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# Find the least five digit number which leaves a remainder 9 in each case when divided by 12,40 and 75

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## 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 12:22×3for 40 prime factors are 40:=23×5For 75 prime factors are 75=52×3So, 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=23×3×52=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.  Suggest Corrections  0      Similar questions  Related Videos   Area of Any Polygon - by Heron's Formula
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