a diff reminder qs
My Question
If N = 1 + 11 + 111 + 1111 + . . . . . . . 11 terms
Then the reminder of N when divided by 12 is
I ‘ll post the answer and diff ways of solvin it after 9pm
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I am not at ease with remainder problems but i
tried this in a very crude way.
The remainders of individual terms when divided by 12 are as follows
1 + 11 + 3 + 7 + 11 + 3 + 7 + 11 + 3 + 7 + 11 = 75
when 75 is divided by 12 the remainder is '3'
pls post any easy and refined way of finding the remainder
thank u
shahid
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Using Chinese Remainder Theorem
12 = 4*3 such that 4*r - 3*t =1, where r =1 and t =1
Let x be the remainder when N is divided by 3 and y be the remainder when N is divided by 4.
So, x = (1-1+0) + (1-1+0) + (1-1+0) + (1-1) = 0
and y = 1 - 1 - 1 - 1 - 1.. = -9 = 3.
S remainder of N when divided by 12 = 4rx - 3ty = 4*1*0 - 3*1*3 = -9 = 3mod12.
Kamal Lohia
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1mod 12=1
10 mod 12=(-2)
nd (10)^n mod 12=4 ,where n>1
there r 11 --1's
10--10's
n so on
going this way rem=3
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Thanks guys. . I can’t think of any other method. For those who still have not solved the question here is a stepwise explanation
Sum of First two terms is divisible by 12
Reminder of 3rd + 4th + 5th term = 6rd + 7th + 8th term = 9th + 10th + 11th term = 9 = (-3)
Reminder of 3 one’s = 3
Reminder of 4 one’s = 7 ( 3 + 4)
Reminder of 5 one’s = 11 (7 + 4)
Reminder of 6 one’s = 15 = 3 (11 + 4)
Reminder of 7 one’s = 7 (3 + 4)
Reminder of 3 one’s = 3
Reminder of 4 one’s = 7 ( 3 + 4)
Reminder of 5 one’s = 11 (7 + 4)
moral : most of the reminder qs can be solved in many diff ways . .
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I am getting the answer as 3
is it correct