mod. arithmetic

[demoskopico]

http://www.manhattanreview.com/forum/showthread.php?t=141

the link above is referred to a remainder problem solved with modular arithmetic.

look at the following easy problem:

if x+2 is a multiple of 10, what is the remainder when x is divided by 5?

it is easy to say at a glance that the result is 3. anyway, is there any scientific method which can be used for more difficult problems? For example, can we use mod arith. and how?

[kevin]

http://www.manhattanreview.com/forum/showthread.php?t=141

the link above is referred to a remainder problem solved with modular arithmetic.

look at the following easy problem:

if x+2 is a multiple of 10, what is the remainder when x is divided by 5?

it is easy to say at a glance that the result is 3. anyway, is there any scientific method which can be used for more difficult problems? For example, can we use mod arith. and how?

x+ 2 = 10k

x = 10k - 2 = 2(5k) - 2

x + 5 = 5(2k) + 3

Remember that x + 5m (m is an integer) and x have the same remainder when divided by 5. Thus x+5 (and thus x) is 3 greater than a multiple of 5

[demoskopico]

x+ 2 = 10k

x = 10k - 2 = 2(5k) - 2

x + 5 = 5(2k) + 3

Remember that x + 5m (m is an integer) and x have the same remainder when divided by 5. Thus x+5 (and thus x) is 3 greater than a multiple of 5

interesting, I have never followed this method. can you provide me with a couple of specific exercises, also made by yourself, do that I can practice with this way? thanks.

[kevin]

interesting, I have never followed this method. can you provide me with a couple of specific exercises, also made by yourself, do that I can practice with this way? thanks.

If k is an integer so that 2k is 6 more than a multiple of 20, what is the remainder when k - 12 is divided by 5?

(A) 0
(B) 1
(C) 2
(D) 3
(E) cannot be determined with the information provided

[demoskopico]

If k is an integer so that 2k is 6 more than a multiple of 20, what is the remainder when k - 12 is divided by 5?

(A) 0
(B) 1
(C) 2
(D) 3
(E) cannot be determined with the information provided


is it 3?

2k-6=20q
k-10q=3
we know that k-5q=12

let's have a system

-5q=-8
q=8/5 remainder is 3

I am not sure about this approach and, anyway, I see it is different from yours. Where am I wrong?


maybe this one is the correct:

2k=20m+6 --> k=10m+3 --> (k-12)=10m-9 --> (k-12)=5*[2*(m-1)]+1 --> (k-12)=5*n+1