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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
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B
5199960
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C
5199950
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D
5199940
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Solution

The correct option is A 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 =1×4!=24
No. of times rest digits come at these places except the first place =96244=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=964=24
Now add the digits column-wise as normal addition.
On last 4 places the sum of digits in each column=18×(2+4+6+8)+24×0=360
On the first place the sum of digits =24×(2+4+6+8)=480
Hence keeping in the mind the carry ,
the last digit of sum =0 with carry 36 (360)
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.

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