# NUMBER SENSE: Factors of Whole Numbers

** Least Common Multiple
**One way to find LCM(30, 96) is to write lists of

multiples of 30 and of 96.

The least common multiple is the **smallest** number

common to both lists of multiples.

Multiples of 30:

30, 60, 90, 120, 150,…450, 480,

510,…,2880, 2910,…

Multiples of 96:

96, 192, 288, 384, 480, 776,…,2880, 2976,…

From these lists we see that LCM(30, 96) = 480.

However, LCM(30, 96) can also be found

using prime factorizations.

30 = 2^{2} • 3^{1} • 5^{1}

96 = 2^{5} • 3^{1} • 5^{0}

Notice that 5^{0} = 1, so the prime factorization of

96 is really 2^{5} • 3.

To find LCM’s, make sure that each prime

occurs in both lists.

To find the LCM, choose each prime raised to

the highest power from either list.

LCM(30, 96)

= 2^{5} • 3^{1} • 5^{1}

= 480

**Least Common Multiple
Worksheet**

Use prime factorization to find

LCM (24, 64)

LCM(32, 48)

The characterization of GCF and LCM in

terms of prime factorization leads to:

GCF(a, b) • LCM(a, b) = a • b .

Example: GCF(30, 96) • LCM(30, 96) = 30 • 96

GCF(30, 96) = 6

LCM(30, 96) = 480

6 • 480 = 30 • 96

2880 = 2880

Using GCF(30, 96) = 6, we could have found

LCM (30, 96) as

Why does

GCF(a,b ) • LCM(a,b ) = a • b ?

Consider an example:

a = 4500 = 2^{2} • 3^{2} • 5^{3}

b = 4050 = 2^{1} • 3^{4} • 5^{2}

**GCF(a, b ) =** 2^{1} • 3^{2} • 5^{2}

(Choose the smallest powers)

**LCM(a, b ) =** 2^{2} • 3^{4} • 5^{3}

(Choose the largest powers)

But a • b = 2^{2} • 3^{2} • 5^{3} • 2^{1} • 3^{4} • 5^{2},

and this is the same as

**GCF(a , b ) • LCM (a , b )
**Both are 18,225,000

**Least Common Multiples
Worksheet**

Find LCM(24, 64) by applying

GCF(a, b) • LCM(a, b) = a • b.

Find LCM(32, 48) by applying

GCF(a, b) • LCM(a, b) = a • b

**LEAST COMMON MULTIPLE AND
ADDITION OF FRACTIONS **

Least common multiples help with the

addition of fractions.

For example

We rewrite this problem in terms of

equivalent fractions with a common

denominator.

One possible choice for a common

denominator is 8 • 12,

Or,

by using LCM(12, 8) = 24

**Addition of Fractions
Worksheet**

Find the sums for the following fractions

by using the least common denominator.

**Sample Word Problems**

Try these problems:

1. Two joggers are running around a track.

One jogger runs a lap in 6 minutes and

the other takes 10 minutes. If they start at

the same time and place, how long will it

take for them to meet at the starting place

at the same time.

2. A 48 member band follows a 54 member

band in a parade. If the same number of

players must be in each row of both

bands, what is the largest number of

players that can be placed in each row?

Assume each row has the same number of

players.

Answers:

1. LCM(6, 10) = 30

2. GCF(48, 54) = 6

**Enrichment
Determining Whether a Number is Prime
**For a number that is not a perfect square,

factors occur in pairs. To determine whether a

number is prime, it is not necessary to check

all smaller numbers as possible factors.

Is 137 prime?

For any pair of factors whose product is 137,

one of the factors would be less than
.

It is sufficient to confirm that the numbers

2, 3, 4, 5, 6, 7, 8, 9, 10, and 11

are not factors of 137

to show that 137 is prime.

In fact, it is not necessary to check all whole

numbers between 1 and 11.

E.g., if 6 were a factor of 137,

its prime factors (2 and 3)

would also be factors of 137.

We only need to check that the **primes <**

2, 3, 5, 7, and 11

are not factors of 137.

To establish that a is prime,

it is sufficient to confirm that

all primes less than or equal to

fail to be factors of a.

**Checking Whether a Number is Prime
Worksheet**

What is the smallest list of numbers that must be

checked as possible factors of 223?

Is 223 prime?

What is the smallest list of numbers that must be

checked as possible factors of 179?

Is 179 prime?

** Divisibility Rules
**Basic ingredients:

“A|B” means “A divides B or in other words,

there is an integer k such that

B = A • k.

Example: 2|10 because 10 = 2 • 5

Theorem: If A|B and A|C , then A|(B+C).

Example: If 2|10 and 2|8, then 2|18

Proof: If A|B , there is an integer k_{1}, such that

B = A • k_{1}. If A|C, there is an integer

k_{2} such that C = A • k_{2}.

Therefore:

B + C

= A • k_{1} + A • k_{2}

= A (k_{1} + k_{2})

So, A|(B + C)

Here is another result

Theorem 2: If A|(B + C) and A|C, then A|B

Example: If 2|18 and 2|10, then 2|(18-10) or

2|8.

Proof: If A|(B+C), B + C = A • k_{1} , for some

integer k_{1} . If A|C, C = A • k_{2} for some integer

k_{2}. Then

B = (B + C) – C

= A • k_{1} – A • k_{2}

= A (k_{1} – k_{2})

Therefore A|B.

**Divisibility by 2
**2|(a •10

^{2}+ b • 10 + c ) if and only if 2|c.

Proof: Let B = a• 10^{2} + b • 10. Then 2|B

because B = 2 • (A • 50 + 6 • 5). If 2|c , then

2|(B + c) by Theorem 1. If 2|(B + c ), then 2|c

by Theorem 2.

The only single digit numbers divisible by 2

are 0, 2, 4, 6, and 8.

Therefore, 2|(a • 10^{2} + b • 10 + c) if and only

if c = 0, 2, 4, 6, or 8.

**Divisibility by 3
**3|(a • 10

^{2}+ b • 10 + c ) if and only if

3|(a + b + c )

Proof:

a • 10^{2} +b • 10 + c

= a • (99 + 1) + b • (9 + 1) + c

= (a • 99 + b • 9) + (a + b + c)

Let B = a • 99 + b • 9. Then 3|B

Because B = 3 • (a • 33 + b • 3).

Let C = a + b + c . If 3|C , then 3|(B + C) by

Theorem 1. If 3|(B +C ) then 3|C by Theorem

2.

Example: 3|129 because 3|(1+2+9)

Prev | Next |