Question

The sum of all numbers greater than $10,000$ by using the digits $0,2,4,6,8$ considering no digit being repeated in any number is

• A. $5199970$
• B. $5199960$
• C. $5199950$
• D. $5199940$

Answer 1

Answer: (B)

$5199960$

Total number of integers that can be made from these numbers without repetition $=5!-4!=96$
Now we write down all these $96$ numbers one below another.
We see that no. of times $0$ comes at (once, tenth, hundredth etc )places except the first place $=1\times 4!=24$
No. of times rest digits come at these places except the first place $=\frac{96-24}{4}=18$
In the first place no. of times 0 comes $=0$ since the number has to be of $5$ digits.
In the first place no of times $2,4,6,8$ come$=\frac{96}{4}=24$
Now add the digits column-wise as normal addition.
On last $4$ places the sum of digits in each column$=18\times (2+4+6+8)+24\times 0=360$
On the first place the sum of digits $=24\times (2+4+6+8)=480$
Hence keeping in the mind the carry ,
the last digit of sum $=0$ with carry $36$ $\left(360\right)$
second last digit of sum$=6$ with carry $39$ $(360+36=396)$
3rd last digit of sum$=9$ with carry $39$ $(360+39=399)$
2nd digit of sum$=9$ with carry $39$ $(360+39=399)$
and 1st number of sum$=480+39=519$
Therefore, the sum$=5199960$
Hence, option 'B' is correct.