Question

find the least number of 5 digit , which being divided by 40,60 and 75 leaves 31 ,51 and 66as remainders respectively.

Answer 1

Difference, 40-31 = 9;
60-51 = 9;
75-66 = 9;
Difference between numbers and remainder is same in each case.
Then,
The answer = {(LCM of 40, 60, 75)-9};

40 = 2*2*2*5;
60 = 2*2*3*5;
75 = 3*5*5;
LCM = 2*2*2*5*5*3 = 600;
But, the least number of 5 digits = 10000;
10000/600, we get remainder as 400.
Then, the answer = 1000-(600-400)-9; = 10191.

*mark it as BRAINLIEST answer.

Answer 2

Number = N

N = 40 * a + 31   -- (1)
         As N > 10, 000,  a >= 251
N = 60 * b + 51   -- (2)
         As N > 10,000,  b >= 166
N = 75 * c + 66    --- (3)
          So c >= 133
Also a > b > c.

So from (1) and (2), 40 (a - b) = 20 (b + 1)
       =>  2 (a - b ) = b + 1        or     2 a = 3 b + 1   --- (4)
           or  3 (a - b) = a + 1
       So b is an odd number.   a+1 is divisible by 3.

From (2) and  (3) ,  we get  :  60(b-c) = 15 (c+ 1)
           =>  4 (b-c) = c + 1    or    4 b = 5 c + 1   --(5)
               or  5 (b - c) = b + 1

     So  b = 5 x - 1       for some x.
        =>     5 c = 4b  - 1 = 20 x - 5
                c = 4 x - 1
        =>   2a = 3 b + 1 = 15 x - 3 + 1 
            So  x must be a multiple of 2.  Let x = 2 y.
 
          So c = 8 y - 1,     b = 10 y - 1,    a = 15 y - 1.
  As c >= 133,   y >= 17

N = 75 c + 66 = 75*8 y - 75 + 66 > 10000
           y >= 17.
 For y = 17,  c = 135,  b = 169,  c = 254,  N = 10191.

Check if N = 10,191  is good...