Question

find the least 5 digit number which is exactly divisible by 20,25 and30

Answer 1

20 = 2*10 = 2*2*5
 25 = 5*5
 30 = 2*15 = 2*3*5

 First we decide the least number of 2's we must have:
              a. 20 contains two 2's
 b. 25 contains no 2's
c. 30 contains one 2
  So to contain 20 as a factor, our number must contain two 2's as factors.

  Next we decide the least number of 3's we must have:
a. 20 contains no 3's
b. 25 contains no 3's
c. 30 contains one 3
 So to contain 30 as a factor, our number must contain one 3 as a factor

 First we decide the least number of 5's we must have:
 a. 20 contains one 5
b. 25 contains two 5's
c. 30 contains one 5 So to contain 20 as a factor, our number must contain two 2's as factors

  The smallest integer to contain all those as factors is 2*2*3*5*5 = 300.
 Trouble is, 300 is only a 3-digit number, and we want a 5-digit number.
so we must multiply 300 by the smallest integer that will make it be at least 10000, which is the smallest 5-digit number.

  We divide 10000 by 300 and get 33.33333... So the smallest integer greater than that is 34. So our number = 300*34 or 10200. 

Answer 2

LCM= 300
Multiples of 300 =300,600,900.............3000,6000,9000..............9300,9600,9900, 10200,10500,10800
10200 is the answer
300*34=10200