Question

Find the least 5 digit number which leaves remainder 9 in each case when divided by 20 40 75

Answer 1

Answer:

Step-by-step explanation:

Point to remember:

Successive numbers which are exactly divisible by some numbers are multiples of LCM of those numbers.

This is the reason LCM is least common multiple ,that means before that number there is no such number and after that number there are infinitely such numbers.

Given numbers,

20,40 and 75.

Prime factorization of:

20=2²×5,

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.

Hope it helps.