1
The divisibility tests worth knowing.
These let you factor without trial and error.
| Divisible by | Test | Example |
|---|
| 2 | last digit is even | 134 ✓ |
|---|
| 3 | digits add to a multiple of 3 | 231 → 6 ✓ |
|---|
| 4 | last two digits divide by 4 | 1,316 → 16 ✓ |
|---|
| 5 | ends in 0 or 5 | 745 ✓ |
|---|
| 6 | passes both 2 and 3 | 132 ✓ |
|---|
| 9 | digits add to a multiple of 9 | 459 → 18 ✓ |
|---|
| 10 | ends in 0 | 890 ✓ |
|---|
| 11 | alternating digit sum divides by 11 | 913 → 9−1+3 = 11 ✓ |
|---|
The 3 and 9 tests are the ones that earn their keep. Adding digits is far faster
than dividing, and it works on any size of number.
2
Break both numbers into primes.
Keep dividing by the smallest prime that fits: 24 → 12 → 6 → 3 → 1.
Once both numbers are written as prime powers, the GCF and LCM are just two
different ways of reading the same two lines.
3
GCF: the lowest power of each shared prime.
2: lower of 2³ and 2² → 2²
take the smaller
3: lower of 3¹ and 3² → 3¹
same rule
GCF = 4 × 3 = 12
multiply them
The GCF must divide both numbers, so it can never exceed the smaller one.
4
LCM: the highest power of every prime that appears.
2: higher of 2³ and 2² → 2³
take the larger
3: higher of 3¹ and 3² → 3²
same rule
LCM = 8 × 9 = 72
multiply them
GCF = 12 · LCM = 72
A free check
GCF × LCM = the two numbers multiplied
12 × 72 = 864, and 24 × 36 = 864 ✓
Rebuilding the numbers from GCF and LCM: if the GCF is 5 and the LCM is 60, write the
numbers as 5a and 5b, where a and b share no factor. Then 5ab = 60, so ab = 12, and the
pairs a, b are 1 and 12 or 3 and 4 (2 and 6 share a 2). The numbers are 5 and 60, or
15 and 20.
5
Which one does the word problem want?
This is what the exam actually tests.
GCF — splitting up
cutting into equal groups
the answer is smaller than the numbers
LCM — coming round again
events repeating together
the answer is bigger than the numbers
| Question | Use |
|---|
| largest equal groups from 24 and 36 items | GCF = 12 |
|---|
| two buses leave every 24 and 36 min — when together again? | LCM = 72 min |
|---|
| biggest square tile for a 24 by 36 floor | GCF = 12 |
|---|
| how many of those tiles? | (24 ÷ 12)(36 ÷ 12) = 2 × 3 = 6 tiles |
|---|
| smallest number divisible by both | LCM = 72 |
|---|
Ask whether the answer should be big or small
Cutting things up gives a smaller number — that is the GCF. Waiting for things to
line up gives a bigger one — that is the LCM.
For a clock time, turn the LCM into hours and add it on: lights that flash every 15,
18 and 30 minutes flash together at 6:00 a.m.; LCM(15, 18, 30) = 90 minutes = 1 h 30 min,
so they next flash together at 7:30 a.m.
6
Variations: no shared prime, three numbers, and different remainders.
The two extremes: 8 and 15, then 18 and 54.
8 = 2³, 15 = 3 × 5
no prime in common
GCF = 1, LCM = 8 × 15 = 120
nothing shared, nothing to remove
54 = 3 × 18
one divides the other
GCF = 18, LCM = 54
the smaller and the larger number themselves
What is different: no factoring needed once we spot the pattern. GCF × LCM = product still holds: 1 × 120 = 120 and 18 × 54 = 972.
Three numbers: the GCF and LCM of 30, 45 and 75.
30 = 2 · 3 · 5, 45 = 3² · 5, 75 = 3 · 5²
all three in primes
GCF = 3 · 5 = 15
only primes in all three; 2 is missing from 45
LCM = 2 · 3² · 5² = 450
every prime, highest power
What is different: GCF × LCM = product only works for two numbers. Here 15 × 450 = 6,750, but 30 × 45 × 75 = 101,250.
Different remainders: the smallest number that leaves remainder 4 when divided by 5, remainder 5 when divided by 6, and remainder 7 when divided by 8.
each remainder is 1 short of its divisor
5 − 4 = 6 − 5 = 8 − 7 = 1
n + 1 is a multiple of 5, 6 and 8
add the shortfall instead
LCM(5, 6, 8) = 120, so n = 119
check: 5·23 + 4, 6·19 + 5, 8·14 + 7 ✓
What is different: the remainders are not equal, so there is nothing to subtract. What is equal is the gap to each divisor, and that gets added.
7
When there is a remainder, shift by it first.
A remainder of r means the number is r more than an exact multiple. Move the r out of the way, use the LCM or GCF, then deal with r.
The smallest number that leaves remainder 4 when divided by 6, by 8 and by 9?
n − 4 is a multiple of 6, 8 and 9
take the remainder away
LCM(6, 8, 9) = 2³ × 3² = 72
highest powers
n = 72 + 4 = 76
check: 76 = 6·12 + 4 = 8·9 + 4 = 9·8 + 4 ✓
The largest number that divides 52 and 76 leaving remainder 4 in each case?
it divides 52 − 4 = 48 and 76 − 4 = 72 exactly
take the remainder away
GCF(48, 72) = 24
check: 52 = 2·24 + 4, 76 = 3·24 + 4 ✓
The planted choices
72 forgets to add the remainder back; GCF(52, 76) = 4 uses the numbers before the
remainder was removed. For “the smallest such number above 100”, keep adding the LCM:
76, 148, … The divisor must also be bigger than the remainder — 24 > 4 ✓.
On the exam
GCF and LCM items come as divisibility checks, “GCF × LCM = product”, bells and tiles,
and the remainder twist. Decide first whether the answer should be smaller (GCF) or
bigger (LCM) than the numbers, and subtract any remainder before you factor.
✓
Check: list a few multiples and factors.
factors of 24
1, 2, 3, 4, 6, 8, 12, 24
—
factors of 36
1, 2, 3, 4, 6, 9, 12, 18, 36
biggest shared is 12 ✓
multiples of 24
24, 48, 72, 96
—
multiples of 36
36, 72, 108
first shared is 72 ✓
Remember the flow
Primes · Lowest powers for GCF · Highest for LCM
GCF × LCM = the product of the two numbers.
