Question

In maths

how many 5digit number divisible by 11are three containing each of the digits 2,3,4,5,6​

Answer 1

Answer:

Step-by-step explanation:

Our question is: how many 5 digit number divisible by 11are three containing each of the digits 2,3,4,5,6​.

We know that a number that is divisible by 11 has a sum of its digits a multiple of 11.

Now the only multiple of 11 we can possibly have is 0, since the next multiple is 11 and it is impossible to have because the greatest possible sum is 6 + 5 + 4 = 15, so we only need to focus on a+c+e =b+d.

That's five variables. We need a smart approach.

Note that the right hand side needs ‘big guys',there're only two numbers that need to be equal in sum to the three on the left hand side.

So we set b=6,and d as 5,but in this case, the “big guys “ “overkill “ that is, the RHS is 11 and the LHS, sum of remaining digits, is9.

So we change d to 4,and it works perfectly. Both sides 10.

Assured that we have at least one solution, we can proceed to arrange the digits to find all such numbers . I mean, I could have chosen to set b as 4 and d as 6, and that would have done the trick as well. So for the right side, we have two choices.

Similarly we have three choices for a and 2 for c. e takes the remaining digit .

Thus, in all, we have 2*3*2=12 ways of choosing a b c d and e. (the product is a standard outcome of expressing the total number of ways of doing a task that can be decomposed into independent choices)

I hope this helps your studies!!

Keep it up!

Answer 2

Answer:

For getting a 5 digit no to contain each of the 5 require digits, each digit can only be used once.

Thus there are 5! = 120 5 digit numbers containing 2,3,4,5 & 6  

Step-by-step explanation:

The smallest number is clearly 23456

The next smallest is found by swapping the 5 & 6 so 23465

Similarly the 3rd smallest is found by swapping 4 and 5, that is 23546

and again we swap the 5 & 6 to generate the 4th smallest, 23465

After continuing the swapping 3 &4 then regenerating more get the lowest 8 numbers

23456,

23465,

23546,

23564,

24356,

24365,

24536,

24563

Divide each by 11, lets start with the lowest and find the first of these divisible by 11 is 24365, the 6th lowest of the 120 5 digit numbers containing 2,3,4,5 & 6.