What’s the common factor of 36 and 90

Mathematics

2 answers:

zepelin [54]3 years ago

6 0

Answer:

1, 2, 3, 6, 9, and 18.

Step-by-step explanation:

gtnhenbr [62]3 years ago

6 0

Answer:

1, 2, 3, 6, 9, 18.

Step-by-step explanation:

factors of 36:

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

factors of 90:

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

common factors of 36 and 90 means factors that belongs to both 36 and 90

common factors:

1, 2, 3, 6, 9, 18.

You might be interested in

An=an-1+6an-2 n>2 a0=3 a1=6

Svetach [21]

$\begin{cases}a_0=3\\a_1=6\\a_n=a_{n-1}+6a_{n-2}&\text{for }n>2\end{cases}$

Let the generating function for $a_n$ be

$A(x)=\displaystyle\sum_{n\ge0}a_nx^n$

Multiplying both sides by $x^{n-2}$

$a_nx^{n-2}=a_{n-1}x^{n-2}+6a_{n-2}x^{n-2}$

Summing both sides over the non-negative integers greater than or equal to 2 gives

$\displaystyle\sum_{n\ge2}a_nx^{n-2}=\sum_{n\ge2}a_{n-1}x^{n-2}+6\sum_{n\ge2}a_{n-2}x^{n-2}$
$\displaystyle\frac1{x^2}\sum_{n\ge2}a_nx^n=\frac1x\sum_{n\ge1}a_nx^n+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(\sum_{n\ge0}a_nx^n-a_0-a_1x\right)=\frac1x\left(\sum_{n\ge0}a_nx^n-a_0\right)+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(A(x)-3-6x\right)=\frac1x\left(A(x)-3\right)+6A(x)$
$A(x)=\dfrac{3x+3}{1-x-6x^2}$
$A(x)=\dfrac3{5(1+2x)}+\dfrac{12}{5(1-3x)}$

For $|x|$ , the two series converge to

$\displaystyle A(x)=\frac35\sum_{n\ge0}(-2x)^n+\frac{12}5\sum_{n\ge0}(3x)^n$
$\implies A(x)=\displaystyle\sum_{n\ge0}\dfrac{3(-2)^n+12(3)^n}5x^n$
$a_n=\dfrac{3(-2)^n+12(3)^n}5=\dfrac{3(-2)^n+4(3)^{n+1}}5$