Find the smallest 5 digit number which leave a reminder 9 in each case when divided by 12,40 and 75
Answers
Answer:
Given numbers,
12,40 and 75.
Prime factorization of
12=2²×3
40=2³×5
75=3×5²
LCM=2³×3×5²=600
Now first let us find greatest four digit number exactly divisible by given numbers.
greatest four digit number divisible by given numbers=9999-remainder when 9999 divided by LCM of given numbers.
greatest four digit number divisible by given numbers=9999-399=9600
Therefore 9600 is the greatest four digit number divisible by given numbers.
Then,
if we add 600 to 9600 then we can get least five digit number exactly divisible by given numbers.
Therefore,
least five digit number exactly divisible by given numbers=9600+600=10200
but given that,
required number when divided by given numbers leaves remainder 9
Therefore required number=10200+9=10209
Hence 10209 is the least five digit number when divided by 12,40 and 75 leaves remainder 9.