Finite Math Examples

$20$ , $0 - 160 x$

Step 1

Since $20, - 160 x$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $20, - 160$ then find LCM for the variable part $x^{1}$ .

Step 2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 3

Since the LCM is the smallest positive number, $LCM (20, - 160) = LCM (20, 160)$

Step 4

The prime factors for $20$ are $2 \cdot 2 \cdot 5$ .

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Step 4.1

$20$ has factors of $2$ and $10$ .

$2 \cdot 10$

Step 4.2

$10$ has factors of $2$ and $5$ .

$2 \cdot 2 \cdot 5$

Step 5

The prime factors for $160$ are $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5$ .

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Step 5.1

$160$ has factors of $2$ and $80$ .

$2 \cdot 80$

Step 5.2

$80$ has factors of $2$ and $40$ .

$2 \cdot 2 \cdot 40$

Step 5.3

$40$ has factors of $2$ and $20$ .

$2 \cdot 2 \cdot 2 \cdot 20$

Step 5.4

$20$ has factors of $2$ and $10$ .

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 10$

Step 5.5

$10$ has factors of $2$ and $5$ .

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5$

Step 6

Multiply $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5$ .

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Step 6.1

Multiply $2$ by $2$ .

$4 \cdot 2 \cdot 2 \cdot 2 \cdot 5$

Step 6.2

Multiply $4$ by $2$ .

$8 \cdot 2 \cdot 2 \cdot 5$

Step 6.3

Multiply $8$ by $2$ .

$16 \cdot 2 \cdot 5$

Step 6.4

Multiply $16$ by $2$ .

$32 \cdot 5$

Step 6.5

Multiply $32$ by $5$ .

$160$

Step 7

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 8

The LCM of $x^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x$

Step 9

The LCM for $20, - 160 x$ is the numeric part $160$ multiplied by the variable part.

$160 x$