Question 1089879
find the least positive integer that leaves a remainder of 1,2 and 3 when divided by 3,5 and 7
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I am assuming you want the least n for  n = 1 mod 3,   n = 2 mod 5, and  n = 3 mod 7.   
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One way to solve it is to apply the Chinese Remainder Theorem (see https://brilliant.org/wiki/chinese-remainder-theorem/  for a description on how it works.  It even has an example similar to this problem.)
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<pre>
Let n be the number.

Start with largest modulus and write as a regular equation:
n = 7j + 3     (1)

Next write that equation in terms of the next-highest modulus :  
7j + 3 = 2 mod 5
7j = -1 mod 5 
7j = 4 mod 5     (-1 and 4 are the same, mod 5)

Solve for j:   
  j = 2 mod 5   ( find this by listing multiples of 7:  7, 14, 21, etc, look for the one that gives remainder 4 
                          when divided by 5)

Re-write as equation: 
   j = 5k + 2    (2)

Now substitute  for j from (2)  into (1):
   n = 7(5k + 2) + 3     
   n = 35k + 17        (3)

Finally, write (3) as a congruence for the 3rd modulus:

     35k + 17  = 1 mod 3
        35k = -16 mod 3
        35k = 2 mod 3

Solve for k:
        k = 1 mod 3 
 
Re-write as regular equation:
         k = 3m + 1     (4)

Substitute for k from (4) into (3):
            
          n = 35(3m+1) + 17
          n = 105m + 52
  
So  n = 52 mod 105    and 52 is the answer.
</pre>
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Ans:  {{{ highlight(52) }}}
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<pre>
Check: 

  x mod 3 = 1:   1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, <b> 52</b>, 55, 58, ...
  x mod 5 = 2:   2, 7, 12, 17, 22, 27, 32, 37, 42, 47, <b>52</b>, 57, 62, …
  x mod 7 = 3:   3, 10, 17, 24, 31, 38, 45, <b>52</b>, 59, 66, …
 
We can see that 52 is the smallest number that appears on all 3 lists.