Question
Question: The smallest natural number of the form\(123X43Y\), which is exactly divisible by 6 is……...
The smallest natural number of the form123X43Y, which is exactly divisible by 6 is……
Solution
HINT-Number is exactly divisible by 6 if the number is divisible by 2 and 3.
We have a number of the form123X43Ywhich is divisible by 6.
By the divisibility test of 6 it should be divisible by 2 and 3.
For 123X43Yto 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.
∴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..............(1) x=2,y=0...............(2) x=1,y=1...............(3)
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 forx=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.