Question

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

Answer 1

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 $12:2^2\times 3$

for $40$ prime factors are $40:=2^3\times 5$

For $75$ prime factors are $75=5^2\times 3$

So, now let us find out the greatest four digit number that which is exactly divisible by given numbers

$\therefore$ the greatest four digit number divisible by given numbers $=9999$

So, LCM of the given numbers is $LCM=2^3\times 3\times 5^2=600$

So, to find out the greatest four digit divisible by given numbers $\Rightarrow$ $9999-remainder$

$\Rightarrow$ $9999-399=9600$

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, $10200+9=10209$

$\therefore$ $10209$ is the least 5-digit number which when divided by $12,40$ and $75$ leaves a remainder $9$.