Possible remainder when multiple of a number is divided by multiple of the same dividor

Question

I have a couple of questions regarding which I am confused.

$1)$ What is the Greatest, Positive Integer $n$ such that $2^n$ is a factor of $12^{10}$

$(3\cdot 2^2)^{10}$ So, my guess is $n = 12$? Please correct me if this is wrong.

Now to the important question

$2)$ When the positive integer $m$ is divided by $4$, the remainder is $3$, What is the remainder when $22m$ is divided by $8$.

I have tried all possible logic that comes in my mind but i can't figure this problem out.

How can i solve this...

Answer 1

• $12^{10}=(2^2\times3)^{10} =2^{20}\times 3^{10}$

• $m=4k+3$ for some integer $k$ so $22m=88k+66 = 8(11k+8)+2$

Answer 2

$1)$

$n=20$.

Note that $4$ divides $12$. i.e. $2^2$ divides $12$. So $2^{2n}$ divides $12^n$.

So $n=20$ is an answer. (why not anything more than $20$?, since $2$ is the greatest $k$ such that $2^k$ divides $12$)

Edit: Both questions have now been answered by Henry. :)

Answer 3

1) I think n=20. 12^10=4^10 . 3^10 = 2^20 . 3^10

as 3^10 only has factors of 3 but not 2, so 2^20 is the greatest factor. so n=20.

2) when m is divided by 4, remainder is 3. when 22m is divided by 4, remainder=66 mod 4 =2 As all multiples of 8 are divisible by 4, then if 22m is divided by 8, remainder=2. therefore, remainder =2