says the same thing, since the remainder 66 is how far 5454 is past the last landing spot.

Example
Listing multiples and checking membership
0816243240485654
Jumps of 88 land on 4848 and 5656, never on 5454.

List the first five multiples of 88, then decide whether 5454 is a multiple of 88.

Multiples of 8:8, 16, 24, 32, 40.Is 54 a multiple?54÷8=6 remainder 6 ⇒ No.\begin{aligned} \text{Multiples of } 8:\quad & 8,\ 16,\ 24,\ 32,\ 40. \\ \text{Is } 54 \text{ a multiple?}\quad & 54 \div 8 = 6\text{ remainder } 6 \ \Rightarrow\ \text{No.} \end{aligned}

Because the division leaves a remainder, 5454 is not a multiple of 88.

Side note
Where it goes wrong

Confusing "multiple of" with "factor of": saying 44 is a multiple of 2020. Multiples are the bigger numbers in the times table: 2020 is a multiple of 44, and 44 is a factor of 2020.

Tip

“Factor” and “multiple” are opposite viewpoints of the same fact: since 4×5=204 \times 5 = 20, we can say 44 is a factor of 2020 and 2020 is a multiple of 44. Factors are smaller (or equal); multiples are larger (or equal).

Try it
Try it: multiples

List the multiples of 77 between 5050 and 100100. Is 9191 one of them?

Show the answer

The first multiple of 77 past 5050 is 7×8=567 \times 8 = 56. Counting up by 77: 56,63,70,77,84,91,9856, 63, 70, 77, 84, 91, 98. Yes, 91=7×1391 = 7 \times 13 is on the list.

Prime and Composite Numbers

Concept
Two Kinds of Whole Numbers
  • A prime number has exactly two factors: 11 and itself. The first primes are 2,3,5,7,11,13,17,19,23,…2, 3, 5, 7, 11, 13, 17, 19, 23, \dots
  • A composite number has more than two factors — it can be broken into smaller factors.
  • The number 11 is neither prime nor composite: it has only one factor.
  • 22 is the only even prime; every other even number has 22 as an extra factor.

Arrange nn dots in a rectangle. Every number can form a single row, 1×n1 \times n; a composite number can also form a genuine rectangle with at least two rows, and those rows and columns are its other factors. A prime cannot: its dots only ever make one long row. The square-root test works because any rectangle for nn has a side no longer than n\sqrt{n}.

Example
Prime or composite?
23: only one rectangle, 1 × 23 (prime)39: also a 3 × 13 rectangle (composite)
Dots in rectangles: a prime makes only one.

Classify 2323 and 3939.

23:no number from 2 to 4 divides it evenly ⇒ prime.39:39=3×13⇒ composite.\begin{aligned} 23:\quad & \text{no number from } 2 \text{ to } 4 \text{ divides it evenly } \Rightarrow\ \textbf{prime}. \\ 39:\quad & 39 = 3 \times 13 \Rightarrow\ \textbf{composite}. \end{aligned}

To test a number, try dividing by the small primes 2,3,5,7,…2, 3, 5, 7, \dots You only need to check primes up to its square root.

Side note
Where it goes wrong

Calling 11 prime ("it is only divisible by 11 and itself"). A prime has exactly two different factors; 11 has only one. Keeping 11 out is what makes prime factorizations unique: otherwise 6=2×3=1×2×36 = 2 \times 3 = 1 \times 2 \times 3 would be two different ones.

Tip

To check whether a number is prime, you do not have to test every smaller number — only primes up to its square root. For 5353, checking 2,3,5,72, 3, 5, 7 is enough (since 72=49<53<64=827^2 = 49 < 53 < 64 = 8^2), and none divide it, so 5353 is prime.

Try it
Try it: test with square roots

Is 9191 prime? Is 9797 prime? Which primes do you need to test?

Show the answer

92=819^2 = 81 and 102=10010^2 = 100, so testing the primes 2,3,5,72, 3, 5, 7 is enough for both. 91=7×1391 = 7 \times 13, so it is composite. 9797 is not divisible by 22 (odd), 33 (digit sum 1616), 55 (ends in 77) or 77 (7×13=917 \times 13 = 91, 7×14=987 \times 14 = 98), so 9797 is prime.

Divisibility Rules

Concept
Quick Tests for Even Division

These shortcuts tell you if a number divides evenly without doing the long division.

  • 2: the last digit is even (0,2,4,6,80, 2, 4, 6, 8).
  • 3: the sum of the digits is divisible by 33.
  • 4: the number formed by the last two digits is divisible by 44.
  • 5: the last digit is 00 or 55.
  • 6: it passes the tests for both 22 and 33.
  • 9: the sum of the digits is divisible by 99.
  • 10: the last digit is 00.

Each rule comes from place value. Every power of 1010 is a multiple of 22 and of 55, so only the ones digit can spoil divisibility by 22 or 55. Every 10,100,1000,…10, 100, 1000, \ldots is one more than a multiple of 99 (9,99,9999, 99, 999), so a number leaves the same remainder on division by 99 (or 33) as its digit sum does. And 100100 is a multiple of 44, so only the last two digits matter for 44.

Example
Testing 780 against every rule

Which of 2,3,4,5,6,9,102, 3, 4, 5, 6, 9, 10 divide 780780?

  • 2: last digit 00 is even ✓
  • 3: digits 7+8+0=157+8+0 = 15, divisible by 33 ✓
  • 4: last two digits 80÷4=2080 \div 4 = 20 ✓
  • 5: ends in 00 ✓
  • 6: passes 22 and 33 ✓
  • 9: digit sum 1515 is not divisible by 99 — no
  • 10: ends in 00 ✓

So 780780 is divisible by 2,3,4,5,6,2, 3, 4, 5, 6, and 1010, but not 99.

Side note
Where it goes wrong

Using the digit-sum test for 44: "1+3+6=101 + 3 + 6 = 10 is not divisible by 44, so 136136 is not." The digit sum only works for 33 and 99. For 44, check the last two digits: 36=4×936 = 4 \times 9, so 136=4×34136 = 4 \times 34 is divisible by 44.

Tip

The rule for 66 is really the rules for 22 and 33 working together. In the same way, a number divisible by 99 is automatically divisible by 33 — but not the other way around.

Try it
Try it: every rule

Which of 2,3,4,5,6,9,102, 3, 4, 5, 6, 9, 10 divide 2,3402{,}340?

Show the answer

Last digit 00: divisible by 22, 55 and 1010. Digit sum 2+3+4+0=92 + 3 + 4 + 0 = 9: divisible by 33 and 99. Last two digits 40=4×1040 = 4 \times 10: divisible by 44. Passes 22 and 33, so divisible by 66. All seven divide 2,3402{,}340.

Prime Factorization

Concept
Breaking a Number into Primes

The prime factorization of a number writes it as a product of prime numbers only. Every composite number has exactly one prime factorization (in some order). A factor tree is the easiest way to find it: split the number into any two factors, then keep splitting until every branch ends in a prime. Finally, write repeated primes using exponents.

A factor tree always ends in the same primes because of the Fundamental Theorem of Arithmetic: every whole number greater than 11 is a product of primes in exactly one way, apart from order. Splitting 180180 as 18×1018 \times 10 or 4×454 \times 45 just visits the primes in a different order. That uniqueness is what makes the factorization useful: it is a fingerprint of the number.

Example
Factor tree for 180
1801829331025
Every branch ends in a prime: 180=22×32×5180 = 2^2 \times 3^2 \times 5.

Split 180180 step by step:

180=18×10=(2×9)×(2×5)=2×3×3×2×5.180 = 18 \times 10 = (2 \times 9) \times (2 \times 5) = 2 \times 3 \times 3 \times 2 \times 5.

Group and order the primes, then use exponents:

180=22×32×5.180 = 2^2 \times 3^2 \times 5.

Check: 4×9×5=180.4 \times 9 \times 5 = 180. ✓

Side note
Where it goes wrong

Stopping at a composite leaf: "180=2×9×10180 = 2 \times 9 \times 10." Neither 99 nor 1010 is prime; keep splitting until every leaf is prime: 9=3×39 = 3 \times 3 and 10=2×510 = 2 \times 5.

Tip

No matter which factors you split off first, an honest factor tree always ends with the same set of primes. Starting 180=4×45180 = 4 \times 45 or 180=6×30180 = 6 \times 30 leads to the identical answer 22×32×52^2 \times 3^2 \times 5.

Try it
Try it: factor

Write the prime factorization of 252252 with exponents.

Show the answer

Split repeatedly: 252=4×63=(2×2)×(7×9)=2×2×7×3×3252 = 4 \times 63 = (2 \times 2) \times (7 \times 9) = 2 \times 2 \times 7 \times 3 \times 3. So 252=22×32×7252 = 2^2 \times 3^2 \times 7. Check: 4×9×7=2524 \times 9 \times 7 = 252.

Greatest Common Factor (GCF)

Concept
The Largest Shared Factor

The greatest common factor of two numbers is the biggest number that divides both. There are two reliable methods:

  • Listing: list the factors of each number, then pick the largest one they share.
  • Prime factorization: factor each number into primes, then multiply the primes they have in common (using the smaller exponent of each shared prime).

Any factor of 2424 is built from some of 2424's primes (2,2,2,32, 2, 2, 3), and any factor of 3636 from some of its primes (2,2,3,32, 2, 3, 3). A common factor can only use primes that both numbers have, so the largest one uses all of the shared primes, 2⋅2⋅3=122 \cdot 2 \cdot 3 = 12, and nothing else. That is the "smaller exponent of each shared prime" rule in pictures.

Example
GCF of 24 and 36, two ways
22 23324 = 2·2·2·336 = 2·2·3·3overlap: 2 · 2 · 3 = 12
Shared primes sit in the overlap; their product is the GCF.

Listing method:

24: 1,2,3,4,6,8,12,2436: 1,2,3,4,6,9,12,18,36\begin{aligned} 24:\ & 1, 2, 3, 4, 6, 8, 12, 24 \\ 36:\ & 1, 2, 3, 4, 6, 9, 12, 18, 36 \end{aligned}

Shared factors: 1,2,3,4,6,121, 2, 3, 4, 6, 12. The greatest is 12\boxed{12}.

Prime-factorization method:

24=23×3,36=22×32.24 = 2^3 \times 3, \qquad 36 = 2^2 \times 3^2.

Take the shared primes at their smaller powers: 22×3=12.2^2 \times 3 = 12.

Side note
Where it goes wrong

Taking the larger power of shared primes: 23×32=722^3 \times 3^2 = 72 for the GCF of 2424 and 3636. That is bigger than both numbers, so it cannot divide either one. The GCF uses the smaller powers: 22×3=122^2 \times 3 = 12.

Tip

The prime-factorization method wins for big numbers where listing every factor is slow. For the GCF, keep only primes that appear in both numbers, each raised to the lower power.

Try it
Try it: GCF

Find the GCF of 8484 and 126126 using prime factorizations.

Show the answer

84=22×3×784 = 2^2 \times 3 \times 7 and 126=2×32×7126 = 2 \times 3^2 \times 7. Shared primes at their smaller powers: 21×31×71=422^1 \times 3^1 \times 7^1 = 42. Check: 84=42×284 = 42 \times 2 and 126=42×3126 = 42 \times 3.

Least Common Multiple (LCM)

Concept
The Smallest Shared Multiple

The least common multiple of two numbers is the smallest number that both divide into — the first multiple they share. Again, two methods:

  • Listing: list the multiples of each number until one appears in both lists.
  • Prime factorization: factor each number, then multiply every prime that appears in either number, using the larger exponent of each.

A common multiple of 8=238 = 2^3 and 12=22×312 = 2^2 \times 3 must contain all of 88's primes and all of 1212's, so it needs at least three 22s and one 33. The smallest number with exactly that much is 23×3=242^3 \times 3 = 24. After 2424, the lists keep matching every 2424 (48,72,…48, 72, \ldots), because every common multiple is a multiple of the LCM.

Example
LCM of 8 and 12, two ways
8:08162432404812:012243648first match: 24
The two lists of multiples first meet at 2424.

Listing method:

8: 8,16,24,32,…12: 12,24,36,…\begin{aligned} 8:\ & 8, 16, 24, 32, \dots \\ 12:\ & 12, 24, 36, \dots \end{aligned}

The first shared multiple is 24\boxed{24}.

Prime-factorization method:

8=23,12=22×3.8 = 2^3, \qquad 12 = 2^2 \times 3.

Take every prime at its larger power: 23×3=24.2^3 \times 3 = 24.

Side note
Where it goes wrong

Multiplying the numbers and calling it the LCM: 8×12=968 \times 12 = 96. That is a common multiple, but not the least; it double-counts the shared factor 44. The product equals the LCM only when the GCF is 11.

Tip

For the LCM, use the higher power of every prime; for the GCF, use the lower power of only the shared primes. A handy check: GCF×LCM=\text{GCF} \times \text{LCM} = the product of the two numbers. Here 4×24=96=8×12.4 \times 24 = 96 = 8 \times 12. ✓

Try it
Try it: LCM

Find the LCM of 1818 and 3030, and check that GCF ×\times LCM equals 18×3018 \times 30.

Show the answer

18=2×3218 = 2 \times 3^2 and 30=2×3×530 = 2 \times 3 \times 5. Every prime at its larger power: 2×32×5=902 \times 3^2 \times 5 = 90. The GCF uses the smaller powers of shared primes: 2×3=62 \times 3 = 6. Check: 6×90=540=18×306 \times 90 = 540 = 18 \times 30.

Word Problems: GCF or LCM?

Concept
Deciding Which to Use

The hardest part of a word problem is choosing the right tool. Watch for these signals:

  • Use the GCF when you split or group things into the largest equal piles, rows, or teams — the numbers get smaller. Clues: “greatest,” “largest,” “equal groups,” “split evenly.”
  • Use the LCM when events repeat and you want the next time they line up, or the smallest amount that works for both — the numbers get bigger. Clues: “at the same time,” “again,” “least,” “smallest,” “every.”

The GCF answers "into how many identical groups can both amounts be split?", because the number of groups must divide both amounts, and the largest such number is the greatest common factor. The LCM answers "when do two repeating cycles line up again?", because the meeting time must be a multiple of both cycle lengths. So the test is simple: dividing things up points to factors, repeating in time points to multiples.

Example
A GCF problem

A teacher has 2424 pencils and 3636 erasers. She wants to make identical goody bags using all the supplies, with the greatest number of bags. How many bags, and what is in each?

We are splitting into the largest equal groups ⇒\Rightarrow GCF.

GCF(24,36)=12.\text{GCF}(24, 36) = 12.

So 1212 bags, each with 24÷12=224 \div 12 = 2 pencils and 36÷12=336 \div 12 = 3 erasers.

Side note
Where it goes wrong

Using the LCM for the goody bags: LCM(24,36)=72\text{LCM}(24, 36) = 72 bags. There are only 2424 pencils, so 7272 bags cannot each get one. Splitting into equal groups calls for a common factor, and the greatest is 1212.

Example
An LCM problem

One lighthouse flashes every 88 seconds and another every 1212 seconds. They just flashed together. When will they next flash at the same time?

Repeating events lining up again ⇒\Rightarrow LCM.

LCM(8,12)=24.\text{LCM}(8, 12) = 24.

They flash together again after 2424 seconds.

Tip

Quick gut-check: if the answer should be smaller than the numbers in the problem (fewer, larger groups), it is a GCF. If the answer should be larger (a later time, a bigger shared total), it is an LCM.

Try it
Try it: which one?

(a) Buses leave every 1515 minutes and trains every 2020 minutes; both left at 8:008{:}00. When do they next leave together? (b) A 4848 cm by 6060 cm board is cut into equal squares with no waste. What is the largest square possible?

Show the answer

(a) Repeating cycles lining up: LCM. 15=3×515 = 3 \times 5 and 20=22×520 = 2^2 \times 5, so LCM=22×3×5=60\text{LCM} = 2^2 \times 3 \times 5 = 60 minutes: at 9:009{:}00. (b) The square's side must divide both 4848 and 6060: GCF. 48=24×348 = 2^4 \times 3 and 60=22×3×560 = 2^2 \times 3 \times 5, so GCF=22×3=12\text{GCF} = 2^2 \times 3 = 12 cm squares (4×5=204 \times 5 = 20 of them).

Going Deeper: Advanced Number Theory

Concept
Counting Divisors from the Prime Factorization

Once a number is written in prime-power form, you can count how many factors it has without listing a single one. If

n=pa×qb×rc×⋯ ,n = p^{a} \times q^{b} \times r^{c} \times \cdots,

then the number of divisors is

d(n)=(a+1)(b+1)(c+1)⋯d(n) = (a+1)(b+1)(c+1)\cdots

The idea: when building a divisor, the prime pp can appear 0,1,2,…,0, 1, 2, \dots, or aa times — that is a+1a+1 independent choices. Multiplying the choices for each prime counts every divisor exactly once.

Example
How many divisors does 360 have?

First factor into primes:

360=8×45=23×32×5.360 = 8 \times 45 = 2^3 \times 3^2 \times 5.

The exponents are 3,2,3, 2, and 11. Add one to each and multiply:

d(360)=(3+1)(2+1)(1+1)=4×3×2=24.d(360) = (3+1)(2+1)(1+1) = 4 \times 3 \times 2 = 24.

So 360360 has exactly 24\boxed{24} divisors — far quicker than hunting for all 2424 factor pairs by hand.

Tip

A number is a perfect square exactly when every exponent in its prime factorization is even — which forces every (a+1)(a+1) to be odd, so d(n)d(n) is odd. In fact, perfect squares are the only numbers with an odd number of divisors, because the “middle” pair n×n\sqrt{n} \times \sqrt{n} is counted just once.

Concept
The Sum of All Divisors

You can also add up all the divisors straight from the prime factorization. For each prime power, sum the powers from p0p^0 up to pap^a, then multiply those sums together:

σ(n)=(1+p+⋯+pa)(1+q+⋯+qb)⋯\sigma(n) = \bigl(1 + p + \cdots + p^{a}\bigr)\bigl(1 + q + \cdots + q^{b}\bigr)\cdots

Multiplying the parentheses out reproduces every divisor of nn exactly once, so their grand total falls out automatically.

Example
Adding up the divisors of 360

Using 360=23×32×5360 = 2^3 \times 3^2 \times 5, sum each prime's powers separately:

20+21+22+23=1+2+4+8=15,30+31+32=1+3+9=13,50+51=1+5=6.\begin{aligned} 2^0 + 2^1 + 2^2 + 2^3 &= 1 + 2 + 4 + 8 = 15, \\ 3^0 + 3^1 + 3^2 &= 1 + 3 + 9 = 13, \\ 5^0 + 5^1 &= 1 + 5 = 6. \end{aligned}

Multiply the three totals:

σ(360)=15×13×6=1170.\sigma(360) = 15 \times 13 \times 6 = 1170.

All 2424 divisors of 360360 add up to 1170\boxed{1170}.

Concept
The GCF–LCM Product Theorem

For any two whole numbers aa and bb, the greatest common factor and least common multiple are locked together by

GCF(a,b)×LCM(a,b)=a×b.\text{GCF}(a,b) \times \text{LCM}(a,b) = a \times b.

This makes finding the LCM effortless once you know the GCF (or vice versa): just rearrange.

LCM(a,b)=a×bGCF(a,b).\text{LCM}(a,b) = \frac{a \times b}{\text{GCF}(a,b)}.

Why it works: for each prime, the GCF takes the smaller exponent and the LCM takes the larger one — together they use up exactly the two exponents that appear in aa and bb.

Example
Using the theorem on 18 and 24

We have GCF(18,24)=6\text{GCF}(18, 24) = 6. Instead of listing multiples, use the theorem:

LCM(18,24)=18×246=4326=72.\text{LCM}(18, 24) = \frac{18 \times 24}{6} = \frac{432}{6} = 72.

Check: 18=2×3218 = 2 \times 3^2 and 24=23×324 = 2^3 \times 3, so the LCM is 23×32=72.2^3 \times 3^2 = 72. ✓

Concept
The Euclidean Algorithm

For large numbers, factoring can be slow. The Euclidean algorithm finds the GCF using only repeated division with remainder. Repeatedly replace the larger number by the remainder when it is divided by the smaller:

GCF(a,b)=GCF(b, a mod b),\text{GCF}(a, b) = \text{GCF}(b, \ a \bmod b),

and stop when the remainder hits 00. The last nonzero remainder is the GCF. It works because any common factor of aa and bb is also a factor of their remainder.

Example
Euclidean algorithm for GCF(252, 105)
1051054242the smallest squares, 21 × 21, tile everything
Cut off the largest squares, then repeat on the leftover strip.

Divide, keep the remainder, repeat:

252=2×105+42,105=2×42+21,42=2×21+0.\begin{aligned} 252 &= 2 \times 105 + 42, \\ 105 &= 2 \times 42 + 21, \\ 42 &= 2 \times 21 + 0. \end{aligned}

The last nonzero remainder is 21\boxed{21}, so GCF(252,105)=21\text{GCF}(252, 105) = 21. Check: 252=21×12252 = 21 \times 12 and 105=21×5105 = 21 \times 5. ✓

Concept
Perfect, Abundant, and Deficient Numbers

Compare a number to the sum of its proper divisors (all its divisors except itself):

  • Perfect: the proper divisors add up to the number exactly. The smallest is 6=1+2+36 = 1 + 2 + 3; the next is 28=1+2+4+7+1428 = 1 + 2 + 4 + 7 + 14.
  • Abundant: the proper divisors sum to more than the number. Example: 1212, since 1+2+3+4+6=16>121 + 2 + 3 + 4 + 6 = 16 > 12.
  • Deficient: the proper divisors sum to less than the number. Example: 1010, since 1+2+5=8<101 + 2 + 5 = 8 < 10. Every prime is deficient (its only proper divisor is 11).
Tip

Two numbers are relatively prime (or coprime) when their only common factor is 11, i.e. GCF=1\text{GCF} = 1. They need not be prime themselves: 88 and 99 are relatively prime because 8=238 = 2^3 and 9=329 = 3^2 share no primes. When two numbers are coprime, their LCM is simply their product, since GCF×LCM=a×b\text{GCF} \times \text{LCM} = a \times b becomes 1×LCM=a×b1 \times \text{LCM} = a \times b.

Why It Works

For 12=22⋅312 = 2^2\cdot3 and 18=2⋅3218 = 2\cdot3^2: a common factor can use at most 212^1 and 313^1 (the smaller exponents), so gcd⁡=6\gcd = 6; a common multiple must contain 222^2 and 323^2 (the larger exponents), so lcm=36\text{lcm} = 36. Notice 6×36=216=12×186\times36 = 216 = 12\times18 — in general gcd⁡(a,b)⋅lcm(a,b)=ab\gcd(a,b)\cdot\text{lcm}(a,b) = ab, because each prime's exponent is min + max == the sum of the two exponents.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
GCF and LCM are linked
gcd⁡(a,b)⋅lcm⁡(a,b)=a b\gcd(a,b)\cdot \operatorname{lcm}(a,b) = a\,b

What it means. Once you know the greatest common factor, the least common multiple is free.

Example. gcd⁡(12,18)=6\gcd(12,18)=6, so lcm⁡(12,18)=12⋅186=36\operatorname{lcm}(12,18)=\dfrac{12\cdot 18}{6}=36.

Why it works. Write each number as a product of primes. For each prime, gcd⁡\gcd takes the smaller exponent and lcm⁡\operatorname{lcm} takes the larger. Smaller ++ larger == the two exponents added, which is exactly the exponent in abab.

Tip. Find the (easier) gcd⁡\gcd first, then use lcm⁡(a,b)=abgcd⁡(a,b)\operatorname{lcm}(a,b)=\tfrac{ab}{\gcd(a,b)}.