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The smallest natural number of the form$123X43Y$, which is exactly divisible by 6 is……

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HINT-Number is exactly divisible by 6 if the number is divisible by 2 and 3.
We have a number of the form$123X43Y$which is divisible by 6.
By the divisibility test of 6 it should be divisible by 2 and 3.
For $123X43Y$to be exactly divisible by 2 the last digit $'y'$should be divisible by 2 so, last digit $'y'$should be multiple of 2 which is$\left\{ {0,2,4,6,8} \right\}$.
Now, to be divisible by 3, the sum of the digits should be divisible by 3.
$\therefore 1 + 2 + 3 + x + 4 + 3 + y = 13 + x + y$
So as we see that $13 + x + y$ is divisible by 3 if the lowest possible value of $x + y$ be 2.
From here the possible cases for $x$ and $y$ is
$
  x = 0,y = 2..............\left( 1 \right) \\
  x = 2,y = 0...............\left( 2 \right) \\
  x = 1,y = 1...............\left( 3 \right) \\
$
So from the following cases the third case is eliminated because $'y'$should be divisible by 2.
So for $x = 0,y = 2$ the number will be$ = 1230432$
And for $x = 2,y = 0$ the number will be$ = 1232430$
Now you have to find out the smallest number in the form $123X43Y$, which is exactly divisible by 6
So, it is clear from the above two numbers that for$x = 0,y = 2$, the number is minimum
So the possible number is$ = 1230432$.
So, this is the required answer.

Note: - In such types of questions the key concept we have to remember is that always remember the divisibility rule of 2, 3 and 6 which is stated above, then according to the divisibility rule find the numbers, then check which number is smaller we will get the required answer.

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