Question 1103463
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Let N be the number under the question, and let "r" be the remainder.


Then the number  N-r is multiple of all numbers 95, 112, 214 and 231.


95 = 5*19;

112 = {{{2^4*7}}};

214 = 2*107;

231 = 3*7*11.


Hence, N-r is a multiple of the LCM (Less Common Multiple) of the numbers 95, 112, 214  and 231, which is  {{{2^4*3*5*7*11*19*107}}} = 37569840.


So, the answer is: any number of the form  N = 37569840*k + r  

with the arbitrary  integer "k" and "r"  (0 <= r < 95)  satisfies to the given condition.


The least such a positive integer number is  1 (one; ONE).


The next such a positive integer number is  2 (two; TWO).
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