Question

Find the smallest natural number of the form 123x 43y which is exactly divisible by 6

Answer 1

We have a number of the form $123x43y$ which is divisible by $6$.Now since it is divisible by $6$,by the divisibility test of $6$ it should be divisibility by $2$ and $4$.For $123x43y$ to be exactly divisible by $2$, the last digit ${}^{\prime }{y}^{\prime }$ could assume the following values $(0,2,4,6,8)$.

Now to divisible by $3$ ,the sum of the digits should be divisible by $3$.Now the sum of the digits of the number $123x43y=13+x+y$.

However,we are asked to find the smallest number of the above form.For that ${}^{\prime }x+y^{\prime }$ should be minimum and that is $2$. In that case the sum of the digits are $15$ and hence is divisible by $3$ .Now for $x+y=2$,we can have $x=0,2$ and $y=2,0$.

So,for,$x=0$ & $y=2$ the number will be $1230432$,and

for,$x=2$ & $y=0$ the number will be $1232430$

clearly,$1230432<1232430$

Hence the smallest natural number of the form $123x43y$ exactly divisible by $6$ is $123430$.