SOLUTION: Find the number of solutions to N &\equiv 2 \pmod{5}, \\ N &\equiv 2 \pmod{6}, \\ N &\equiv 2 \pmod{7} in the interval 0 \le N < 1000.

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Question 1207720: Find the number of solutions to
N &\equiv 2 \pmod{5}, \\
N &\equiv 2 \pmod{6}, \\
N &\equiv 2 \pmod{7}
in the interval 0 \le N < 1000.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
Find the number of solutions to
N = {2 mod 5},
N = {2 mod 6},
N = {2 mod 7}.
~~~~~~~~~~~~~~~~~~~~~~~ 

Consider the number N-2.


   N-2 = {0 mod 5},

   N-2 = {0 mod 6} 

and

   N-2 = {0 mod 7}.


It means that N-2 is a multiple of 5, 6 and 7, at the same time.


The integer numbers  N-2  in the interval  [-2,998]  divisible by 5, 6 and 7  
at the same time are those multiple of 210,  i.e. 0, 210, 420, 630, 840.

Hence, the solutions N under the problem's question are the numbers 2, 212, 422,  632, 842.    


In all, there are 5 solutions.    ANSWER

Solved.



Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The numbers are 2 more than a multiple of either 5, 6, or 7.

Since 5, 6, and 7 have no common factors, the numbers have to be 2 more than a multiple of 5*6*7=210.

The numbers N satisfying that requirement in the interval [0,1000] are 2, 212, 422, 632, and 842.

ANSWER: 5 solutions
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