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1. When divided by 7, remainder is 3 2. When divided by 4, remainder is 2
Applying Chinese modulus theorem, we get n= 28s+10 ( s is non-negative integer)
There're plenty of values of m ---> there're infinitively many values of n satisfying the two statements --> E it is!
How to apply Chinese modulus theorem:
from 1: n= 3( mod 7) (1)
from 2: n= 2 ( mod 4) (2)
from 2: n= 4v+2 , substitute this to (1), we get: 4v+2= 3(mod 7)
---> 4v= 1( mod 7) ---> v= 2 (mod 7) ( because 2*4= 8 which is divided by 7 has a remainder of 1) --> v= 7s +2 --> n= 4v+2= 4(7s+2)+2 --> n = 28s + 10
1. When divided by 7, remainder is 3 2. When divided by 4, remainder is 2
Applying Chinese modulus theorem, we get n= 28s+10 ( s is non-negative integer) There're plenty of values of m ---> there're infinitively many values of n satisfying the two statements --> E it is!
How to apply Chinese modulus theorem:
from 1: n= 3( mod 7) (1) from 2: n= 2 ( mod 4) (2)
from 2: n= 4v+2 , substitute this to (1), we get: 4v+2= 3(mod 7) ---> 4v= 1( mod 7) ---> v= 2 (mod 7) ( because 2*4= 8 which is divided by 7 has a remainder of 1) --> v= 7s +2 --> n= 4v+2= 4(7s+2)+2 --> n = 28s + 10
Yes it is E....
I find another set of values
when y =9 and x=5 in my solution but your looks more elegant, laxieqv
1. When divided by 7, remainder is 3 2. When divided by 4, remainder is 2
Applying Chinese modulus theorem, we get n= 28s+10 ( s is non-negative integer) There're plenty of values of m ---> there're infinitively many values of n satisfying the two statements --> E it is!
How to apply Chinese modulus theorem:
from 1: n= 3( mod 7) (1) from 2: n= 2 ( mod 4) (2)
from 2: n= 4v+2 , substitute this to (1), we get: 4v+2= 3(mod 7) ---> 4v= 1( mod 7) ---> v= 2 (mod 7) ( because 2*4= 8 which is divided by 7 has a remainder of 1) --> v= 7s +2 --> n= 4v+2= 4(7s+2)+2 --> n = 28s + 10
Yes it is E....
I find another set of values
when y =9 and x=5 in my solution but your looks more elegant, laxieqv
Thanks!
LAXI can u explain how you got this
---> 4v= 1( mod 7) ---> v= 2 (mod 7)
For people who are a little weary of the whole sequence of the Chinese modulus theorem, I hope this will help: You don't have to memorize it.
Basically you write the two conditions:
n=7a+3
n=4b+2
Therefore 4b=7a-1=8a-a-1
b=2a-(a-1)/4
We know that b is an integer so (a-1)/4 must be an integer.
We can write as a-1=4s, or a=4s+1.
Now we know
a=4s+1,
b=2(4s+1)-s=7s+2
You can solve for n here from either a or b, although you don't have to. As long as you know that when s takes different values, you'll have different values for a and b and thus different values of n. In other words you'll know that the solution is not unique and the answer would be E. _________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
Basically you write the two conditions: n=7a+3 n=4b+2
Therefore 4b=7a-1=8a-a-1 b=2a-(a-1)/4 We know that b is an integer so (a-1)/4 must be an integer. We can write as a-1=4s, or a=4s+1.
Now we know a=4s+1, b=2(4s+1)-s=7s+2
You can solve for n here from either a or b, although you don't have to. As long as you know that when s takes different values, you'll have different values for a and b and thus different values of n. In other words you'll know that the solution is not unique and the answer would be E.
The method is neat! And I must say it's the first time I've heard of it. But for people who are stuck, working out with sample numbers present a very quick approach as well.
1) n = 7q1 + 3
n could be 10
n could be 15
etc
Insufficient.
2) n = 4q2 + 1
n could be 5
n could be 9
etc
Insufficient.
Using 1) and 2)
n = 7q1 + 3
n = 4q2 + 1
7q1+3 = 4q2+1
2 = 4q2-7q1
Possible sets: (q1,q2) = (4,2) (11,6) --> means many possible values of n