Modular arithmetic query

[TerryM]

In some recent studying I have been doing I noticed that the following seems to hold true, but can find no mention of it anywhere as a rule. Would someone be able to point me in the right direction?

if x = m mod n

[edit]

Sorrry, stated that incorrectly...

if x = m mod n
then bx = bm mod bn

I knew that bx = bm mod n but the bn doesn't seem so intuitive

Many thanks

 

[Deleted member 4993]

In some recent studying I have been doing I noticed that the following seems to hold true, but can find no mention of it anywhere as a rule. Would someone be able to point me in the right direction?

if x = m mod n
then bx = b mod bn

Many thanks

Hint: x = m mod n → x/n = K + m/n

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[TerryM]

Hint: x = m mod n → x/n = K + m/n

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

Hi Subhotosh;

I edited the OP as you were replying as I had stated my problem incorrectly.
This isn't school work (I'm 51).

I had been working on a problem involving numbers where

x = 1 mod 24

when I noticed that not only did

3x = 3 mod 24

but

3x = 3 mod 72

It's just that I found this last part strange

Thanks

 

[Deleted member 4993]

In some recent studying I have been doing I noticed that the following seems to hold true, but can find no mention of it anywhere as a rule. Would someone be able to point me in the right direction?

if x = m mod n

[edit]

Sorrry, stated that incorrectly...

if x = m mod n → x/n = K + m/n → (bx)/(bn) = K + (bm)/(bn) → bx = (bm) mod (bn)
then bx = bm mod bn

I knew that bx = bm mod n but the bn doesn't seem so intuitive

Many thanks

.

 

[TerryM]

x/n = K + m/n → (bx)/(bn) = K + (bm)/(bn) → bx = (bm) mod (bn).

Thanks for the reply but I'm not sure I understand it.

Is K the quotient and m/n the remainder?

If so, when I put some figures to what you have written I get this

17/3 = 5 + 2/1 -> (3*17)/(3*3) = 5 + (3*2)/(3*1) -> 51/9 = 5 + 6/3

which is wrong since the remainder is 6, not 2

 

[Deleted member 4993]

Thanks for the reply but I'm not sure I understand it.

Is K the quotient and m/n the remainder?

If so, when I put some figures to what you have written I get this

17/3 = 5 + 2/1 -> (3*17)/(3*3) = 5 + (3*2)/(3*1) -> 51/9 = 5 + 6/3 .... No

17/3 = 5 + 2/3 -> (3*17)/(3*3) = 5 + (3*2)/(3*3) -> 51/9 = 5 + 6/9

Remainder does not remain same

x = m mod n → when you divide 'x' by 'n' we have remainder m (quotient some integer 'K')

b*x = (b*m) mod (b*n) → remainder (b*m) not m (original remainder)

so your example could be written as:

17 = 2 (mod 3) → (17*3) = (2*3) [mod (3*3)] → 51 = 6 (mod 9) → everything agrees

which is wrong since the remainder is 6, not 2

.

 

[Ishuda]

In some recent studying I have been doing I noticed that the following seems to hold true, but can find no mention of it anywhere as a rule. Would someone be able to point me in the right direction?

if x = m mod n

[edit]

Sorrry, stated that incorrectly...

if x = m mod n
then bx = bm mod bn

I knew that bx = bm mod n but the bn doesn't seem so intuitive

Many thanks

To put it a slightly different way than Subhotosh Khan did, what does
x = m mod n
mean? It means there is an integer k so that
x = k n + m.
So what happens if you multiply through by b?

 

[TerryM]

Thanks Subhotosh and Ishuda. I think I have it now.