Find the least five-digit number which when divided by 19 it gives remainder 5.
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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 75
Let us see factors for the given numbers 12,40,75
For 12 prime factors are 12:2
2
×3
for 40 prime factors are 40:=2
3
×5
For 75 prime factors are 75=5
2
×3
So, 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 =9999
So, LCM of the given numbers is LCM=2
3
×3×5
2
=600
So, to find out the greatest four digit divisible by given numbers ⇒ 9999−remainder
⇒ 9999−399=9600
in order to get 5 digit number exactly divisible by the given numbers , we get
9600+600=10200
but given in the question that when 5-digit number is divided by the given numbers we get a remainder 9
as it is greatest number we are finding, here we subtract remainder to 5-digit number, for least number we add the remainder from it
So, we get, 10200+9=10209
∴ 10209 is the least 5-digit number which when divided by 12,40 and 75 leaves a remainder 9.
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