Find the number of 10-digit numbers where the sum of the digits is divisible by 5.
There is another post about this, but the answers were wrong.
Don't you think that you should interact with answering people, when the answer is wrong? What is it that you want? Can you clarify it? Is it ONLY the sum of 10 digits 1234567890 rearranged? Can you repeat ANY digit such as 1122345678??. Clarify your question PLEASE !! Thank you.
Then why don't you try working it out, instead of trying to cheat on your homework?
You need to add a link to the original question.
The link to the question is https://web2.0calc.com/questions/find-the-number-of-10-digit-numbers-where-the-sum
You can repeat all digits.
There are 2*10^8 = 200,000,000 10-digit numbers that work.
OK, I believe that I have what is very close, if not exact number of terms that sum up to multiples of 5, or are all divisible by 5 for all 10-digit numbers. The sequence is somewhat like arithmetic series, but not exactly due to sums that are multiples of 5. For example, your series begins like this:
1000000004 , 1000000013 , 1000000022 , 1000000031 , 1000000040 , 1000000049 , 1000000059 , 1000000068 , 1000000077 , 1000000086 , 1000000095 , 1000000103 , 1000000108, 1000000117........etc.
Notice that the difference is 9 up to the 7th term. But the 7th term itself has a difference of 10 from the
previous term due to the peculiarity that the sum of the digits must be a multiple of or divisible by 5. However, that is not going to change the total number of terms of the entire sequence. And that total is the difference between the last term and the first term, or:
9,999,999,999 - 1,000,000,004 =8,999,999,995 / 5 + 1=1,800,000,000 terms, which is the best I came up with.
The sum total of 1,800,000,000 terms came out to =
P.S. I wrote a short computer code to try and verify the above number and it came up with: 1,799,999,999 !!!, or 1 short of theoretical calculation!