Question

# The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

- 93324
- 66666
- 84844
- None of these

Hint:

### We start solving the problem by finding the total possibilities of getting numbers by fixing each digit in unit place. We then find the sum of all the numbers present in the unit place. Similarly, we multiply 10 for the sum of digits in tenth place, 100 for the sum of digits in tenth place and 1000 for the sum of digits in thousandth place. We then add all these sums to get the required answer.

## The correct answer is: 93324

### Complete step-by-step answer:

According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.

Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.

We need to arrange the remaining three places with three digits.

We know that the number of ways of arranging n objects in n places is n! ways.

So, we get 3!=6 numbers on fixing the unit place with a particular digit.

Now, let us find the sum of all digits. We get sum as 2+3+4+5=14.

Now, we get a sum of digits in units place for all the numbers as 14×6=84.

We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.

i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.

So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is

(84×1000)+(84×100)+(84×10)+(84×1)

Sum = 84000+8400+840+84

Sum = 93324

We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.

Alternatively, we can use the formula for the sum of numbers as

(n - 1)! (sum of digits) (11111 ............ntimes). We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.

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We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.

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