Question

Find the last 5-digit number which is exactly divisible by 20, 25, 30

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 of 20,25,30 is 300

the largest 5 digit number is 99999

when 99999 is divided by 300, we get 99 as remainder.

Subtracting 99 from 99999, we get 99900. So 99900 is the last 5 digit number that is exactly divisible by 20,25 and 30 (you can verify it yourself :D)