To find the least 5-digit number which leaves a REMAINDER 9 in each case when they are divided by 12,40 and 75Let US SEE factors for the given numbers 12,40,75For 12 prime factors are 12:2 2 ×3for 40 prime factors are 40:=2 3 ×5For 75 prime factors are 75=5 2 ×3So, now let us find out the greatest four digit number that which is exactly divisible by given numbers∴ the greatest four digit number divisible by given numbers =9999So, LCM of the given numbers is LCM=2 3 ×3×5 2 =600So, to find out the greatest four digit divisible by given numbers ⇒ 9999−remainder⇒ 9999−399=9600in order to get 5 digit number exactly divisible by the given numbers , we get9600+600=10200but given in the question that when 5-digit number is divided by the given numbers we get a remainder 9as it is greatest number we are FINDING, here we subtract remainder to 5-digit number, for least number we add the remainder from itSo, we get, 10200+9=10209∴ 10209 is the least 5-digit number which when divided by 12,40 and 75 leaves a remainder 9.