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How many times does the number 9 appear in total in all the even numbers from 2 to 9999? (This was the question that I came across in one of the Cracku questions set, but not able to find it in the book marks now,) Could you please help here with the solution?
Hi Aravind,
If we consider all even numbers from 0 to 9999, we get (9999 + 1)/2 = 5000 such numbers.
In these 5000 numbers, the unit's digit is always 0, 2, 4, 6 or 8.
Hence, if we consider the number to be ABCD D can not be 9.
A, B, and C can take any values from 0 to 9.
So, the total number of digits comprising A, B and C in all 5000 appearances = 5000 x 3(for A, B and C) = 15000
Now, among these 15000 digits, digits 0 to 9 appear an equal number of times.
Hence, 9 will appear 1/10 th of the times.
Hence, total number of appearances of 9 = 15000/10 = 1500.
Hope this helps!
