Question #99831

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Answer 1 · 2017-12-19T02:26:06

The LCM is $80$ $80$

Factor each number:

$16 = 2 \cdot 2 \cdot 2 \cdot 2 = 2^{4}$ $16 = 2*2*2*2 = 2^4$

$20 = 2 \cdot 2 \cdot 5 = 2^{2} \cdot 5$ $20 = 2*2*5=2^2*5$

Every multiple of $16$ $16$ has factors that include $2^{4}$ $2^4$

Every multiple of $20$ $20$ has factors that include $2^{2} \cdot 5$ $2^2*5$ .

The Least Common Multiple must include factors of $2^{4}$ $2^4$ , $2^{2}$ $2^2$ and $5$ $5$ and no extra factors (otherwise it wouldn't be "least")

The least number that includes these is $2^{4} \cdot 5 = 80$ $2^4*5 = 80$

Answer 2 · 2017-12-19T08:28:57

$L C M = 2 \times 2 \times 2 \times 2 \times 5 = 80$ $LCM = 2xx2xx2xx2xx5 = 80$

The LCM must be divisible by $16 and 20$ $16 and 20$

Write each number as the product of its prime factors:

$16 = 2 \times 2 \times 2 \times 2$ $" "16 =2xx2xx2xx2$
$" "ul(20= 2xx2color(white)(xxxxxx)xx5)$
$LCM = 2xx2xx2xx2xx5 = 80$

$80$ now contains all the factors of $16 and 20$ but without any duplicates,