This is the second post on remainders. Earlier one focused on standard important theorems that are used in remainder problems. Now we will be delving into details of some other important rules,generalizations,patterns,etc. I'll discuss these as pointers and provide examples along the way if it seems suitable. Let's start then :)
Point 1 : (a^n + b^n) mod (a + b) = 0 when n is odd
(a^n - b^n) mod (a + b) = 0 when n is even
Point 1 : (a^n + b^n) mod (a + b) = 0 when n is odd
(a^n - b^n) mod (a + b) = 0 when n is even
(a^n - b^n) mod (a - b) = 0 for all natural numbers
(a^n - b^n) mod (a^k - b^k) = 0 when k is a factor of n
example : 18^2000 + 12^2000 - 5^2000 - 1 is divisible by ?
323,221,299,237
323,221,299,237
(18^2000 - 5^2000) + (12^2000 - 1^2000) = B1 + B2
B1 is divisible by 13 by result 3 and B2 is divisible by 13 by result 2. Hence the expression is divisible by 13.
Also,writing it as :
(18^2000 - 1^2000) + (12^2000 - 5^2000) = B1 + B2
Also,writing it as :
(18^2000 - 1^2000) + (12^2000 - 5^2000) = B1 + B2
B1 is divisible by 17 by result 3 and B2 is divisible by 17 by result 2. Hence the expression is divisible by 17.
Hence the expression is divisible by 13*17 = 221
Point 2 :
Hence the expression is divisible by 13*17 = 221
Point 2 :
Any number of the abcabc form is divisible by 1001 and hence by extension by 7,11 and 13 as
1001 = 7*11*13
1001 = 7*11*13
Point 3 :
If 10^n mod x = 1, then to find remainder for any number upon division by x, you can take n digits at a time, find individual remainders and simply add them up.
example : Find the remainder when 123123123123123...upto 300 digits is divided by 27.
example : Find the remainder when 123123123123123...upto 300 digits is divided by 27.
Since 999 = 27*37. So, 1000 mod 27 = 1
Hence we can take three digits at a time and find individual remainders.
123 mod 27 = 15 and for 300 digits we'll have 300/3 = 100 such sets.
So, finally we have 15*100 mod 27 = 15 = Remainder.
Hence we can take three digits at a time and find individual remainders.
123 mod 27 = 15 and for 300 digits we'll have 300/3 = 100 such sets.
So, finally we have 15*100 mod 27 = 15 = Remainder.
Point 4 :
(a^n + b^n + c^n + ...) mod (a + b + c + ...) = 0 if and only if a,b,c,.. are in AP and n is ODD
example : 16^3 + 17^3 + 18^3 + 19^3 mod 70 = ?
Answer : Zero. As 16,17,18,19 are in AP and 3 is odd. Hence the expression is divisible by sum of the base terms : (16 + 17 + 18 + 19) = 70.
Point 5 :
Any digit or a set of digits repeated (p - 1) times is divisible by p.
example : 121212121212 mod 7 = ?
Answer : Zero. Since 12 12 12 12 12 12 is a repeated set of 12 coming 6(=7 - 1) times. Hence is divisible by 7.
NOTE : The topic of remainders becomes problematic as there is no end to the generalizations that one can make out of various set of questions. So you always end up finding a new concept or pattern or a generalization. Best way is to keep in mind these important and common ones and add up plus make note of what you keep finding by practicing more and more.
These cover majority of remainder questions. Other techniques are specific to question type at hand or involve different concepts like cyclicity,divisibility rules or base system for example. We will see them someday later when we come to those discussions individually.
Till then, hope you enjoyed learning from my blog and share it wherever you can. Rate, like and share the posts through the social network links at the end of each post. Leave comments for any doubts or personal reviews of how you felt about the article.
Cheers!
AS
No comments:
Post a Comment