SOLUTION: Which number who divided by 7,9,11and remainder 1,2,3?

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Question 1052975: Which number who divided by 7,9,11and remainder 1,2,3?
Found 3 solutions by jim_thompson5910, ikleyn, advanced_Learner:
Answer by jim_thompson5910(35256) About Me  (Show Source):
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Let x be this unknown number that satisfies all three conditions given. For the sake of simplicity, we'll assume x is positive.

Use the Chinese Remainder Theorem

I will use the same basic steps and notation as that PDF is showing.

We have these 3 congruences given
x = 1 (mod 7)
x = 2 (mod 9)
x = 3 (mod 11)

The a%5Bi%5D values are the right hand side values (1, 2 and 3 respectively) where 'i' refers to the modulus value
a%5B7%5D+=+1
a%5B9%5D+=+2
a%5B11%5D+=+3


Multiply out the modulus values
N = 7*9*11 = 693

Calculate the M%5Bi%5D values
M%5B7%5D+=+N%2F7+=+693%2F7+=+99
M%5B9%5D+=+N%2F9+=+693%2F9+=+77
M%5B11%5D+=+N%2F11+=+693%2F11+=+63

Then finally we need to compute the multiplicative inverses y%5Bi%5D values of each M%5Bi%5D value

y%5B7%5D+=+%28M%5B7%5D%29%5E%28-1%29+=+%2899%29%5E%28-1%29+=+1 (mod 7)
Note: 99 = 1 (mod 7) so the solution to 1y = 1 (mod 7) is y = 1

y%5B9%5D+=+%28M%5B9%5D%29%5E%28-1%29+=+%2877%29%5E%28-1%29+=+2 (mod 9)
Note: 77 = 5 (mod 9). The solution to 5y = 1 (mod 9) is y = 2

y%5B11%5D+=+%28M%5B11%5D%29%5E%28-1%29+=+%2863%29%5E%28-1%29+=+7 (mod 11)
Note: 63 = 8 (mod 11) and the solution to 8y = 1 (mod 11) is y = 7


Now we use the information computed above to get the solution for x

(mod 693)

x+=+1%2A99%2A1+%2B+2%2A77%2A2+%2B+3%2A63%2A7 (mod 693)

x+=+1730 (mod 693)

x+=+344 (mod 693)

The smallest positive solution for x is 344

The set of all solutions for x is the set of x values such that x+=+693t%2B344 where t%3E=0 and t is an integer. In other words, x = 344 (mod 693)


Answer by ikleyn(52785) About Me  (Show Source):
You can put this solution on YOUR website!
.
Which number highlight%28cross%28who%29%29 when divided by 7,9,11 highlight%28cross%28and%29%29 gives the remainders 1,2,3 highlight%28respectively%29?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

The numbers that give the remainder 1 when divided by 7, form an arithmetic progression 

1, 8, 15, 22, 29, . . . (1 + 7(n-1)), . . . 

Of these numbers, the smallest number that gives the remainder 2 when divided by 9, is highlight%2829%29.


Next, the numbers that give the remainder 1 when divided by 7 and the remainder 2 when divided by 9, form an arithmetic progression 

29, 29+63 = 92, 29+2*63 = 155, 29+3*63 = 218,  . . . (29+(n-1)*63), . . . 
  (notice: 63 = 7*9)

Among them, the smallest number that gives the remainder 3 when divided by 11, is highlight%28344%29.

Answer.  Your number is 344.


Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
Which number who divided by 7,9,11and remainder 1,2,3?
............................................................................
let the number be x then
x%2F7 remainder is 1
x%2F9 remainder is 2
x%2F11 remainder is 3
2x%2F7 remainder is 2
2x%2F9 remainder is 4
2x%2F11 remainder is 6
%282x%2B5%29%2F7 remainder is 0
%282x%2B5%29%2F9 remainder is 0
%282x%2B5%29%2F11 remainder is 0
meaning 2x+5 is divisble by 7%2A9%2A11
%282x%2B5%29=%28693%29
x=344